Voltage from PWM circuit

[kshuler]

Just curious about a wiring diagram, and thought you might be able to help me out. I am trying to run 12 volt motor, via a PWM circuit, but wanted to get the power in terms of the initial voltage supplied. I have little volt meters but with a PWM circuit, no matter how short the pulses are, they always read 12v. I thought about trying the following circuit diagram to see if this would work:

Any thoughts?

 

[passedpawn]

The power in a PWM ckt is V * I * DC, where DC is the duty cycle from 0 to 1.

That diode and capacitor at the bottom of that ckt attempt to average that.

So, would it work? Yes. There probably needs to be a resistor in parallel with that cap C1, and the combination of them need to be sized according to the time constant of your PWM frequency.

 

[BrunDog]

If it's a brushed motor, as most DC are, and the PWM is fast enough, you don't really need an RC filter.

 

[kshuler]

The power in a PWM ckt is V * I * DC, where DC is the duty cycle from 0 to 1.

That diode and capacitor at the bottom of that ckt attempt to average that.

So, would it work? Yes. There probably needs to be a resistor in parallel with that cap C1, and the combination of them need to be sized according to the time constant of your PWM frequency.

Thank you. What is the point of the resistor, exactly? Does it go from the end of the diode before the capacity to the common lead (1st picture), or inline between the diode and the takeoff of the capacitor (like in 2nd picture)? Is it just because the resistance across the voltmeter is so high that the capacitor would just never discharge, and it would still just read a constant 12v?

If the resistor is in the 2nd configuration, do I even need D1... I would think the leakage current flowing backward while the PWM is in the off portion of the cycle would be minimal with enough resistance and would probably not have much effect on the motor, right?

Thanks!

or

 

[kshuler]

Another question... how would one go about sizing a capacitor and resistor like this. The PWM frequency is 21kHz. I want, for instance, 80% duty cycle to read 80% of 12v (or 9.6V). Is there a chart someone knows about to look this up?

 

[beermanpete]

Unless the PWM circuit is rather lossy you should be able to measure the power into the PWM and use it as a representation of the power consumtion of the motor. Another thought is to measure the current draw of the motor. This is easier than trying to calibrate the peak-hold circuit you have in your diagam.

 

[doug293cz]

Unless the PWM circuit is rather lossy you should be able to measure the power into the PWM and use it as a representation of the power consumtion of the motor. Another thought is to measure the current draw of the motor. This is easier than trying to calibrate the peak-hold circuit you have in your diagam.

A peak hold circuit (OP's original drawing) will never measure at anything other than 12V (or whatever the supply voltage is) anyway, no matter what the PWM is doing, unless the motor itself is an effective averaging filter.

Brew on

 

[augiedoggy]

Forgive me for the basic question but my understanding was pwm just varies the speed and timing of the pulses of voltage and not the actual voltage level itself? I though something like an ssvr would be needed to actually adjust the voltage level?

I use a cheap $2 bridge rectifier and a regular ac knob style wall dimmer to convert 120v ac to a Dc voltage I can control with pwm for my 110v DC mill motor. (the googled directions and you tube video said capacitors for smoothing the voltage would be nice but I also couldnt find anything specing size.)
it has worked well to control the motor speed for about 2 years now.

 

[passedpawn]

Another question... how would one go about sizing a capacitor and resistor like this. The PWM frequency is 21kHz. I want, for instance, 80% duty cycle to read 80% of 12v (or 9.6V). Is there a chart someone knows about to look this up?

GO with the second drawing. Use the calculator at the following site to set your cutoff frequency to your PWM frequency (21kHz). You should end up with about 10k and 0.001uF.

http://www.2pif.com/high-low-pass-filter.php

A 1st order ckt like this (one R and one C) will still have some ripple, but at 21kHz you shouldn't see any effect on the voltmeter. You might need to tweak one of the values a little to get exactly what you want. Maybe use a pot instead of a fixed resistor, then just trim it to get the exact number you want.

 

[passedpawn]

Forgive me for the basic question but my understanding was pwm just varies the speed and timing of the pulses of voltage and not the actual voltage level itself? I though something like an ssvr would be needed to actually adjust the voltage level?

