# P04 - UC Davis Physics

```Physics 116A Notes
Fall 2004
David E. Pellett
Draft v.0.9
• Notes Copyright 2004 David E. Pellett unless stated otherwise.
• References:
– Text for course:
Fundamentals of Electrical Engineering, second edition, by Leonard
S. Bobrow, published by Oxford University Press (1996)
– Others as noted
1
Physics 116A, BJT Amplifiers: Outline
• Small signal AC analysis (review)
• A simple small signal AC BJT model and alternatives
• Emitter follower small signal AC analysis
– Voltage gain
– Input impedance
– Output impedance
• Common emitter amplifier small signal AC analysis
– Voltage gain
– Effect of emitter resistor, RE
2
Simple BJT Amplifiers
Simple one-BJT amplifier analysis:
• Specify amplifier configuration (one BJT node is shared (common) between the input and output circuit)
– Common collector (emitter follower)
– Common emitter
– Common base
• Specify BJT operating point (quiescent point or “Q point”)
– BJT biasing
• Small signal AC analysis of resulting circuit
– Linearized small signal AC models of BJT
– Small signal AC equivalent circuit
– Use to calculate desired quantities such as Av , Rin.
3
Small Signal AC Analysis
At Q point, VBQ = 1.7 V, IBQ = 20 &micro;A, IEQ = 2.0 mA, VEQ = 1.0 V, etc.
R1 = 40 kΩ.
• Introduce signals, vi or ii, which are assumed to be small, first-order linear
variations about the Q point
– E.g., VB = VBQ + vb, where vb is the small signal AC base voltage.
– Like X + ∆X in the calculus ⇒ f (X0 + ∆X) ≈ f (X0 ) +
df
| ∆X
dX X0
– Keeping just first-order terms allows us to “linearize” non-linear circuit
elements like diodes and transistors
– The Q-point DC values (including supply voltages) are ignored in the
small signal model: keep variations from Q-point only.
4
Small Signal AC Analysis: re for BJT
5
A Simple Small Signal AC BJT Model
• Note that vbe can be positive or negative (if negative, the currents are
opposite).
• The same small signal AC model suffices for npn or pnp transistors (just
use |IE | in finding re ).
• The BJT transconductance is defined as gm ≡ ic /vbe ≈ 1/re
6
Alternative Small Signal AC BJT Models
7
Small Signal AC Equivalent Circuit
• We use superposition to separate the small signal AC circuit response
due to vin from the DC Q point values.
• If a voltage at some point does not change due to vin (e.g., VCC ), it will
be zero in the small signal equivalent circuit. Thus, R1 and the BJT
collector connect to ground.
• The BJT is replaced by a linear small signal model.
8
Small Signal AC Analysis: Av
Calculate the voltage gain, Av = vout/vin
• Since C1 , C2 → ∞ they are short circuits for AC, so they are replaced with
wires: vin = vb and vout = ve
• vb = ie re + ie RE , ve = ie RE , so Av = ve /vb = RE /(re + RE ).
• In this case, Av = 500 Ω/(13 Ω + 500 Ω) = 0.975.
(Note that Av is not very sensitive to re here since RE is large.)
9
Small Signal AC Analysis: Rin
Calculate the input (AC) resistance, Rin = vin/iin.
First calculate the resistance “looking into the BJT base,” Rb ≡ vb/ib:
vb = ie (re + RE )
ie = ib + ic = ib + hf eib = (hf e + 1)ib
Rb = vb/ib = (hf e + 1)(re + RE ) = (101)(513 Ω) = 52 kΩ.
|| R0 s ⇒ Rin−1 = R1 −1 + R2 −1 + Rb−1 ⇒ Rin = 6.0 kΩ.
10
Small Signal AC Analysis: Rout
Rout = v/i for a signal applied at the output (think Thévenin).
We also connect a signal source to the input (with vs = 0) since the source
impedance affects the answer. Let R = Rs ||R1 ||R2 . Then vb = ibR. Now find
the resistance looking into the emitter, Re ≡ ve /ie : ve = ie re + vb = ie re + ibR =
ie re + ie R/(hf e + 1) ⇒ Re = re + R/(hf e + 1). Rout = Re ||RE = Re RE /(Re + RE ).
Put in numbers: if Rs = 600 Ω, Re = 18.5 Ω and Rout = 18 Ω. Suitable for
driving a low impedance load.
11
Common Emitter Amplifier: Q Point
• Add RC in the collector circuit and take the output from the collector to
get a CE amplifier.
• Check that BJT in active region with this choice of RC . The base and
emitter circuits are the same as before, so assume VBQ = 1.7 V, IBQ = 20 &micro;A,
IEQ = 2.0 mA, VEQ = 1.0 V. Find VC and VCE : VC = VCC − IC RC .
IC = IE − IB ≈ 2.0 mA. ⇒ VC = 10 V − 2.0 mA &times; 2.5 kΩ = 5 V.
VCE = 5 V − 1 V = 4 V(&gt; .2 V).
• The BJT is in the active region with VCQ = 5 V.
12
CE Small Signal AC Analysis: Av
Av = vout/vin
vout = −ic RC
vin = ie (re + RE )
ic /ie = hf e/(hf e + 1)
Av = −hf eRC /((hf e + 1)(re + RE )) ≈ −RC /(re + RE ).
• In this case, Av = −4.8 and the output resistance, Rout = RC = 2.5 kΩ.
13
Effect of RE and Emitter Bypass Capacitor
• RE stabilizes amplifier operation and improves linearity for large signals
at the expense of gain.
– Characteristic of negative feedback
• You can bypass all or part of RE for AC with a large capacitor (CE ||RE )
– Keeps Q point stability since DC characteristics are the same
– If RE fully bypassed, gain is maximum (Av = −RC /re ≈ −190) but
expect nonlinearity and gain variations for different BJTs
14
```