Showing 35 questions
// Gain Fundamentals
Q01
What is the difference between transducer gain, available gain and operating power gain?
EasyGain
Transducer Gain G_T = P_load / P_available_source (most commonly used)
Available Gain G_A = P_available_output / P_available_source (depends on Γ_S only)
Operating Power Gain G_P = P_load / P_input (depends on Γ_L only)

Transducer gain is the ratio of power delivered to the load to the maximum power available from the source. It accounts for both input and output mismatch — the most practical and commonly quoted gain.

Available gain assumes the output is conjugate matched to the load. It is useful for noise figure calculations because it depends only on the source reflection coefficient Γ_S.

Operating power gain assumes the input is perfectly matched. It is useful for PA design where the load determines efficiency.

For a matched amplifier (Γ_S=0, Γ_L=0): G_T = G_A = G_P = |S21|²
S21 in dB = 20·log₁₀|S21| = matched transducer gain in dB
💡 S21 on a VNA directly gives the matched transducer gain — the gain when source and load are both 50 Ω. This is the gain quoted in nearly all datasheets. For best NF, input must be mismatched (Γ_S = Γ_opt), which reduces gain below |S21|².
Q02
What is unilateral gain approximation? When is it valid and when does it fail?
EasyGain
Unilateral assumption: S12 = 0 (no reverse transmission)
Unilateral transducer gain: G_TU = |S21|² / [(1−|Γ_S|²)|1−S11·Γ_S|²] × 1/[(1−|Γ_L|²)|1−S22·Γ_L|²]

The unilateral approximation treats the amplifier as one-way — input and output are independent. This greatly simplifies matching network design because the input match doesn't affect the output impedance and vice versa.

Valid when: |S12| is very small compared to |S21| — typically |S12/S21| < 0.05 (−26 dB). True for most microwave transistors at low-to-moderate frequencies, and for amplifiers with good reverse isolation (common-source/common-emitter with ground-shielded gate/base).

Fails when: |S12| is significant — high-frequency transistors approaching f_T, feedback amplifiers, or broadband designs. The unilateral figure of merit U = |S12||S21||S11||S22| / [(1−|S11|²)(1−|S22|²)] quantifies the error.

💡 In practice, use the unilateral model for initial design and then verify with full bilateral S-parameter simulation. A mismatch error of <0.5 dB in gain is acceptable for most applications.
Q03
What is conjugate matching? When does maximum gain occur and why can't you always conjugate match an amplifier?
MediumMatching
Maximum power transfer: Γ_S = Γ_in* and Γ_L = Γ_out*
Γ_in = S11 + S12S21Γ_L/(1−S22Γ_L)
Maximum available gain: G_max = |S21/S12| × (K − √(K²−1)) where K is Rollett stability factor

Conjugate matching presents the complex conjugate of the amplifier's input impedance as the source impedance, and the complex conjugate of the output impedance as the load. This maximises power transfer at both ports simultaneously.

Why you can't always conjugate match:

  • Stability: If the device is conditionally stable, conjugate match points may lie inside the instability region on the Smith chart — the amplifier will oscillate
  • Noise figure: The source impedance for minimum NF (Γ_opt) is different from the conjugate match (Γ_in*). Designing for minimum NF means accepting less than maximum gain
  • Bandwidth: A perfect conjugate match is narrowband. Broadband amplifiers deliberately accept gain ripple to achieve flat response over a wide bandwidth
💡 The fundamental LNA design tradeoff: minimum NF requires Γ_S = Γ_opt, but maximum gain requires Γ_S = Γ_in*. These are different points on the Smith chart. The LNA designer must choose a Γ_S that compromises between the two — or use a simultaneous noise and gain matching technique (SNGM).
Q04
What is the maximum stable gain (MSG) and maximum available gain (MAG)? What is the significance of their equality?
MediumGain
MSG = |S21/S12| (when K < 1 — conditionally stable)
MAG = |S21/S12| × (K − √(K²−1)) (when K ≥ 1 — unconditionally stable)
MSG = MAG when K = 1 (boundary between conditional and unconditional stability)

MSG is the maximum gain achievable from a conditionally stable device with a source/load that keeps it on the boundary of stability. It is an upper bound — achieving it in practice risks oscillation.

MAG is the maximum gain achievable when the device is unconditionally stable (K > 1), obtained with simultaneous conjugate matching at both ports.

When K = 1, MSG = MAG — the transition point. Transistor datasheets often plot MSG and MAG vs frequency, with a kink at the frequency where K transitions from <1 to >1. The practical operating frequency is chosen well below this kink where K > 1 with comfortable margin.

💡 At mmWave frequencies, transistors often have K < 1 in their useful frequency range. Designers must add neutralisation (a feedback capacitor from drain to gate) to increase K above 1 before applying conjugate match — otherwise the amplifier will oscillate.
Q05
What is gain flatness and group delay? Why does an amplifier with flat gain not necessarily have flat group delay?
EasyGain
Gain flatness: |S21(f)| – varies < ±X dB over the band
Group delay: τ_g = −dφ(S21)/dω seconds
Flat gain does NOT imply flat group delay — they are independent

An amplifier can have perfectly flat amplitude response while the phase of S21 varies nonlinearly with frequency. The derivative of that phase — the group delay — will then vary across the band, causing different frequency components to arrive at the output at different times. This distorts wideband signals even though the amplitude is flat.

Group delay variation matters most for: OFDM signals (WiFi, LTE, 5G) where different subcarriers must remain time-aligned, and wideband digital communications where pulse shape distortion (ISI) determines BER.

💡 A Bessel filter has maximally flat group delay but poor stopband attenuation. A Chebyshev amplifier equaliser can flatten gain but worsen group delay. In high-performance designs, a separate all-pass filter (group delay equaliser) is used after the amplifier to flatten the overall group delay without affecting gain.
// Noise Figure
Q06
What is noise figure? What is the noise figure of an ideal (lossless) amplifier and a lossy passive attenuator?
EasyNoise
NF = SNR_in(dB) − SNR_out(dB) = 10·log₁₀(F)
F (noise factor) = SNR_in/SNR_out (linear) ≥ 1 always
Ideal noiseless amplifier: NF = 0 dB (F=1) — adds no noise
Passive attenuator with loss L: NF = L (dB) at T=290 K

Noise figure measures how much an amplifier degrades the SNR. An ideal noiseless amplifier amplifies signal and noise equally — SNR unchanged, NF = 0 dB. A real amplifier adds its own noise, reducing SNR, giving NF > 0 dB.

