An operational amplifier has a slew rate of 2 V / µs. If the peak output is 12 V, what is the power bandwidth?
Solution:
The slew rate of an operational amplifier is
As for output free of distribution, the slews determines the maximum frequency of operation fmax for a desired output swing.
so 
So bandwidth = 26.5 kHz.
For the given circuit in fig. 1. Iin(off) = 20 nA. If Vin(off) = 0, what is the differential input voltage?. If A = 105, what does the output offset voltage equal?

Fig. 1
Solutin:
Iin(off) = 20 nA
Vin(off) = 0
(i) The differential input voltage = Iin(off) x 1k = 20 nA x 1 k = 20µ V
(ii) If A = 105 then the output offset voltage Vin(off) = 20 µ V x 105 = 2 volt
Output offset voltage = 2 volts.
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