6
\$\begingroup\$

I have to choose an OP-AMP mainly based on slew rate. I am using the op-amp as a differential comparator with dynamic (time) hysteresis with a time-constant 3*tau=500µs.

My upper threshold of the hysteresis |Uth|=±5V while my bottom is the minimal possible

enter image description here

How do I calculate the necessary slew rate given this?

Please keep in mind I am yet to add a voltage stabilizer in Uout with a zener.

\$\endgroup\$
  • \$\begingroup\$ Remember that slewrate is more significant for large signals. The units are in Volts per second or similar and with small signals you can achieve very high frequency operation. Large squarewaveish signals are the most demanding if linear amplification is required. If you offer a bit more circuit and application detail you will get better answers. \$\endgroup\$ – KalleMP Dec 19 '17 at 22:16
  • \$\begingroup\$ Opamps have a SPEC of slewrate. Do you trust that value? \$\endgroup\$ – analogsystemsrf Dec 20 '17 at 3:46
  • \$\begingroup\$ Why do people insist on using op-amps as comparators? Why not just use a comparator? They are not compensated and so they slew much faster. \$\endgroup\$ – John D Dec 20 '17 at 18:47
1
\$\begingroup\$

There are too many problems with this design so the question becomes irrelevant until you become clear on the overlooked assumptions that nmatter.

Slew rate, dV/dt is limited by output current
- i.e. dV/dt=I/C for some given capacitance load.
- usually a comparator is used rather than an Op Amp. - the hysteresis and thresholds are directly affected by output swing and positive feedback ratio of that swing. - your design is referenced to ground so a bipolar supply is needed if using an Op Amp.
- the R3 value is assumed to be much larger than output impedance of driver

Thus if 3*tau=500µs then R3C1=500us/3=166us and R3 >> Rout and if Vout=5Vpp then the slew rate is always much faster and a transition time much less ( <10% ?) than your RC time constant. Slew rate and latency are thus affected by your system requirements as well.

But without that input, I would suggest a slew rate requirement of >5V/16us or >0.3 V/us which is easily doable by many IC's. ( keep in mind Analog IC's are designed for many purposes with bandwidths from 25kHz to 500 MHz)

\$\endgroup\$
0
\$\begingroup\$

To calculate the minimum required slew rate for the opamp, you have to start with a spec. You need to decide what exactly you want out for a certain specific input. This is not clear from your verbal description.

Since slew rate is something that happens over time, this spec must necessarily have a time component. One way to do this is to show the input signals and the desired output signal on a plot of voltage as a function of time. Ideally, the input is a step so that the rest is only about delay of the circuit itself.

After that, you can probably see how far the output needs to move within what time. The slew rate is the ratio of those. Note that this is one thing that is amplitude dependent. If you want a larger output, then you need a faster slew rate to transition between the two levels over the same time. Therefore, your example inputs and outputs should be for the largest output swing you want.

You mentioned time constant, but I'd look at this dV/dt across a capacitor, which causes a fixed current. During slew-limited transition, you probably don't have a exponential decay anyway.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.