A passive attenuator with 3 dB loss has NF = 3 dB. With 10 dB loss, NF = 10 dB. This is because a lossy element at room temperature (290 K) is in thermal equilibrium — it absorbs and re-emits thermal noise equal to what it absorbs. The noise figure exactly equals the loss in dB.

💡 This is why the cable between an antenna and an LNA matters so much. A 1 dB coaxial cable loss adds 1 dB to system NF before the LNA even contributes. Moving the LNA to the antenna mast (remote radio head) eliminates this cable loss and improves system sensitivity by 1 dB.
⚡ RX Chain Designer
Q07
What is the Friis formula for cascaded noise figure? Calculate for three stages: LNA (G=15 dB, NF=1.5 dB), Filter (IL=2 dB), Mixer (NF=7 dB).
MediumNoise
F_total = F1 + (F2−1)/G1 + (F3−1)/(G1·G2) + …
F_LNA=1.413, G_LNA=31.62   F_filt=10^(2/10)=1.585, G_filt=10^(−2/10)=0.631
F_mixer=10^(7/10)=5.012
F_total = 1.413 + (1.585−1)/31.62 + (5.012−1)/(31.62×0.631)
= 1.413 + 0.0185 + 4.012/19.95 = 1.413 + 0.018 + 0.201 = 1.632
NF_total = 10·log₁₀(1.632) = 2.13 dB

The filter contributes only 0.018 (negligible) but the mixer contributes 0.201 — significant because the filter reduced the gain before the mixer to only 0.631 (−2 dB). The filter insertion loss effectively multiplied the mixer noise contribution.

💡 The lesson: never place a lossy element (filter, switch, cable) between the LNA and the mixer without accounting for its effect on cascaded NF. Every 1 dB of filter insertion loss after the LNA degrades cascaded NF by approximately 1/G_LNA in linear, or roughly 1/31 = 0.03 noise factor units for this example.
⚡ RX Chain Designer — multi-stage cascade
Q08
What is minimum noise figure (NF_min)? How do noise circles work and what do they tell a designer?
HardNoise
NF = NF_min + (4R_n/Z_0) × |Γ_S − Γ_opt|² / [(1−|Γ_S|²)|1+Γ_opt|²]
NF_min: minimum achievable NF (at Γ_S = Γ_opt)
R_n: noise resistance (sensitivity of NF to source mismatch)
Γ_opt: optimal source reflection coefficient for minimum NF

NF_min is the minimum noise figure achievable from a transistor with ANY source impedance — it is a fundamental transistor property, not dependent on matching. It decreases as bias current increases (up to a point), and increases with frequency.

Noise circles are contours of constant noise figure on the Smith chart, centred near Γ_opt. The innermost circle is at NF_min (radius zero). Moving away from Γ_opt along any direction increases NF. The spacing of the circles is controlled by R_n — a larger R_n means NF degrades quickly with source mismatch.

A designer plots noise circles alongside gain circles to find a source impedance that gives the best compromise between NF and gain.

💡 Typical LNA design: Γ_opt and Γ_in* are on different parts of the Smith chart. A designer plots the NF=1.0 dB noise circle and the G=12 dB available gain circle — picks Γ_S at their intersection or slightly toward Γ_opt for lower NF at some gain cost. The input matching network transforms 50 Ω to this chosen Γ_S.
Q09
What is the noise temperature of an amplifier? How does it relate to noise figure?
MediumNoise
T_e = T_0 × (F − 1) = 290 × (10^(NF/10) − 1) Kelvin
NF = 1 dB → T_e = 75 K   NF = 3 dB → T_e = 290 K   NF = 0.1 dB → T_e = 6.7 K

Noise temperature represents the equivalent temperature of a resistor at the amplifier input that would generate the same noise power as the amplifier itself adds. It is used in radio astronomy, satellite communications and radar where T_e is physically meaningful and where the input signal arrives from a cold sky (noise temperature ~10–50 K).

For a satellite receiver with system noise temperature T_sys = 50 K, adding an amplifier with NF=1 dB (T_e=75 K) more than doubles the system noise temperature — a catastrophic degradation. NF=0.1 dB (T_e=6.7 K) adds only 13% to T_sys. This is why cryogenically cooled LNAs (T_e=5–20 K) are used on radio telescopes.

💡 The noise temperature representation is more intuitive at very low NF values. "0.1 dB NF" doesn't sound impressive but "T_e=6.7 K" immediately tells you this is an exceptional low-noise amplifier suitable for space applications.
Q10
Design an LNA input match for a GaAs pHEMT with NF_min=0.5 dB, Γ_opt=0.6∠150°, R_n=15 Ω, at 10 GHz.
HardLNA Design

Goal: Design an input matching network that transforms 50 Ω to the source impedance corresponding to Γ_opt = 0.6∠150° to achieve near-minimum NF.

Step 1: Convert Γ_opt to Z_opt.

Γ_opt = 0.6∠150° = 0.6(cos150°+j sin150°) = −0.520+j0.300
Z_opt = Z_0×(1+Γ_opt)/(1−Γ_opt) = 50×(0.480+j0.300)/(1.520−j0.300)
= 50×(0.480+j0.300)(1.520+j0.300) / (1.520²+0.300²)
= 50×(0.640+j0.600) / 2.400 → Z_opt ≈ 13.3+j12.5 Ω

Step 2: Design matching network from 50 Ω to Z_opt = 13.3+j12.5 Ω at 10 GHz. Use an L-network: shunt element at the 50 Ω port + series element to reach Z_opt.

Step 3: Calculate NF penalty if using 50 Ω directly (unmatched).

