0
\$\begingroup\$

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?

\$\endgroup\$
  • 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 Nov 11 '18 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 Nov 11 '18 at 21:48
  • \$\begingroup\$ Are you mixing up op amps and transistor amplifiers? \$\endgroup\$ – Hearth Nov 11 '18 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 Nov 11 '18 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 Nov 11 '18 at 23:53
0
\$\begingroup\$

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).

\$\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.