HW2 · Small-Signal Gain & Frequency Response (Baker 21.7, 21.26, 21.30, 21.36)

Discussion prompt

Before we open HW2 · Small-Signal Gain & Frequency Response (Baker 21.7, 21.26, 21.30, 21.36): without looking back, what was the main idea of HW2 · DC Bias: Currents & Saturation (Baker 20.25, 20.45), and what could you do by the end of it that you could not do before?

Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.

Discussion prompt

Replace every saturated MOSFET by a transconductance gm (a controlled current gm*vgs) in parallel with an output resistance ro. Two formulas generate every number today:

That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.

Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.

Put in order

These are the steps of Recipe: any single-stage gain or corner, scrambled. Put them back in order before the next slide shows you.

1. DC first: get ID, then gm and ro for every device (done - all 20 uA).
2. Find Gm: the input device's transconductance (or gm-ratio for a buffer).
3. Find Rout: combine the resistances at the output node (1/gm, ro, or cascode gm*ro^2).
4. Add caps last: each capacitor makes one pole at 1/(2*pi*R*C) - coupling caps set low corners, load caps set high corners.

Why: This is the order the recipe itself gives. Recalling the sequence without the slide in front of you is the difference between recognising the method and being able to run it — most of what goes wrong in practice is a step done out of turn.

Discussion prompt

Recipe: any single-stage gain or corner works on the cases you have just seen. Push it to the edge: what is the most degenerate input it still handles — empty, zero, one item, everything equal — and what is the first case where it stops being true? Name the case, not just "it breaks".

Hint: Try the smallest legal input, then the largest, then the one where two things collide. Methods are specified at their edges; the middle takes care of itself.

Discussion prompt

Explain The circuit: a common-gate stage by analogy to something with no Analog CMOS IC Design in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.

Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.

Discussion prompt

Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.

Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.

Discussion prompt

Explain Common gate = a current buffer with a load to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.

Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.

Predict first

With the gate at AC ground, the resistance seen looking into M2's source is the device's 1/gmn in parallel with its own ron:

Commit before you compute: what does Step 1 - resistance looking into the source come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.

Correct: Put numbers in

Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. 1/gmn = 1/155uA = 6.45 k, in parallel with ron = 833 k -> 6.40 k.

Animation

Shows: Each line of the worked example "Step 1 - resistance looking into the source", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: 1/gmn = 1/155uA = 6.45 k, in parallel with ron = 833 k -> 6.40 k. The 1/gmn dominates because ron is huge.

Discussion prompt

At the drain, three things sit in parallel: the diode load 1/gmp, the load's own rop, and M2's ron looking down:

What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.

Hint: Anything you would have to invent to get started is a thing the problem must supply.

Animation

Shows: Each line of the worked example "Step 2 - resistance at the output node", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: 1/gmp = 6.45 k dominates the two 833 k resistors -> Rout = 6.35 k. The diode-connected PMOS is a low-resistance load.

Predict first

A student writes this, and it looks reasonable:

Seeing an NMOS with a PMOS load and assuming gate input, inverting gain.

It is wrong. Say what breaks — and say it before you turn the page.

Correct: That is the common-source result.

Identify the input node first: signal enters the source, gate is AC ground -> common gate.

Why: That is the common-source result. Here the gate is AC-grounded - it is not the input - so this model is the wrong topology.

Eliminate the wrong options

In the large-ro limit, vout/vin is closest to:

3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.

• A. +1 (about 0 dB), non-inverting
• B. -gmn*ron ~ -129, inverting
• C. +gmn*ron ~ +129
• D. 0 - the gate is grounded so there is no output

Survives elimination: A

Why: Common-gate gain is Rout/Rin,source = (1/gmp||rop||ron)/(1/gmn||ron) ~ gmn/gmp. With gmn = gmp here, that is ~ +1, and it is non-inverting because the input is at the source.

