Just one more Exam question

Expalin the term slew rate when applied to an operational amplifer.

A 1V r.m.s sinusoidal signal is applied to an non inverting amplifier having a voltage voltage gain of 5. the operational amplifer used a gain- bandwidth product of 1Mhz and a slew rate of 0.5V/us. at the frequency will slew rate distortion occur.

Please Help

 

What is the freq. of the sinusoidal wave that is applied? The question never states it, but asks if it will matter to the operation of the circuit?

??

 

Input freq 200kHz (Gain of 5, GBP of 1Mhz is max).

Find the slew rate of an "ideal" op amp (V/uS form) required to reproduce the signal faithfully.

Simulate or draw the output of the non-optimal opamp in your question to see where the slew rate of 0.5 V/uS is the limiting factor.

slew rate is the maximum rate the output signal can change. For lower frequency signals, or for lower gain, the slew rate doesn't produce as much distortion.

Is the answer supposed to be in graph form, simulation, or THD %?

 

Key to solving this type of exercises is to understand that the slew rate describes the how fast the OpAmp to increases or decreases its output.
If an input sine of 2Vpp, 1MHz needs to be amplified 10 times, the moment the output needs to change the faster, is when the signal's derivative (slope) is maximum (in magnitude). In a sine wave, this is when it crosses the x-axis and the rate of change equals the derivative. Therefore, in this case, the output wave will be \(10\cdot \sin (2\cdot \pi \cdot 1M \cdot t)\) and has a maximum rate of change of \(\frac{d}{dt}10\cdot \sin (2\cdot \pi \cdot 1M \cdot t) | _{\tiny{t=0}}=10 \cdot 2 \cdot \pi \cdot 1M=62.83 \cdot 10^6\).
As a result, in order for the signal to be indistorted, the OpAmp must have a slew rate of at least 62.83MV/s, or 62V/μs as it is usually written. For example, the 741 OpAmp has typically 0.5V/μs slew rate.

 

I think it's time we offered the answer here.

Vin=A*sin(ωt)
Vout=Av*[A*sin(ωt)]

d(Vout)/dt=Av*A*ω*cos(ωt)
The maximum slew rate each cycle occurs at the maximum, i.e., when cos(ωt)=1.
d(Vout)/dt(max)=Av*A*ω
ω=[d(Vout)/dt]/(Av*A)

GIVEN:
A=1V rms = 1.414V peak
Av=5 (this is the gain)
d(Vout)/dt=0.5V/μsec=5e5V/sec

ω=5e5/(5*1.414)
ω=70.7krad/sec
F=ω/2π
F=11.25kHz This is the frequency where slew rate limiting will occur.

Note that the GBW is sort of a red herring. It just lets you know that GBW will not be an issue, since the bandwidth at a gain of 5 will be 200kHz for very small amplitude signals.

 

Just for the sake of completion, Ron, maybe you should correct the
\(\frac{dV_{out}}{dt}=Av\cdot A \cdot \omega\) into \( \frac{dV_{out}}{dt}=Av\cdot A \cdot \omega \cdot \cos (\omega t)\)
and state that you will use the maximum of the two parts of the equation, that is:
\(max \left( \frac{dV_{out}}{dt} \right)=Av \cdot A \cdot \omega\).

By the way, what's your new picture this time supposed to talk about?

 

Just for the sake of completion, Ron, maybe you should correct the
\(\frac{dV_{out}}{dt}=Av\cdot A \cdot \omega\) into \( \frac{dV_{out}}{dt}=Av\cdot A \cdot \omega \cdot \cos (\omega t)\)
and state that you will use the maximum of the two parts of the equation, that is:
\(max \left( \frac{dV_{out}}{dt} \right)=Av \cdot A \cdot \omega\).

Thanks, George. I've been doing slew rate by rote for so long that I forgot to include that.

By the way, what's your new picture this time supposed to talk about?

No message. I ran across it when I was browsing through My Pictures, and I just thought it was funny. I'm not sure where I got it. I hope it's not copyrighted.

Pretty radical, though, don't you think?

 

If they start copyrighting errors, we 're in serious trouble.
...radical, eh?...heh...