I have LM358p op-amps Datasheet

I'm trying to base my design off this: enter image description here

retrieved from here. My control voltage will vary between 0 and 3.5v. I want the output pulses to vary between 2.5 and 0.5ms and at a frequency of about 50Hz. This is to control a servo using Vcontrol.

My clock signal is from a 555 timer in astable mode. The square wave voltage is 0 to 8.4v.

I understand that this diagram is based on the ideal op-amp, so I'll need a resistor on the non-inverting input equal to the inverting input resistor. I'll also need a high value resistor parallel to the capacitor. The output will probably not reach 0v, but that's ok as long as the triangle is at least 2.5ms wide at its widest point.

I only have a single rail power supply, 0 to +12v to power the op-amp. The clock signal also never goes negative. I don't quite understand how the op-amp integrates. Is the non-inverting input the reference line, above which the integral begins to rise and below the integral falls?

enter image description here

What should my input pulses look like and how do I calculate the R and C values for the op-amp?

  • 1
    \$\begingroup\$ You don't need a resistor on the non-inverting input. What you do need is a resistor in parallel with the capacitor, or else your op amp will just saturate and not do what you want it to. The non-inverting input does indeed act as the zero point for the integration. \$\endgroup\$
    – Hearth
    Commented Nov 11, 2018 at 20:50
  • \$\begingroup\$ I've been looking through The Art of Electronics and it looks like that input resistor is a necessary component. The output is amplified by -1/RC, so if R were close to 0, the output would just be a square wave \$\endgroup\$
    – griffin175
    Commented Nov 11, 2018 at 21:48
  • \$\begingroup\$ Are you mixing up op amps and transistor amplifiers? \$\endgroup\$
    – Hearth
    Commented Nov 11, 2018 at 22:08
  • \$\begingroup\$ @griffin175 Given your writing, perhaps this is off the table so to speak... but have you eliminated any thinking along the lines of simply using the 555 to generate a range of frequencies based on adjusting a lower impedance driver on pin 5's voltage value? Just curious, mostly, if you've already excluded such thoughts. Not complaining at all. \$\endgroup\$
    – jonk
    Commented Nov 11, 2018 at 23:46
  • \$\begingroup\$ @jonk The pulse width needs to be voltage controlled. I don't have any more practical way to create some kind of voltage controlled impedance. \$\endgroup\$
    – griffin175
    Commented Nov 11, 2018 at 23:53

3 Answers 3


You don`t need the opamps, you only need the 555. This is from the amazing book The Art of Electronics

enter image description here

If you are bent on using opamps then you need to BIAS your inputs when using single rail supplies. Otherwise they won`t oscilate at all. You can bias them with a resistive divider.


I don't quite understand how the op-amp integrates.

The left op-amp is used as an integrator, and is probably best left for you to research the theory behind how it works. But for simplicity sake, you can consider it as the active equivalent of a series RC circuit. For more information on how the integrator works, simply search "Op-Amp Integrator."

how do I calculate the R and C values for the op-amp?

You need to choose an RC value such that you obtain your desired triangle wave at 50 Hz.

As others have suggested, you will need a feedback resistor parallel with your capacitor, and that RC combination will determine your "cutoff" frequency, where any frequencies above this "cutoff" will be an integral output of the input.

Below is your overall transfer function of the integrator alone:

$$ \frac{V_o}{V_{in}}=-\frac{R_F}{R_i}\times\frac{1}{1+j\omega(R_FC)} $$

Take the magnitude of the above at 50Hz (or 314.16 rad/s), and that will be your gain at that frequency.

A resistor on the non-inverting input isn't entirely necessary, but is typically used to balance the bias currents and is set equal to the equivalent resistance seen by the inverting input. It doesn't not affect your gain or cutoff frequency.

What should my input pulses look like

If you're talking about the clock, it should be a 50Hz with a 50% duty cycle, since the frequency of your pulses is 50Hz. If you're talking about the voltage control input, then they shouldn't be pulses. The voltage control determines the duty cycle of your output PWM.

For example, suppose your triangle wave oscillates between 0V and 8.4V. For a 50% duty cycle (10ms pulse width in your case), set the voltage control to half of the max voltage of your triangle wave. (4.2V).

If you want 2.5ms, then 2.5/20ms = 12.5% duty cycle. Set your voltage control to 12.5% of your max voltage (1.05V).


I'm trying to base my design off this:

Sorry, but that circuit will never work as shown. It is not intended to be a "real" circuit - it's what is called a "notional" circuit.

Where to start?

The waveforms as shown are simply impossible. The integrator input is a positive signal, yet so is the output. Nope. Never happen without some extra biasing, since the integrator is an inverting integrator referenced to ground. Next, it requires a perfect 50% duty cycle on the input clock, or or the output will drift at a rate determined by the input phase mismatch and the input voltage swing. The drift will only stop (be limited by) the excursion limits of the integrator op amp output stage. If the input clock duty cycle is very close to 50% (and a 555 output should be close), you'll get one tip or the other of the triangle blunted where it hits the stop. This may or may not be an issue, depending on accuracy requirements and how stable the offset circuitry is.

The fact that you're limited to a single power supply will only make things worse.

I suppose you could do something like this:


simulate this circuit – Schematic created using CircuitLab

Here, R1/C1 are chosen to get a decent voltage swing centered around the voltage at C3. This voltage is set by adjusting R2/R3, and should be in the center of the voltage swing of the input clock. I've assumed that this would 6 volts.

OA2 serves as a low-pass averager for the integrator output, and has R4/C2 much larger than R1/C1. The output is fed back to OA1 to keep the average voltage at the C3 voltage.

Note that loop stability is not guaranteed.


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