Check your understanding

In the large-ro limit, vout/vin is closest to:

• A. +1 (about 0 dB), non-inverting (correct)
• B. -gmn*ron ~ -129, inverting
• C. +gmn*ron ~ +129
• D. 0 - the gate is grounded so there is no output

Answer: A

Why: Common-gate gain is Rout/Rin,source = (1/gmp||rop||ron)/(1/gmn||ron) ~ gmn/gmp. With gmn = gmp here, that is ~ +1, and it is non-inverting because the input is at the source.

Why B tempts people

That is the common-source gain with an ro load. The gate is AC-grounded here, not the input, and the load is the low 1/gmp of a diode-connected PMOS - not ro.

Why C tempts people

Right magnitude idea for an ro load but wrong load: the diode-connected M3 presents 1/gmp ~ 6.45 k, not rop, so the gain is gmn/gmp ~ 1, not gmn*ro.

Why D tempts people

AC-grounding the gate is exactly how a common-gate stage works - the source is the input. The stage amplifies normally; it does not zero the output.

Discussion prompt

Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.

Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.

Predict first

M1B is a common-source device whose load is the resistance looking into M1T's source, ~1/gm1T. So:

Commit before you compute: what does Gain 1 - vd1b/vin (the inner node) come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.

Correct: Both 10/2 at 20 uA

Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. gm1B = gm1T = 155 uA/V, so the ratio is -1.

Animation

Shows: Each line of the worked example "Gain 1 - vd1b/vin (the inner node)", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: gm1B = gm1T = 155 uA/V, so the ratio is -1. The inner node mirrors the input, inverted, with ~unity magnitude.

Fill in the blanks

From Gain 2 - vout/vd1b (the cascode device) — finish the line. Write what belongs on the right of the equals sign before you look.

R_R_{ocasn}\,\|\,R_{ocasp},\quad R_{ocas}\approx g_m r_o^2 = ___

Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Rocas ~ 155uA x (833k)^2 ~ 108 Mohm each; in parallel Rout ~ 53.8 Mohm.

Animation

Shows: Each line of the worked example "Gain 2 - vout/vd1b (the cascode device)", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: Rocas ~ 155uA x (833k)^2 ~ 108 Mohm each; in parallel Rout ~ 53.8 Mohm.

Draw it

Translate both ways. First write the expression above as a sentence with no symbols in it at all. Then cover it, and write your sentence back as notation. If the two versions disagree, the disagreement is the thing to fix.

Predict first

A student writes this, and it looks reasonable:

Treating the output node like a simple stage with Rout = ro.

It is wrong. Say what breaks — and say it before you turn the page.

Correct: Uses a single ro. That ignores the whole point of cascoding both the device and the load.

Cascoding multiplies output resistance by ~gm*ro.

Why: Uses a single ro. That ignores the whole point of cascoding both the device and the load.

Discussion prompt

The rule this trap just fixed:

Both the NMOS cascode and the PMOS cascode load present ~108 Mohm; in parallel ~53.8 Mohm.

Now break it on purpose. Build a case that violates it and follow the consequences until something visibly fails. Where does the failure first show up — and would you have noticed it if you had not been looking?

Hint: The dangerous rules are the ones whose violation still produces an answer. If yours fails loudly, try to find one that fails quietly.

Check your understanding

Which statement matches the three gains?

• A. vd1b/vin ~ -1, vout/vd1b ~ +8300, vout/vin ~ -8300 (correct)
• B. vd1b/vin ~ -8300, vout/vd1b ~ -1, vout/vin ~ -8300
• C. All three ~ -129 (the intrinsic gain)
• D. vd1b/vin ~ -1, vout/vd1b ~ +1, vout/vin ~ -1

Answer: A

Why: The inner node is loaded by 1/gm1T (looking into the cascode source), so vd1b/vin ~ -gm1B/gm1T ~ -1. The cascode device then drives the full Rout, giving vout/vd1b ~ gm1T*Rout ~ +8300. Their product is vout/vin ~ -8300.