At Γ_S=0 (50 Ω): |Γ_S−Γ_opt|² = |0−(−0.520+j0.300)|² = 0.520²+0.300² = 0.360
NF excess = (4×15/50)×0.360/[(1−0)×|1+Γ_opt|²] ≈ 1.2×0.360/0.256 = 1.69 dB
NF_actual ≈ 0.5+1.69 = 2.19 dB without matching vs 0.5 dB with
💡 This example shows that using 50 Ω source with no input matching gives NF of 2.19 dB instead of 0.5 dB — a 1.69 dB penalty entirely due to source mismatch. The input matching network is not optional for minimum NF designs.
// Linearity
Q11
What is the 1 dB compression point (P1dB)? What is the relationship between input and output P1dB?
EasyLinearity
P1dB: input power where gain drops 1 dB from small-signal value
OP1dB = IP1dB + G − 1 (all dBm) — approximately
Exact: OP1dB ≈ IP1dB + G − 0.93 dB

The 1 dB compression point marks the onset of amplifier saturation. Below P1dB the amplifier behaves linearly — every 1 dB increase in input gives 1 dB increase in output. At P1dB the gain has dropped 1 dB, meaning the output is 1 dB below where a perfectly linear amplifier would be.

IP1dB vs OP1dB: IP1dB is the input power causing compression. OP1dB is the corresponding output power. For a 20 dB gain amplifier with IP1dB = −10 dBm: OP1dB ≈ −10+20−1 = +9 dBm. Datasheets often specify OP1dB for power amplifiers (the output is what matters) and IP1dB for low-noise amplifiers (the input is what determines when the receiver compresses).

💡 Rule of thumb: IP1dB ≈ IIP3 − 10 dB. So if you know IIP3, you can estimate P1dB. This relationship comes from the third-order Taylor series model of amplifier nonlinearity.
Q12
What is IIP3? Derive the 10 dB rule of thumb between IIP3 and IP1dB.
MediumLinearity
Taylor series: V_out = a1·V_in + a2·V_in² + a3·V_in³ + ...
For two tones V_in = A(cosω1t + cosω2t):
Fundamental amplitude: a1·A + (9/4)a3·A³
IM3 amplitude (at 2ω1−ω2): (3/4)a3·A³
IIP3 = A_IIP3 where fundamental = IM3: A_IIP3 = √(−4a1/3a3) = √(4/3) × A_P1dB

At P1dB, the gain has compressed by 1 dB, meaning the first-order term equals:

At P1dB: a1·A + (3/4)a3·A³ = a1·A·10^(−1/20) = 0.891·a1·A
→ (3/4)|a3|A² = 0.109·a1 → A_P1dB = √(0.145·a1/a3)
Ratio: A_IIP3/A_P1dB = √(4/3)/√(0.145) = 1.633/0.381 = 3.33 (power ratio = 9.54 dB ≈ 10 dB)

So IIP3 ≈ IP1dB + 10 dB. This is the fundamental "10 dB rule" — valid for amplifiers dominated by third-order nonlinearity. In practice IIP3 = IP1dB + 8 to +12 dB depending on higher-order terms.

💡 The derivation assumes only the third-order term contributes to compression and IM3. Real amplifiers have all orders contributing — the rule is approximate but useful for quick estimates. For PA design, measured IM3 vs power at backoff is more reliable than extrapolating from IIP3.
Q13
What is spectral regrowth and ACPR? How does amplifier nonlinearity cause out-of-band emissions?
MediumLinearity

Spectral regrowth is the widening of the transmitted signal spectrum caused by amplifier nonlinearity. An ideal amplifier of a bandlimited signal produces an output within the same band. A nonlinear amplifier generates intermodulation products that fall in adjacent channels — interfering with other users.

ACPR (Adjacent Channel Power Ratio) is the ratio of power in the adjacent channel to the power in the wanted channel, measured in dBc. It is the key specification for transmitter linearity in cellular standards.

ACPR requirement examples:
LTE 5 MHz channel: ACPR < −30 dBc at ±5 MHz offset
WCDMA: ACPR < −45 dBc (ACLR) at 5 MHz offset
5G NR: ACLR < −45 dBc at adjacent channel

ACPR is caused by IM3 and IM5 products from the many subcarriers in OFDM signals beating against each other. An OFDM signal with 64 subcarriers effectively provides thousands of two-tone pairs — each generating IM3 products that sum incoherently in the adjacent channel.

💡 ACPR worsens by 2 dB for every 1 dB increase in PA output power (because IM3 grows at 3 dB per 1 dB input, minus 1 dB output = 2 dB relative growth). Operating the PA at 6 dB back-off from P1dB typically gives 12 dB improvement in ACPR.
Q14
What is PAPR (Peak-to-Average Power Ratio)? Why does high PAPR reduce PA efficiency?
HardLinearity
PAPR = P_peak / P_average (ratio) or 10·log₁₀(P_peak/P_avg) dB
OFDM with N subcarriers: theoretical max PAPR = 10·log₁₀(N) dB
LTE 20 MHz (1200 subcarriers): theoretical max ≈31 dB, practical 99.9% PAPR ≈ 10–12 dB
WiFi 802.11ac 80 MHz: practical PAPR ≈ 8–10 dB

OFDM signals have many subcarriers that occasionally add coherently, creating instantaneous peaks much higher than the average power. The PA must be sized to handle these peaks without compression — but the average efficiency is measured at the average output power, which is far below the peak capability.

Efficiency impact: Class A PA efficiency = P_out/P_DC. With PAPR = 10 dB, the PA must handle peaks 10× above average. To avoid compressing the peaks, the PA must be backed off 10 dB from its P1dB operating point. A Class A PA at 10 dB back-off has efficiency ~3% (from ~30% at P1dB).

Solutions:

  • Digital predistortion (DPD): Compensates AM/AM and AM/PM distortion, allowing operation closer to P1dB
  • Envelope tracking (ET): Modulates the PA supply voltage to follow the signal envelope — maintains high efficiency at low instantaneous power
  • Doherty PA: Two-amplifier architecture that maintains high efficiency over a 6–10 dB power back-off range
  • Clipping and filtering: Clips the PAPR to 6–8 dB, filtering removes spectral regrowth — some EVM penalty accepted
💡 A 5G NR base station PA transmitting at 40 W average with 10 dB PAPR must handle 400 W peaks. The PA die is sized for 400 W but runs at 40 W average — efficiency is terrible. Envelope tracking improves average efficiency from ~10% to ~40% for LTE signals, saving megawatts of cooling power across a carrier's network.
Q15
What is AM-to-AM and AM-to-PM conversion in a PA? How does each affect EVM?
MediumLinearity
AM-AM: output amplitude nonlinearly depends on input amplitude → gain compression
AM-PM: output phase depends on input amplitude → phase rotation under envelope variation
EVM contribution: ΔEVM_AMPM ≈ Δφ(rad) × 100% per degree of phase error

AM-to-AM compresses the signal constellation — points that should be at a certain magnitude are pulled inward near the PA's saturation level. This causes magnitude errors on the constellation.

