Outline
• Transistors/FET’s
Transistors/FET’s
• They are used to:
– Switches
– Amplifiers
Emitter Follower (Common Collector)
Common Emitter
Differential
– Analog switches (FET)
How does a transistor work? (Emitter follower)
• Use the npn transistor for explanation. Everything is opposite for pnp.
• When a transistor receives current into the base, IB, it produces current
into the collector, IC, which is β~102±0.5 times larger (current source) [as
long as it is connected to a modestly positive voltage (VCE ~ 1 V) relative
to the emitter.]
– If VCE comes down below 1 V, it is harder for electrons to move across the base to
the collector, making more electrons to flow to the base rather than to the collector,
and β will decrease.
VCC
in
+15V
2N3904
270
IB
IC
out
3.3 k
VEE
• For a typical amplifier usage, some amount of IB and IC flow (bias
currents) so that there is room for them to decrease when the input signal
requires them to. Note that they cannot flow in the opposite direction!
• In many cases, how much average currents flow through the transistor
[with no signal] is determined so that
1. maximum dynamic range is obtained – most positive and negative
parts of the signal start to be distorted at the same time.
2. Many transistors optimum IC to produce minimum noise.
• So how does this circuit function without a signal?
• In this circuit, if Vin = 0, what is VB, VE and VC, assuming that
VEE = – 15 V?
• So equations that we can use to figure out the unknowns IB, VB, IE and VE
are the following.
–
–
–
–
IB = – VB/270 from Ohm’s Law about 270Ω resister.
VE = VB – 0.6: diode forward current voltage drop.
IE = (1+β) IB: characteristics of transistor
IE = (VE – VEE)/3.3k: Ohm’s law about 3.3k resister.
• Solving this algebra is easy if tedious. Substituting everything in terms of
VB you would obtain
• (1+β)(–VB/270) = (VB – 0.6 +15)/3.3k
• So, VB ~ – (14.4/3.3k)/[(1+β)/270+1/3.3k] = – (14.4/3.3k)/(101/270) = –
0.012 V,
VE = VB – 0.6 = –0.612 V,
• VC = 15 V.
• IB ~ (14.4/3.3k)[1/(1+β)] = 43 µA.
• What conclusions can we draw from these computations?
IB is so small, one can almost ignore it.
Then VB is the same as the input voltage – the 270Ω resister may actually be the
internal resistance of the input device such as microphone or photosensor. This
means that the input impedance of this amp is large enough (under normal
circumstances) that little current flow into it, and the voltage decrease due to
the current is minimal.
VE is the same as the input except the –0.6 V offset.
• Now, when signal comes in, what will happen?
• Suppose that the input voltage changes by ∆v (can this be considered
known?), the only change in the 4 equations above is this:
• (1+β)(∆v –VB)/270 = (VB – 0.6 +15)/3.3k,
• then the solution changes to:
• VB ~ [– (14.4/3.3k)+ {(1+β)/270)} ∆v]/[(1+β)/270+1/3.3k]
= ∆v – (14.4/3.3k)(270/101) = ∆v – 0.012 V ~ ∆v; the base voltage is
basically the same as input because IB is so small, there is hardly any
voltage drop between them.
• VE = VB – 0.6 = –0.612 V ~ ∆v – 0.6V; VE is also the same as the input
except –0.6V voltage offset.
• Gain is 1: No amplification! What good is that?
By the way, since we need to know IB later, let’s calculate it; IB =
(14.4/3.3k)(1/β)
• The reason why this unit gain amp is still very useful is that often a signal
source has large output impedance so that it cannot drive a small input
impedance circuit or output device (like a speaker or motor).
• The input impedance of this circuit is relatively large (turns out to be
270 + 3.3k*β ~ 300k. This is often large enough to be connected to a
large-output impedance circuit w/o losing voltage.
• With the above calculation, ignoring 1/3.3k term, signal dependent
part of the input current and voltage are ∆v and 0. So Zin = ∞.
• When the calculation above is done w/o neglecting 1/3.3k, the
answers contain a term with ∆v.
VB contains {(1+β)/270)}/[(1+β)/270+1/3.3k] ∆v ~ (1 – [1/3.3k]/[(1+β)/270])
∆v
IB contains [1/3.3k]/(1+β) ∆v.
• Zin = ∆v/[3.3k*(1+β)]/∆v = 3.3k*(1+β).
• This means that the input to this circuit looks like it is connected to
–15 V via resister of Zin.
• Sometimes, 3.3k is the output load. Then you can interpret this as
the load resistance is increased by (1+β) by inserting the emitter
follower between the source and the load.
• The output impedance of this circuit is 270/(1+β). How can we figure
this out?
• When the output is open (just 3.3k in emitter), Vout = VE ~ ∆v – 0.6V.
• If the load is also 3.3k, it is the same as changing the emitter resistance to
3.3k/2 = 1.65k. This will not change the output voltage significantly. It
would still be ∆v – 0.6V.
• If the output voltage stay the same regardless what resister is connected to
it, then the output impedance is almost zero.
• To the next order approximation, the voltage does change slightly from
∆v[1 – 270/3.3k(1+β)] – 0.6V to ∆v[1 – 270/1.65k(1+β)] – 0.6V. The
change is ∆v[270/3.3k(1+β)]. So it must be that the fractional voltage loss
Zout/(Rout+RL) must equal 270/3.3k(1+β) or
Zout=3.3k*270/[3.3k(1+β)-270] ~ 270/(1+β) ~ 3Ω.
• One can re-interpret this in the following way. Suppose that 270Ω
represents the output impedance of the signal source (mic, for example), it
is reduced by (1+β) when this circuit is inserted after the signal source.