# Op Amp IC slew rate discrepancy

I have been trying to understand slew rate and so I set up a circuit with an op amp (LM741) that would allow me to measure it. The claimed slew rate of the package is about 0.50V/us, but my slew rate turned out to be about 0.3V/us.

For such a small number, this discrepancy seems a little too great. I am positive that I set up the circuit right because my circuit's DC gain is very close to what I expected. So how come the slew rate is so much smaller? Are there any specific external factors that might influence the op amp's measured slew rate (like the particular oscilloscope or the power supply I am using)?

• If they don't give a minimum of a maximum on the datasheet, then the specification isn't even tested in production. Expect it to vary wildly. The variation you describe doesn't surprise me at all. – ChateauDu Sep 25 '17 at 15:58

The slew rate is given as 'typical' 0.5V/us under the following conditions:

• Ta = 25 degrees C
• Power supplies +/-15VDC
• Gain = 1

Minimum is not specified except for the 741A, and it is 0.3V/us.

If any of your conditions are other than as above you can expect some variation, and in any case the minimum is not guaranteed for the general purpose version (though I would expect it to be within 20-30% of that in most cases under the specified test conditions).

One of the many "features" of the 741 is that its current reference is defined by the supply rails and a resistor.

The tolerance of resistor fabrication on an IC isn't great. However if you test your opamp at supply voltages other than what was used in the datasheet for the quoted slew-rate, your results will differ.

You can see the primary reference current is,

$$I_{ref} = \dfrac{ V_{s+} - V_{s-} - 2V_{BE} }{39 \text{k}}$$

The opamp's slew-rate is the result of having finite current to charge/discharge the frequency compensation cap (30 pF). Q13 forms a current mirror Q12. Q13 is the only circuit element that can charge the 30 pF Cap.

Recall the differential equation of a capacitor as,

$$\dfrac{dV}{dt} = \dfrac{I(t)}{C}$$

So, larger charge current yields a larger dV/dt (slew-rate). Larger capacitance, yields a slower slew-rate.

So the amount of current available to charge the 30 pF cap is $M * I_{ref}$. As said above which is derived based on a low tolerance resistor and the potential of the supply rails.