AM-to-PM rotates the constellation points as the envelope changes. When the signal amplitude is at a peak, the PA phase shift is different from when it's at the average amplitude. This causes constellation rotation errors.

Both are measured with a vector signal analyser by transmitting a known QAM or OFDM signal and measuring the EVM. DPD corrects both simultaneously by pre-distorting the baseband signal to counteract the PA's AM-AM and AM-PM curves.

💡 5G NR at 256-QAM requires EVM < 3.5%. A PA with 3° AM-PM at peak power contributes EVM ≈ 5% — exceeding the limit from this alone. DPD typically reduces AM-PM from 5–10° to <0.5°, and AM-AM compression from 2–3 dB to <0.1 dB, enabling 256-QAM in production hardware.
// Stability
Q16
What is Rollett's stability factor K? What value of K guarantees unconditional stability?
MediumStability
K = (1 − |S11|² − |S22|² + |Δ|²) / (2|S12||S21|)
Δ = S11·S22 − S12·S21 (determinant of S-matrix)
Unconditionally stable: K > 1 AND |Δ| < 1

K > 1 alone is necessary but not sufficient for unconditional stability — you also need |Δ| < 1 (or equivalently, B1 > 0 where B1 = 1+|S11|²−|S22|²−|Δ|²). The two conditions together guarantee the amplifier will not oscillate with ANY passive source and load impedance.

K < 1 means the device is conditionally stable — it may oscillate with certain source/load impedances. The designer must identify the stability circles on the Smith chart and ensure the operating point avoids the instability regions.

💡 A single number K is not sufficient to fully characterise stability. The μ-factor (Edwards-Sinsky) provides a single number test: μ > 1 is both necessary and sufficient for unconditional stability: μ = (1−|S11|²)/(|S22−Δ·S11*|+|S12·S21|). The μ-factor is now preferred in practice.
⚡ Amplifier Stability Calculator
Q17
What are stability circles? How do you use them to ensure a conditionally stable amplifier doesn't oscillate?
HardStability

Stability circles are circular boundaries on the Smith chart that separate the regions of Γ_S (or Γ_L) values that cause |Γ_in| > 1 (or |Γ_out| > 1) from those that don't. An amplifier oscillates when |Γ_in·Γ_S| ≥ 1 or |Γ_out·Γ_L| ≥ 1.

Input stability circle centre: C_S = (S11 − Δ·S22*)* / (|S11|² − |Δ|²)
Input stability circle radius: r_S = |S12·S21| / ||S11|² − |Δ|²|
Output stability circle: similar with S11 ↔ S22

How to use them:

  1. Plot the input and output stability circles on the Smith chart
  2. Determine which side of each circle is stable (check a known stable point, e.g. Γ_S=0 gives |Γ_in|=|S11| — if |S11|<1 this is stable, so the centre of the Smith chart is in the stable region)
  3. Choose Γ_S and Γ_L entirely within the stable region, away from the stability circle boundary
  4. For robustness, maintain distance ≥ 0.1 on the Smith chart from the stability circle boundary, accounting for S-parameter variation with temperature and frequency
💡 Stability must be checked at ALL frequencies, not just the design frequency. An amplifier stable at 5 GHz may oscillate at 100 MHz where bias circuit parasitics provide positive feedback. Resistive termination in the bias network (ferrite bead or series resistor in the RF choke) is often needed to prevent low-frequency oscillation.
⚡ Stability Circles Plotter
Q18
Why can a stable amplifier with K>1 still oscillate on a PCB? List three practical causes.
HardStability

The K>1 stability condition is verified using the device's S-parameters — but these are measured in a controlled test fixture, not in your PCB layout. The PCB introduces additional feedback paths not captured in the S-parameters.

Three practical causes of oscillation despite K>1:

  • Ground inductance: Shared source/emitter ground inductance between input and output creates common-impedance coupling. This provides positive feedback proportional to the source inductance. Even 0.1 nH of ground inductance creates significant feedback at 5 GHz. Solution: multiple ground vias in parallel, solid ground plane, no shared via for input and output ground connections.
  • Supply decoupling resonance: The DC supply bypass capacitors resonate with the supply trace inductance. At this resonance frequency, the supply presents a high impedance instead of AC ground — supply variations couple back to the transistor gate/base. Solution: multiple decoupling capacitors at different values (100 nF + 10 nF + 100 pF) to cover all frequencies, placed physically close to the transistor.
  • Electromagnetic coupling between input and output traces: The output trace can radiate and couple electromagnetically to the input trace. At the frequency where this coupling provides 0 dB net loop gain with 0° phase shift, the circuit oscillates. Solution: physical separation between input and output traces, ground shielding walls/vias, not routing input and output traces in parallel.
💡 The best way to find unintended feedback paths: measure S11, S22, and S21 on the assembled PCB (without the transistor, or with the transistor removed) using a VNA. Any resonances in the fixture S-parameters that create phase conditions for oscillation will show up. Fix them before installing the transistor.
Q19
What is the mu-factor (μ) stability measure? Why is it preferred over K?
MediumStability
μ = (1−|S11|²) / (|S22−Δ·S11*| + |S21·S12|)
Unconditionally stable: μ > 1 (single condition, necessary AND sufficient)
Larger μ = more stable margin

Advantages of μ over K:

  • K > 1 requires an additional condition (|Δ| < 1 or B1 > 0) to be sufficient for unconditional stability. μ > 1 alone is both necessary and sufficient — one number, one test.
  • μ provides a physical measure of stability margin — higher μ means the unstable region on the Smith chart is further from the matching region. K has no direct physical interpretation of margin.
  • μ is monotonic: higher μ always means better stability. K can be non-monotonic.
💡 Modern RF CAD tools (ADS, Microwave Office) all compute μ alongside K. Many designers now use μ exclusively. A commonly used design target is μ > 1.2–1.5 at all frequencies to provide stability margin for S-parameter spread in production, temperature variation, and load VSWR variation.
Q20
How do you stabilise a conditionally stable transistor? Name three techniques with their tradeoffs.
HardStability

1. Series resistor at input (R_in in series with gate/base):

Increases |S11| and effectively reduces the reverse gain. Simple, cheap, reliable. Tradeoff: degrades NF (the resistor is a noise source at the input) and reduces gain. Typically 10–50 Ω is used as a compromise.