I use a cheap $2 bridge rectifier and a regular ac knob style wall dimmer to convert 120v ac to a Dc voltage I can control with pwm for my 110v DC mill motor. (the googled directions and you tube video said capacitors for smoothing the voltage would be nice but I also couldnt find anything specing size.)
it has worked well to control the motor speed for about 2 years now.

Right, the voltage isn't changed. PWM is a train of pulses at a fixed frequency, but the pulse width (duty cycle) changes. The duty cycle is directly proportional, then, to the power delivered.

The output from that bridge is DC, but pretty terrible without the caps. It would be full-wave rectified and bounce from 0 to 120V 120 times a second. The motor is highly inductive and likely doing a pretty good job of averaging the current.

 

[ajdelange]

Not to put too fine a point on it but I'm sitting here bored so I will mention that voltage will bounce from 0 to 120*sqrt(2) = 170V 120 times a second. 120 is the rms voltage. Given the quality of some of the posts here I would guess everyone knows that and it was just an "oops" in the post.

 

[BrunDog]

I must be off my rocker. Where are you gents getting 120V? I see a 12V DC supply and motor.

 

[passedpawn]

Not to put too fine a point on it but I'm sitting here bored so I will mention that voltage will bounce from 0 to 120*sqrt(2) = 170V 120 times a second. 120 is the rms voltage. Given the quality of some of the posts here I would guess everyone knows that and it was just an "oops" in the post.

oops

I must be off my rocker. Where are you gents getting 120V? I see a 12V DC supply and motor.

See augie's hijaak

 

[doug293cz]

Forgive me for the basic question but my understanding was pwm just varies the speed and timing of the pulses of voltage and not the actual voltage level itself? I though something like an ssvr would be needed to actually adjust the voltage level?

...

An SSVR is just a PWM that works on half sine waves instead of DC. So, an SSRV doesn't change the peak voltage until you get to less than 50% duty cycle. With an SSVR the average output voltage is much less linear w.r.t. to control signal than a DC chopper PWM.

Brew on

 

[Maylar]

Got an old VOM? An analog voltmeter would show you the average voltage.

 

[kshuler]

Got an old VOM? An analog voltmeter would show you the average voltage.

Yes, I thought about an old analog voltmeter before. Unfortunately, the space on my panel I'm building is very small, and I can't find a very tiny panel mount analog voltmeter. It was an oversight on my part... I just assumed a digital voltmeter would average out the signal like an analog one would... I was surprised to see that this did not occur, and I am trying to fix things.

Thank you everyone for all the help. I think I have now what I need. Thanks for the link, passeddpawn, I'll try that and see how it works!

Klaus

 

[kshuler]

Thinking about the flow of electricity... if I use the second diagram, then don't I need to eliminate the diode? Otherwise there is only one path out for the electrons (through the voltmeter) which is going to be very high resistance. I would think then that the capacitor would never discharge, right?

Klaus

 

[passedpawn]

Thinking about the flow of electricity... if I use the second diagram, then don't I need to eliminate the diode? Otherwise there is only one path out for the electrons (through the voltmeter) which is going to be very high resistance. I would think then that the capacitor would never discharge, right?

Klaus

Yep I think you're right. It's probably still a peak hold ckt with the diode.

 

[Maylar]

Thinking about the flow of electricity... if I use the second diagram, then don't I need to eliminate the diode? Otherwise there is only one path out for the electrons (through the voltmeter) which is going to be very high resistance. I would think then that the capacitor would never discharge, right?

Klaus

Correct. You want the series resistor, it forms a lowpass filter with the cap. With the diode in place you'll also need a parallel resistor else the cap will charge to the peaks. Without the diode the series resistor would also discharge it.

 

[kshuler]

Great, thanks!

Klaus

 

[ajdelange]

Looking at the original circuit I see a train of square pulses of, presumably, 12 V amplitude so it seems the appropriate processing is a simple LPF to pass only the DC component of the pulse train. That can be measured with any meter. At a 21 kHz chop rate 10k and 10 uf would give plenty of high frequency rejection. A dc voltmeter will then read the average voltage. You might also want to put 100k or 1 meg across the cap to make the integrator 'leaky' so that you observe changes in setting faster.