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Amplifier Transfer Function
Note that the ideal amplifier transfer function:
oc vout( t) = A vo v i ( t )
is an equation of a line (with slope = Avo and y -intercept = 0).
v out
Avo < 0
A > 0 vo
v in
This ideal transfer function implies that the output voltage can be very large, provided that the gain Avo and the input voltage vin are large. 4/3/2018 0b889a89cca0d5a93dc4f68c3707e08c.doc 2/7
However, we find in a “real” amplifier that there are limits on how large the output voltage can become. The transfer function of an amplifier is more accurately expressed as:
in L+ vin ( t ) > L +
in in vout( t )= A vo v in ( t ) L- < v in ( t ) < L +
in L- vin ( t ) < L -
This expression is shown graphically as:
vout
L+
in L- L- = Avo
vin in L+ L+ = Avo Avo
L- 4/3/2018 0b889a89cca0d5a93dc4f68c3707e08c.doc 3/7
* This expression (and graph) shows that electronic amplifiers
have a maximum and minimum output voltage (L+ and L-).
* If the input voltage is either too large or too small (too negative), then the amplifier output voltage will be equal to
either L+ or L- .
* If vout = L+ or vout =L- , we say the amplifier is in saturation (or compression).
Amplifier saturation occurs when the input voltage is greater than:
L+ in vin > B L+ Avo or when the input voltage is less than:
L - ∐ in vin < L- Avo
Often, we find that these voltage limits are symmetric, i.e.:
in in L-= - L + and L - = - L +
For example, the output limits of an amplifier might be L+ = 15 V and L- = -15 V.
However, we find that these limits are also often asymmetric
(e.g., L+ = +15 V and L- = +5 V). 4/3/2018 0b889a89cca0d5a93dc4f68c3707e08c.doc 4/7
Q:Q: Why Why do do we we care care ifif an an amplifieramplifier saturates? saturates? Does Does it it causecause any any problems problems, ,or or otherwiseotherwise result result in in performance performance degradationdegradation????
A:A: Absolutely Absolutely! ! If If an an amplifier amplifier saturates—evensaturates—even momentarily— momentarily— thethe unavoidable unavoidable result result will will be be a a distorteddistorted outputoutput signal. signal.
For example, consider a case where the input to an amplifier is a triangle wave:
v (t) in
in L+
t
in L- 4/3/2018 0b889a89cca0d5a93dc4f68c3707e08c.doc 5/7
in in Since L-< vin ( t ) < L + for all time t, the output signal will be within the limits L+ and L- for all time t, and thus the amplifier output will be vout (t) = Avo vin (t):
v (t) out
L+
t
L-
Consider now the case where the input signal is much larger, in in such that vin( t )> L+ and v in ( t ) < L - for some time t (e.g., the in in input triangle wave exceeds the voltage limits L+ and L- some of the time):
v (t) in
in L+
t
in L- 4/3/2018 0b889a89cca0d5a93dc4f68c3707e08c.doc 6/7
This is precisely the situation This is precisely the situation about which I earlier about which I earlier expressed caution. We now expressed caution. We now must experience the palpable must experience the palpable agony of signal distortion! agony of signal distortion!
v (t) out
L+
t
L-
Note that this output signal is not a triangle wave!
in in For time t where vin( t )> L+ and v in ( t ) < L - , the value Avo v in ( t ) is greater than L+ and less than L-, respectively. Thus, the output voltage is limited to vout( t )= L+ and v out ( t ) = L - for these times.
As a result, we find that output vout ( t ) does not equal Avo v in ( t ) — the output signal is distorted! 4/3/2018 0b889a89cca0d5a93dc4f68c3707e08c.doc 7/7
in in In reality, the saturation voltages L+, L - , L + , and L - are not so precisely defined. The transition from the linear amplifier region to the saturation region is gradual, and cannot be unambiguously defined at a precise point.
v out
L +
in L- L- = Avo v in L+ in L+ = Avo A vo
L -
v (t) out
L+
t
L-