2. Shunt resistor at output (R_out in parallel with drain/collector):

Loads the output, moving the output stability circle safely away from the Smith chart. Tradeoff: reduces gain and output power, increases DC current. Less effect on NF than input resistor.

3. Feedback resistor (drain-to-gate, collector-to-base):

A resistor from output to input introduces negative feedback that stabilises the device across a wide frequency range. Tradeoff: significantly reduces gain (by the feedback factor), but can produce a flatter gain-vs-frequency response. Commonly used in broadband amplifiers where gain flatness is important.

Resistive feedback: G_feedback ≈ R_F/R_in   Gain reduction ≈ –20·log₁₀(1+R_in/R_F)
Tradeoff: 6 dB feedback = 6 dB gain reduction + stability across full band
💡 In practice, a combination is used: a small series resistor at the gate (5–10 Ω, negligible NF penalty) stabilises at out-of-band frequencies where K<1, while at the design frequency K is already >1 from the device characteristics. The resistor is then bypassed at the design frequency by the matching network's reactive elements.
// Power Amplifier Classes
Q21
Compare Class A, B, AB and C amplifier efficiency and linearity. What is the theoretical maximum efficiency of each?
EasyPA Classes
Class A: conduction angle Θ=360°, max efficiency=50%, most linear
Class B: Θ=180° (half sine), max efficiency=78.5% (π/4), moderate linearity
Class AB: 180°<Θ<360°, efficiency 50–78.5%, good compromise (most common in phones)
Class C: Θ<180°, max efficiency→100% (as Θ→0), highly nonlinear (used in CW/FM only)

Class A: Transistor conducts for the full RF cycle. Maximum linearity — always in active region. Maximum efficiency 50% (P_out/P_DC). The other 50% is always dissipated as heat even with no signal. Used for: low-power LNAs and driver stages where linearity is paramount.

Class B: Transistor conducts for only half the RF cycle (each transistor in a push-pull pair). More efficient — no idle current. Has crossover distortion at zero crossing — requires push-pull configuration. Used in: audio amplifiers, some RF push-pull stages.

Class AB: Slight positive bias so transistor conducts slightly more than 180° — eliminates crossover distortion while maintaining efficiency closer to Class B. The dominant choice for RF power amplifiers in cellular handsets.

Class C: Biased below threshold — conducts only at input signal peaks. Very high efficiency but severe nonlinearity (clips most of the cycle). Suitable only for constant-envelope signals (FM, CW, FSK).

💡 LTE and 5G NR phones use Class AB PAs at ~30–40% efficiency. A 23 dBm (200 mW) transmit power consumes ~500–700 mW from the battery — about 20–25% of total phone power in a call. Improving PA efficiency by 5 percentage points extends battery life noticeably.
Q22
What is a Doherty power amplifier? How does it maintain high efficiency at power back-off?
HardPA Classes

The Doherty PA (invented 1936, widely adopted for 3G/4G/5G base stations ~2000s) uses two amplifiers — a "main" (carrier) amplifier and an "auxiliary" (peaking) amplifier — combined through a quarter-wave impedance inverter.

Operation:

  • At low power (back-off region): only the main amplifier is active. The auxiliary is off. The main amplifier sees an impedance of 2R_opt (instead of R_opt), allowing it to operate at near-saturation efficiency even at half the peak power.
  • As power increases toward peak: the auxiliary amplifier turns on and its current "load-pulls" the main amplifier's load impedance back from 2R_opt toward R_opt, allowing both amplifiers to simultaneously reach saturation at peak power.
  • At peak power: both amplifiers operate at saturation — maximum efficiency for both.
Efficiency at 6 dB back-off (typical PAPR operation point):
Doherty: η ≈ 57–65% (vs 25–35% for simple Class AB at same back-off)
This is why Doherty dominates base station PA design since 2005
💡 The quarter-wave impedance transformer in the Doherty is narrowband by nature — it works at exactly f₀. For wideband Doherty (covering 700 MHz to 2.7 GHz for multi-band base stations), the transformer must be replaced with a more complex broadband equivalent. This is an active research area in modern PA design.
Q23
What are Class D, E and F amplifiers? What makes them theoretically 100% efficient?
HardPA Classes

Class D, E and F are "switching mode" PAs where the transistor operates as a switch (fully ON or fully OFF) rather than as a linear element. Theoretically, a perfect switch dissipates zero power (either zero voltage across it or zero current through it at any instant) — giving 100% efficiency.

Class D: Two transistors in push-pull switching between supply and ground. The load is connected through an LC filter that extracts only the fundamental frequency. Theoretical efficiency: 100%. Used widely in: audio (Bluetooth speakers, Class D audio amplifiers). Difficult at RF because transistor switching time becomes significant relative to the RF period.

Class E: Single transistor with a load network carefully designed so that when the transistor turns ON, both the voltage across it AND its derivative are zero (zero-voltage switching, ZVS). No switching loss. Theoretical efficiency: 100%. Practical efficiency: 80–95% at RF. Used in: wireless power transfer, ISM-band transmitters, RFID readers.

Class F: Uses harmonic tuning — the load network presents open circuit at odd harmonics and short circuit at even harmonics. This shapes the drain voltage into a square wave and the drain current into a half-sine, minimising overlap. Theoretical efficiency: 100% (ideal square wave). Practical: 70–85%. Used in: UHF and microwave PAs where Class E switching is too fast.

💡 Class E is the dominant choice for wireless power transfer (Qi, MagSafe) and medical implant transmitters where battery life is critical. The ZVS condition is maintained over load variation by an adaptive tuning controller. 5G mmWave PAs are exploring Class F because the very high frequency makes Class E impractical — the transistor cannot switch fast enough relative to the 60 GHz RF period.
Q24
What is envelope tracking (ET)? How does it improve PA efficiency for OFDM signals?
HardPA Classes

Envelope tracking modulates the PA supply voltage in real-time to track the instantaneous envelope of the transmitted signal. Instead of running the PA from a fixed supply, the supply voltage rises and falls with the signal amplitude.

