# Opamp demonstrates slew rate much lower than specified in datasheet

TL,DR: Use a comparator, not an opamp. See Frosty's answer below.

I'm trying to make a pulse waveform generator with variable duty cycle. The 555 timer is used to generate a sawtooth waveform, which is then compared to a constant voltage from a voltage divider. The schematic works in principle, but for some reason the output of the opamp has a very low slew rate.

This is the schematic:

Vcc is 5 volts. R1, R2 and C1 are such that the width of one tooth of the sawtooth waveform is roughly 200 microseconds (I'm seeing that with the scope). However it takes roughly 10 microseconds for the output of the opamp to go from 0 volts to 4 volts.

The opamp in question is TL072, which, according to its datasheet has a typical slew rate of 13V/us. There's also a graph that is showing "Normalized slew rate", which is roughly 1V/us at +/-15V power supply.

Am I misunderstanding what slew rate really is? Shouldn't this amplifier go from 0 to 4 volts in about 0.3 microseconds?

UPDATE: The circuit behaves as described with no load.

• Slewrate depends on output capacitance. The slewrate given is with a 2kOhm resistor and 100pF capacitance. I don't know what your load is on the PWM out pin, but it could be the cause of the lower risetime. Commented Sep 27, 2017 at 16:36
• From DS: "(1) VCC+ and VCC– are not required to be of equal magnitude, provided that the total VCC (VCC+ – VCC–) is between 10 V and 30 V." If you only have a 5V rail, that's your problem Commented Sep 27, 2017 at 16:37
• Minimum supply voltage of TL072 is +/-5V (or 10V single supply). Also, it is not rail-to-rail so getting 4Vout with 5Vin is not realistic even if the chip could run at 5V. Commented Sep 27, 2017 at 16:38
• In addition, you are using the opamp where you should be using a comparitor. The opamp is driving into saturation on each transition. You may be seeing some delay as it recovers. Commented Sep 27, 2017 at 16:38
• @laptop2d from the pinout and circuit connection, definitely a 555. Commented Sep 27, 2017 at 16:58

There are several contributing factors to the poor performance of the circuit:

1. Insufficient supply voltage

Minimum Supply Voltage for the TL072 is listed as 10V. Using only 5V is well below this minimum.