Why B tempts people

Swaps where the gain appears. The inner node sees only 1/gm (low R) so it can't have the big gain; the big gain is at vout where Rout is ~53.8 Mohm.

Why C tempts people

That uses a single ro for Rout. Both the device and the load are cascoded, so Rout ~ gm*ro^2 and the gain is ~65x larger than intrinsic.

Why D tempts people

Correct that the inner node is ~ -1, but vout/vd1b is the cascode gain gm1T*Rout ~ 8300, not 1 - otherwise the stage wouldn't amplify at all.

Discussion prompt

Explain The circuit: common-source with coupling to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.

Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.

Discussion prompt

Explain A coupling cap blocks DC, passes signal by analogy to something with no Analog CMOS IC Design in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.

Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.

Fill in the blanks

From Midband gain — finish the line. Write what belongs on the right of the equals sign before you look.

A_-g_{m2}\,(r_{o2}\,\|\,r_{o1}) = ___

Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. With gm ~ 155 uA/V and ro ~ 833 k (ID = 20 uA), ro2||ro1 ~ 417 k gives |Av| ~ gm*ro/2 ~ 64 (~36 dB).

Animation

Shows: Each line of the worked example "Midband gain", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: With gm ~ 155 uA/V and ro ~ 833 k (ID = 20 uA), ro2||ro1 ~ 417 k gives |Av| ~ gm*ro/2 ~ 64 (~36 dB). Use your section's exact sizes/bias from the table.

Predict first

A student writes this, and it looks reasonable:

Seeing a capacitor and assuming the gain falls at high frequency.

It is wrong. Say what breaks — and say it before you turn the page.

Correct: Confuses a series coupling cap with a shunt load cap.

Ask: is the cap in series (coupling) or to ground (load)?

Why: Confuses a series coupling cap with a shunt load cap. A series cap's impedance falls with frequency - it helps the signal through, not blocks it.

Check your understanding

The Big coupling cap sets:

• A. A low-frequency (high-pass) corner; midband is flat above it (correct)
• B. A high-frequency (low-pass) corner; gain rolls off above it
• C. The midband gain value itself
• D. Nothing - an ideal cap has no effect on frequency response

Answer: A

Why: A series coupling cap is open at DC and a short at high frequency, so it forms a high-pass edge with the series resistance: fL = 1/(2piR*Cbig). Above fL the gain is flat at the midband value. 'Big' makes fL very low.

Why B tempts people

That describes a shunt load cap to ground. A series coupling cap does the opposite - its impedance drops with frequency, so it passes highs and blocks lows.

Why C tempts people

Midband gain is set by gm and the output resistance (gm2*(ro2||ro1)), not by the coupling cap. The cap only sets where the low end rolls off.

Why D tempts people

Even an ideal cap has frequency-dependent impedance (1/jwC); it is exactly that dependence that creates the high-pass corner at fL.

Discussion prompt

M2 (PMOS, 30/2) is the common-source stage. Its drain (vout) is loaded by a cascode NMOS current source (huge Rocasn) and a 10 pF capacitor to ground.

That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.

Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.

Fill in the blanks

From Midband gain — finish the line. Write what belongs on the right of the equals sign before you look.

|A_g_{m2}\,R_{out}\approx g_{m2}\,r_{o2}| = ___

Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. 155 uA/V x 833 k ~ 128, about 42 dB.

Animation

Shows: Each line of the worked example "(a) The high-frequency pole and Bode plot", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: Flat at |Av| = 42 dB up to fH ~ 19 kHz, then a single-pole -20 dB/decade roll-off above it. (The Big input cap's low corner is far below and ignored.)

Discussion prompt

1 MHz is far above fH ~ 19 kHz, so we're on the -20 dB/dec slope, not the flat midband. The gain there is well below 128:

What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.