Fixed supply PA: efficiency = P_out/(V_DD²/(2R_load)) — drops rapidly at back-off
ET PA: supply tracks signal → V_DD(t) = k×|envelope(t)| + V_headroom
PA always operates near saturation → high efficiency maintained at ALL power levels

How it works:

  1. The digital baseband predicts the signal envelope (can be done with a few microseconds look-ahead)
  2. A high-efficiency DC-DC converter (switching at 50–100 MHz) follows this envelope and adjusts the PA supply voltage
  3. The PA always operates near its saturation point regardless of instantaneous output power
  4. For an OFDM signal with 10 dB PAPR: PA efficiency improves from ~10% (fixed supply, 10 dB back-off) to ~40–50% with ET
💡 Envelope tracking is now mandatory in premium 5G handset chipsets (Qualcomm Snapdragon X65, Apple A17). The ET system consists of the PA chip, a separate envelope tracker IC, and digital coordination from the modem. Implementation requires tight co-design between chipsets — a key competitive differentiator in the smartphone supply chain.
Q25
What is power added efficiency (PAE)? Why is it used instead of drain efficiency for PA characterisation?
MediumPA Classes
Drain efficiency η_D = P_out / P_DC (ignores input power)
PAE = (P_out − P_in) / P_DC = η_D × (1 − 1/G)
For G=10 dB (10×): PAE = η_D × (1−0.1) = 0.9×η_D   → small difference
For G=3 dB (2×): PAE = η_D × (1−0.5) = 0.5×η_D   → large difference

Drain efficiency ignores the RF input power consumed. For high-gain PAs this doesn't matter much. But for a driver stage with only 3 dB gain, half the output power came from the input — drain efficiency overstates how efficiently the DC power was converted to RF output net power.

PAE accounts for the RF input power subtracted from the output, giving the net RF power generated from DC power. PAE is always ≤ drain efficiency, and approaches drain efficiency as gain increases.

💡 For a final-stage PA with 10–15 dB gain, PAE ≈ drain efficiency (within 10%). For a driver amplifier with 3–6 dB gain, PAE is significantly lower. The total PA chain efficiency is the product of all stage efficiencies — optimising just the final stage while ignoring driver efficiency is a common suboptimal design approach.
// Bias, Matching and Practical Design
Q26
What is an RF choke? Why is it needed in transistor bias circuits and how do you design one?
MediumBias

An RF choke (RFC) is a high-impedance element that passes DC current to the transistor's drain/collector while blocking RF signal from flowing into the DC supply. Without it, RF energy would flow into the supply, be absorbed, and degrade performance — or worse, cause oscillation through supply coupling.

Design options:

  • Quarter-wave short-circuit stub: A λ/4 open-circuited transmission line stub in shunt with the drain. Presents a short circuit at its resonant frequency (for DC bypass) but high impedance at the design frequency. Simple but narrowband. Works well for single-band amplifiers.
  • Wound RF choke inductor: A ferrite-core or air-core inductor sized so Z=jωL > 500 Ω at the design frequency. Wideband but has self-resonance — choose an inductor whose SRF is well above the operating frequency. Add series resistor to damp the resonance and prevent RF choke from resonating with bypass capacitors.
  • Shunt resonant circuit: A parallel LC resonant at the design frequency presents very high impedance. Wideband near resonance but low impedance far from resonance — good for narrowband amplifiers.
💡 The bias network is often responsible for amplifier oscillation. At low frequencies where the transistor has high gain, the RF choke can resonate with the bypass capacitors to create a feedback path. Add a series resistor (10–100 Ω) in the RF choke to damp this resonance and stabilise the amplifier at out-of-band frequencies.
Q27
What is self-biasing in a GaAs FET amplifier? What happens if the drain voltage is applied before the gate bias?
MediumBias

GaAs pHEMT and MESFET transistors use a depletion-mode device — they are normally ON with zero gate bias and require a negative gate voltage (typically −0.5 to −2 V) to reduce drain current to the desired operating point. They have a Schottky gate junction that can be damaged by excessive positive gate-source voltage (> +0.5 V).

Self-biasing (source resistor bias): A resistor R_S is placed in the source connection. The drain current I_D flowing through R_S creates a positive voltage at the source, which makes the gate negative relative to the source (V_GS = V_G − I_D×R_S). This automatically sets the operating point — no separate negative supply needed.

Sequencing problem: If V_DS is applied before V_GS is set negative, the transistor initially has V_GS = 0 — maximum drain current flows. The transistor may thermally stress or the drain current may exceed the RF choke's current rating. For depletion-mode GaAs FETs, always:

  1. Apply gate bias (negative voltage) first to pinch off the device
  2. Then apply drain voltage
  3. On power-down: remove drain first, then gate
💡 Many GaAs PA failures in the field are caused by power sequencing violations — the drain voltage appears before the negative gate supply is established during power-up. A simple diode clamp and power sequencing IC prevents this. It's often the most important protection circuit in a GaAs MMIC-based design.
Q28
What is a broadband amplifier matching technique? Compare resistive feedback vs reactive multi-section matching.
HardMatching

Broadband amplifiers must provide gain over a wide frequency range (e.g. 1–10 GHz) where the transistor's gain naturally falls at 6 dB/octave (20 dB/decade) with frequency. Two main approaches:

Resistive feedback: A resistor from drain to gate (or collector to base) provides wideband negative feedback. The gain becomes frequency-independent (determined by the feedback resistor ratio), and input/output impedances are flattened. The amplifier is inherently stable. Tradeoff: gain is reduced by the feedback factor, and efficiency is lower because the feedback resistor dissipates power. NF degrades compared to a narrowband design.

Feedback gain: G ≈ R_F / R_source (simplified)
Input impedance: Z_in ≈ R_F / G   Output impedance: Z_out ≈ R_F

Reactive multi-section matching (gain slope equalisation): Design the input matching network to intentionally mismatch the input more at low frequencies (where gain is naturally high) and less at high frequencies (where gain is lower). This creates a "gain slope" in the matching network that compensates the transistor's natural gain rolloff. Higher gain than feedback approach but narrower bandwidth and more complex design.

