Download Lab3Questions

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What is the common-mode voltage, VCM? Does VCM change? Why
or why not?
o The common mode voltage is 2.5Vdc. No, the VCM does not
change because is used as a reference for the AC input signal
to swing around. It is created by the 10k Voltage divider at
the positive input.
What is the ideal closed-loop gain?
o The ideal closed-loop gain is –Rf/Rin which is equal to 5k/5k =
1 or 0dB/Unity Gain.
What is the output swing and what is it centered around?
 The output is centered around VCM which is set at 2.5 V.
o What happens if the input isn't centered around VCM, that is,
2.5 V?
 If the input is not centered around 2.5 V, then the opamp’s open-loop gain is not sufficient to pull Vm up to Vp.
If the input voltage at the Vp terminal increases, the
output voltages will decrease, and vice versa. Vp must be
appropriately within bound of the positive supply rail.
o Provide a detailed discussion illustrating that you understand
what is going on.
 We have a 5V input at VCC which is voltage divided to
2.5 V at the positive input terminal, Vp. This voltage,
then appears at Vm, due to the infinite open-Loop gain
of the nearly ideal Op-Amp. This voltage is then
multiplied by the closed-loop gain of the topology which
is designated by –Rf/Ri. In our case, this would be 1x.
For AC, Vp is at AC ground. Consequently, Vm is then at
AC ground while the input AC voltage is multiplied by the
same closed-loop gain. In conclusion, the output is
centered around 2.5V with an AC swing of +/- 100mV.
What is the maximum allowable input signal amplitude? Why?
o Since our rails are set at 5V, and 0V. We could then swing +/2.5 V. Anything larger, will result in clipping of the signal. The
maximum allowable input signal would then be 2.5V
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What is the maximum allowable input signal if the magnitude of the
gain is increased to 10? Why?
o By increasing the magnitude of the gain by 10, the largest
input signal would be 250mV theoretically.
What is the point of the 0.01 uF capacitors from VCC and VCM to
ground?
 These capacitors are just to decouple the voltages, and
detract noise from the power supply.
o Are these values critical or could 0.1 uF, 1,000 pF, 1 uF, etc.
capacitors be used?
 These values could be anything. They are used to hold
the voltage at a node to a specific value to detract from
sudden swings. They are just charge buckets.
The data sheet shows that this op-amp has an input bias current
that flows out of the op-amp's inputs of typically 20 nA.
o This current flows out of both the non-inverting and inverting
inputs through the resistors connected to these inputs.
o Show how the operation of the circuit can be effected if, for
example, R1 and R2, are much larger. Explain what is going on.
 If we had a 20nA input bias current on each of the opamp terminals, using a large resistor will produce a larger
voltage at Vm, and consequently, Vp. This mismatch of
voltages would then change the output voltage of our opamp.
 What is the input offset current? What does this term
describe?
 The input offset current is the difference
between the input bias currents at Vp and Vm. It
describes the mismatch between the two input
currents of the op-amp.
Explain how the following circuit can be used to measure the op-amp's
offset voltage.
o The following circuit has a specific VCM. This VCM should be
used as the reference for the output voltage. If this VCM is
off when measuring, one can conclude that the Voffset = Vout
– VCM.
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Note that if the output voltage is precisely the same as VCM then
the op-amp has no offset voltage (this is very possible).
o To measure small offset voltages increase the gain by
increasing RF to 100k or larger. Explain what is going on.
Measure the offset voltage of 4 different op-amps and compare
them.