Emitter & source followers
Derive the two canonical non-inverting buffer stages from complete biased circuits and their real small-signal equivalents. Show why their voltage gain approaches but stays below unity, why their input resistance can be high and their output resistance low, how load resistance enters the gain directly, and how emitter/source degeneration is the operating principle rather than an unwanted correction.
- Start from complete DC-biased emitter-follower and source-follower circuits and identify the required Q-point region checks.
- Redraw each stage at midband with the collector/drain supply at AC ground and the actual load attached to the emitter/source node.
- Derive the BJT follower gain vout/vb=[(β+1)RE′]/[rπ+(β+1)RE′] with RE′ the effective emitter load under the declared low-frequency assumptions.
- Derive the corresponding base input resistance rin,b=rπ+(β+1)RE′ and explain why the follower can buffer a high-resistance source from a lower-resistance load.
- Derive the MOS source-follower gain vout/vg=gmRS′/(1+gmRS′) with RS′ the effective source load in the simplified model.
- Explain why both gains are positive and less than unity for passive positive loads in the first-order models.
- Derive a first output-resistance estimate for each follower with the input increment zero and state the assumptions behind the approximation.
- Include bias-network and signal-source loading so the total source-to-load voltage transfer is not confused with the intrinsic follower gain.
- Explain the effect of body transconductance gmb in a MOS source follower and why it reduces gain when the body is not tied to the source.
- Validate limiting cases: load→∞, gm→∞, β→∞, load→0 and source resistance→∞.
A follower has voltage gain near one, so why is it useful? Because voltage gain is only one property: followers trade gain for buffering, high input resistance, low output resistance and the ability to drive a lower-resistance load.
02.1Emitter follower: output taken from the feedback node
Figure 2.1 — Emitter follower: complete circuit
The one question is: why is the output taken from the emitter while the collector remains tied to the supply?
R1–R2 and RE establish Q. The collector is tied to VCC, which becomes AC ground in small signal. The output follows the emitter through Cout.
The collector is tied to the supply, the input controls the base and the output is taken from the emitter. If emitter current rises, emitter voltage rises through RE. That rise reduces vBE relative to the original base perturbation, opposing the current increase.
The stage therefore contains local negative feedback by construction. It is not a common-emitter stage with the “wrong” output node.
02.2Derive the BJT follower gain
Figure 2.2 — BJT follower: small-signal equivalent
The one question is: how does controlled current drive RE′ to create an output with the same sign as the input?
RE′ is the effective emitter load. rπ connects base to emitter; gm·vπ injects current from the AC-grounded collector into the emitter node.
Let the effective emitter load be
and use the low-frequency hybrid-π model. The emitter current increment is
Because vo=ieRE′ and vπ=vb−vo,
so
or equivalently
For positive passive RE′, the gain is positive and strictly below one.
02.3High input resistance is the reflected emitter load
The base current is ib=vπ/rπ while emitter current is (β+1)ib. The base sees
If the base bias divider has equivalent resistance RB, the complete stage input is
A low emitter load can therefore be “reflected” upward by roughly β+1 at the base, which is the core buffering effect.
02.4Worked emitter-follower example
Let gm=40 mS, β=100, rπ=2.5 kΩ, RE=2.2 kΩ, RL=1 kΩ and ro=50 kΩ.
Therefore
The stage delivers nearly the base voltage to a 1 kΩ load while presenting tens of kilohms at the transistor base.
02.5Emitter-follower output resistance
To estimate Rout, disconnect the load, zero the independent input source and retain the bias/source resistance seen at the base. Let RBs be the resistance from base to AC ground after source zeroing.
Neglecting higher-order effects, the emitter sees approximately
This shows the complementary buffering property: resistance connected to the base is divided downward by approximately β+1 when viewed from the emitter.
02.6Source follower: same feedback pattern, field-effect input
Figure 2.3 — Source follower: complete circuit
The one question is: how does the gate set the control point without supplying load current?
RG1–RG2 establishes VG, the drain stays on VDD, and the source supplies current to RS and the load through Cout. Load current comes from the drain–source path, not from the gate.
The MOS drain is tied to the supply and the output is taken from the source. As source voltage rises, vGS=vg−vs falls, reducing the controlled drain current. The local feedback mechanism is therefore directly visible in the control voltage itself.
The gate draws negligible ideal low-frequency current, but the bias network still determines the circuit input resistance.
02.7Derive the source-follower gain
Figure 2.4 — MOS follower: small-signal equivalent
The one question is: why does vo approach vg without becoming exactly equal to it?
The drain is AC ground. gm·vgs drives RS′, but a rise in vo reduces vgs: this is the follower’s intrinsic local negative feedback.
Let
and neglect body effect first. KCL at the source gives
so
Again, a positive passive load gives a positive gain below unity.
02.8Worked MOS follower and body effect
Let gm=5 mS, RS=3.3 kΩ, RL=2 kΩ and ro=100 kΩ. Then
If the body is AC-grounded rather than tied to the source, body effect adds a transconductance gmb that opposes source motion. The simplified gain becomes
Thus body effect lowers follower gain for the same gm and load.
02.9Why followers are buffers rather than voltage amplifiers
The useful “gain” is often power/current-driving capability: a weak source controls a larger transistor current drawn from the DC supply while the signal voltage is reproduced approximately at the output.
02.10Mastery exercises
- For β=99, rπ=2 kΩ and RE′=500 Ω, find emitter-follower Av and rin,b.
- What happens to Av as RE′→∞? What about RE′→0?
- If RB=50 kΩ and rin,b=80 kΩ, find the stage Rin.
- For gm=4 mS and RS′=1.5 kΩ, find source-follower Av.
- If gmb=1 mS is added to exercise 4, recompute Av.
- Why is the follower non-inverting?
- Why can the follower deliver more load current than the signal source despite Av<1?
- State one reason the real high-frequency follower input is not an infinite impedance.
Solutions
1. (β+1)RE′=50 kΩ; Av=50/(52)≈0.9615 and rin,b≈52 kΩ.
2. Av→1 as RE′→∞ and Av→0 as RE′→0.
3. 50 kΩ∥80 kΩ≈30.8 kΩ.
4. gmRS′=6, so Av=6/7≈0.857.
5. Av=6/[1+(0.004+0.001)1500]=6/8.5≈0.706.
6. A positive input increase raises emitter/source voltage; local feedback reduces the difference but does not reverse the sign.
7. The transistor draws energy/current from the DC supply and controls it with the input signal.
8. Gate/base capacitances and parasitics create frequency-dependent input current.
Chapter 03 will use matched transistor behavior to build current mirrors, the first reusable bias blocks that feed later active loads.