💡 Resistive feedback is used in DC-to-microwave broadband amplifiers (MMIC catalogue parts, e.g. GALI, ERA series). Multi-section reactive matching is used in high-performance microwave amplifiers where low NF is required. For the widest bandwidths (DC to 20 GHz), resistive feedback is the only practical choice.
Q29
What is load-pull measurement? What does it reveal that S-parameter measurement cannot?
MediumMeasurement

Load-pull is a large-signal measurement technique where the load impedance presented to a transistor is systematically varied (using a tuner or active load-pull system), and the output power, efficiency, gain and linearity are measured at each impedance point. Contours of constant output power, PAE and IM3 are plotted on the Smith chart.

What S-parameters cannot tell you:

  • S-parameters are small-signal — they assume linear operation. At large signal levels (near P1dB), the transistor impedance changes significantly. The optimum output match for maximum power (Z_opt_load) is completely different from the conjugate match of S22*.
  • Load-pull reveals the P1dB and PAE contours — the load impedance that gives maximum efficiency or maximum power at a specified power level
  • Source-pull (varying source impedance) simultaneously reveals the optimum input match for both gain and NF at large signal
S22 → small-signal conjugate match → maximum available gain at small signal
Load-pull → large-signal optimum load → maximum output power or efficiency at P1dB
These are always different impedances — typically 5–30° apart on Smith chart
💡 PA matching networks are ALWAYS designed using large-signal load-pull data, never from S22 alone. The small-signal S22 match would give the wrong load and significantly sub-optimal P1dB and PAE. S-parameters are used only for small-signal stages (LNA, driver) where the transistor operates linearly.
Q30
What is digital predistortion (DPD)? How does it work and what nonlinearities does it correct?
HardLinearisation

Digital predistortion is a software linearisation technique that applies the inverse of the PA's nonlinear transfer function to the digital baseband signal before it reaches the PA. The combined PA + predistortion block behaves as a linear amplifier.

Ideal: DPD(x) × PA(x) = G·x (linear)
DPD function: f_DPD(x) = PA−¹(x) / G
DPD corrects: AM/AM (gain compression), AM/PM (phase vs amplitude), memory effects

How it works in practice:

  1. A sample of the PA output is fed back (observation receiver) and compared to the input
  2. An adaptive algorithm (LUT-based or Volterra series) computes the predistortion coefficients by minimising the error between the ideal linear output and the actual PA output
  3. The coefficients are updated continuously (every few milliseconds) to track PA aging, temperature change and power level variation
  4. The predistorted signal is sent to the DAC and upconverter before the PA

DPD can improve ACPR by 15–25 dB and reduce EVM from 5–10% to <1%, allowing the PA to operate 3–6 dB closer to compression and improving efficiency.

💡 DPD is now standard in all 4G and 5G base stations. Without DPD, a PA running at 46 dBm with 10 dB PAPR would need to back off 10 dB to meet ACPR specs — giving terrible efficiency. With DPD, the same PA can operate at 40 dBm average (6 dB back-off) while meeting ACPR — 4× better efficiency from software alone.
// Design Scenarios and System
Q31
Design an LNA for a 5 GHz WiFi receiver. Specify transistor requirements and matching strategy.
HardLNA Design

Requirements from system budget: NF < 2 dB, gain > 15 dB, IIP3 > −5 dBm, input VSWR < 2:1.

Transistor selection: GaAs pHEMT or CMOS 65 nm. At 5 GHz, a 0.1 μm GaAs pHEMT gives: NF_min ≈ 0.4 dB, |S21| ≈ 18–22 dB, Γ_opt ≈ 0.55∠170°, R_n ≈ 15 Ω. This gives the budget to achieve NF=1.5 dB with some source mismatch from ideal Γ_opt.

Matching strategy:

  1. Input match: Design for Γ_S near Γ_opt (NF priority). Use an L-network or single-stub match on the Smith chart from 50 Ω to Z_opt. Accept 1–2 dB gain reduction vs maximum available gain.
  2. Output match: Conjugate match to S22* for maximum gain. The input and output are nearly independent (small S12) so design sequentially.
  3. Bias: Use self-bias source resistor for stability. Add RF choke at drain. Decouple with 100 pF + 10 nF capacitors to ground within 2 mm of the transistor.
  4. Stability check: Verify K>1 or μ>1 at all frequencies 100 MHz–20 GHz. Add 10 Ω series resistor at gate if needed at out-of-band frequencies.
Expected results: NF=1.4 dB, G=16 dB, IIP3=−3 dBm, OP1dB=+5 dBm
System NF with preselector (0.5 dB IL): NF_sys = 0.5 + 1.4 = 1.9 dB
💡 Always simulate the complete LNA including matching networks, bond wires (1 nH each), package parasitics and PCB ground parasitics before fabrication. The ideal S-parameter design and the PCB reality typically differ by 0.3–0.5 dB NF and 1–2 dB gain — accounting for this in the simulation prevents expensive board respins.
Q32
A PA shows oscillation at 100 MHz during testing. The design frequency is 2.4 GHz and K>1 at 2.4 GHz. Diagnose and fix.
HardDebug

The transistor has very high gain at 100 MHz (gain falls 6 dB/octave, so if it has 20 dB at 2.4 GHz it has ~40 dB at 100 MHz). K at 100 MHz is almost certainly <1 because S12 effects are much more significant at low frequency. The oscillation is caused by a feedback loop at 100 MHz that was not accounted for in the 2.4 GHz stability analysis.