Hint: Anything you would have to invent to get started is a thing the problem must supply.

Animation

Shows: Each line of the worked example "(b) Waveforms at 1 MHz, 10 mV", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: Common source inverts (180 deg); a pole well past its corner adds another ~90 deg lag. Sketch the output ~25 mV, inverted and phase-shifted relative to the 10 mV input.

Predict first

For a single-pole response, the unity-gain frequency is the gain-bandwidth product: midband gain times the corner.

Commit before you compute: what does (c) Unity-gain frequency come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.

Correct: Notice ro cancels

Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. fu depends only on gm2 and CL - not on ro or lambda.

Animation

Shows: Each line of the worked example "(c) Unity-gain frequency", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: fu depends only on gm2 and CL - not on ro or lambda. That makes it the robust number to quote.

Predict first

A student writes this, and it looks reasonable:

Reporting the corner fH as the unity-gain frequency.

It is wrong. Say what breaks — and say it before you turn the page.

Correct: At fH the gain is still |Av,mid|/sqrt(2) ~ 90, nowhere near 1.

Unity gain is where |Av| falls to 1, a factor |Av,mid| past the corner.

Why: At fH the gain is still |Av,mid|/sqrt(2) ~ 90, nowhere near 1. The corner is where roll-off starts, not where gain hits unity.

Sort into buckets

Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.

Holds up

Replace every saturated MOSFET by a transconductance gm (a controlled current gm*vgs) in parallel with an output resistance ro. Two formulas generate every number today:; M2 (NMOS, 10/2) is the amplifier. Its gate is held at a DC bias by the M1/20 uA mirror through a Big resistor and tied to AC ground by a Big cap.; Note the non-inverting sign: push current into the source and the drain voltage rises. That's the opposite of a common-source stage.

Breaks

Seeing an NMOS with a PMOS load and assuming gate input, inverting gain.; Treating the output node like a simple stage with Rout = ro.

sound

These are stated as this lesson states them — each one survives the edge cases HW2 · Small-Signal Gain & Frequency Response (Baker 21.7, 21.26, 21.30, 21.36) puts it through.

flawed

Each of these is lifted from a trap in this deck: reasonable-sounding, and wrong in a way that only shows up once you rely on it.

Eliminate the wrong options

The unity-gain frequency fu is closest to:

3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.

• A. 2.47 MHz
• B. 19 kHz
• C. 245 kHz
• D. 316 MHz

Survives elimination: A

Why: fu = |Av,mid|fH = gm2/(2piCL) = 155uA / (2pi*10pF) ~ 2.47 MHz. The ro cancels, so fu depends only on gm and CL.

Check your understanding

The unity-gain frequency fu is closest to:

• A. 2.47 MHz (correct)
• B. 19 kHz
• C. 245 kHz
• D. 316 MHz

Answer: A

Why: fu = |Av,mid|fH = gm2/(2piCL) = 155uA / (2pi*10pF) ~ 2.47 MHz. The ro cancels, so fu depends only on gm and CL.

Why B tempts people

That is the corner fH where roll-off begins; the gain there is still ~90, far above 1. Unity gain is |Av,mid| further out in frequency.

Why C tempts people

Off by ~10x - looks like |Av,mid|/sqrt-ish times fH or a decade slip. The clean way is gm/(2piCL), which gives 2.47 MHz.

Why D tempts people

That is gm/CL without the 2pi (an angular frequency in rad/s read as Hz). Dividing by 2pi gives the actual 2.47 MHz.

Comparison matrix

From Verifying in SPICE (.tran and .ac): refill the set up column from what you know. The rest of the table is as it appeared.

Draw it

One page, no notation unless you need it: draw how these connect — Problem 21.26 Common-Gate Gain · Problem 21.30 A Cascode's Three Gains · Problem 21.7 Frequency Response of a CS stage · Problem 21.36 Bode, Waveforms & GBW. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.