Diagnosis:

  1. Measure S-parameters from 10 MHz to 5 GHz — look for K<1 at low frequencies
  2. Identify the feedback path: check if the oscillation frequency corresponds to a resonance in the drain RF choke + bypass capacitor combination
  3. Check if input and output trace coupling provides feedback at 100 MHz

Fixes:

  • Add a series resistor (22–56 Ω) in the gate bias line: This provides resistive loading at low frequencies where gain is highest, reducing the loop gain below oscillation threshold without affecting the 2.4 GHz design
  • Add a ferrite bead in the RF choke: Ferrite beads provide resistive loss above ~100 MHz, damping the RFC resonance that creates the feedback path
  • Reduce RF choke inductance: Smaller L reduces the choke resonance Q and shifts the problematic resonance frequency
  • Add a shunt resistor (100–470 Ω) at the drain: Loads the output at all frequencies, reducing gain and improving low-frequency stability — negligible effect at 2.4 GHz if large enough resistor value
💡 Always simulate and verify stability from DC to 3×f_operating using S-parameters swept over this full range. Most RF simulators default to simulating from 100 MHz — missing exactly the low-frequency instability range where trouble commonly occurs. Set the sweep start to 10 MHz.
Q33
What is the Friis linearity formula for cascaded IIP3? Calculate for: LNA (G=18 dB, IIP3=−5 dBm) and Mixer (IIP3=+8 dBm).
HardSystem Design
Friis linearity cascade (referred to input):
1/IIP3_total = 1/IIP3_1 + G_1/IIP3_2 + G_1·G_2/IIP3_3 + … (all linear)
IIP3_LNA = −5 dBm → 0.316 mW   IIP3_mixer = +8 dBm → 6.31 mW
G_LNA = 18 dB → 63.1 (linear)
1/IIP3_total = 1/0.316 + 63.1/6.31 = 3.16 + 10.0 = 13.16
IIP3_total = 1/13.16 = 0.0760 mW → 10·log₁₀(0.0760) = −11.2 dBm

The mixer dominates the cascaded IIP3 (contributes 10 of the 13.16 total, vs 3.16 from the LNA). The mixer's IIP3 referred through the 18 dB LNA gain is 8−18 = −10 dBm — worse than the LNA's own −5 dBm. To improve: reduce LNA gain by 3 dB → mixer contribution drops from 10 to 5 → total IIP3 improves to ~−8 dBm. NF will degrade slightly.

💡 The cascaded IIP3 formula shows that gain before a nonlinear stage is the enemy of linearity. Every 1 dB of additional LNA gain reduces cascaded IIP3 by 1 dB (mixer contribution dominates). This is why LNA gain is capped at 15–20 dB in most receivers — more gain gives negligible NF improvement but significant IIP3 degradation.
⚡ RX Chain Designer — IIP3 cascade
Q34
A receiver requires −95 dBm sensitivity at 10 MHz bandwidth. What maximum NF is allowed? Design a two-stage LNA to achieve it.
HardSystem Design
Thermal noise floor: kTB = −174 + 10·log₁₀(10×10⁶) = −174+70 = −104 dBm
Required SNR for demodulation: assume SNR_min = 10 dB (QPSK at BER=10−³)
Maximum NF = Sensitivity − kTB − SNR_min = −95−(−104)−10 = −1 dB?
— Check: −104 + NF + SNR = −95 → NF = −95+104−10 = −1 dB?

Wait — NF cannot be negative. Re-examine: if sensitivity = −95 dBm, kTB = −104 dBm, SNR = 10 dB, then NF = −95−(−104)−10 = 9−10 = −1 dB is impossible.

Correct interpretation: sensitivity = kTB + NF + SNR_min → −95 = −104 + NF + 10 → NF = −95+104−10 = −1 dB. This is impossible — the specification cannot be met at 10 MHz bandwidth with QPSK. Either reduce SNR requirement (simpler modulation), reduce bandwidth (narrow channel), or reduce SNR_min (use FEC coding).

With FEC at rate 1/2, effective SNR_min ≈ 3 dB → NF = −95+104−3 = 6 dB — achievable.

Two-stage LNA for NF=4 dB (2 dB margin):
Stage 1: GaAs pHEMT, G=12 dB, NF=1.5 dB
Stage 2: GaAs pHEMT, G=12 dB, NF=3 dB
Friis: F=1.413+(2.0−1)/15.85=1.413+0.063=1.476 → NF=1.69 dB total
💡 This worked example shows the importance of checking system feasibility before designing hardware. A sensitivity requirement that cannot be met with any NF > 0 dB means the link budget is infeasible and must be renegotiated — either increase TX power, reduce path loss, use directional antennas, or use more robust modulation.
⚡ RX Chain Designer
Q35
Complete system: Antenna (0.5 dB loss) + LNA (G=18 dB, NF=1.2 dB, IIP3=−5 dBm) + Filter (IL=2 dB) + Mixer (CL=7 dB, NF=7 dB, IIP3=+10 dBm). Calculate cascaded NF and IIP3. State which stage dominates each.
HardSystem Design
NF cascade (Friis), all 4 stages:
F_ant=10^(0.5/10)=1.122, G_ant=10^(−0.5/10)=0.891
F_LNA=1.318, G_LNA=10^(18/10)=63.1
F_filt=10^(2/10)=1.585, G_filt=0.631
F_mix=10^(7/10)=5.012

F_total=1.122 + (1.318−1)/0.891 + (1.585−1)/(0.891×63.1) + (5.012−1)/(0.891×63.1×0.631)
= 1.122 + 0.357 + 0.0104 + 0.113 = 1.602 → NF = 2.05 dB

Dominant: antenna cable (0.357) and LNA (1.122 base+0.357 antenna = 1.479 combined)
IIP3 cascade:
1/IIP3 = 1/IIP3_LNA_eff + G_LNA_eff/IIP3_mix
IIP3_LNA_eff at input = −5+(−0.5) = −5.5 dBm (cable degrades IIP3 slightly, negligible)
Effective input IIP3_LNA = −5 dBm = 0.316 mW
1/IIP3 = 1/0.316 + (0.891×63.1)/IIP3_mix = 3.16 + 56.2/(IIP3_mix_effective)
IIP3_mix_eff = 10 dBm through 0.631 filter = +10−2 = +8 dBm relative, but need at system input:
IIP3_mix_input_referred = 10−18−(−2) = −6 dBm = 0.251 mW
1/IIP3 = 3.16 + 56.2/0.631/1 ... (simplified) → IIP3_total ≈ −11 dBm
Dominant: mixer through LNA gain

Summary: NF is antenna/cable + LNA dominated. IIP3 is mixer (through LNA gain) dominated. Both problems have the same solution — reduce LNA gain by 3 dB, which improves IIP3 by ~2 dB at the cost of ~0.2 dB NF.

💡 This is the classic receiver design dilemma: the same component (LNA gain) that helps NF hurts IIP3. The gain plan — choosing the gain of each stage — is the most important system-level decision in receiver design. Get it wrong and no amount of component optimisation will fix the overall performance.
⚡ RX Chain Designer — complete cascade