Forums

Filtering a PWM signal

Started by Bob Engelhardt April 2, 2012
I've been watching some of Dave Jones videos (EEVblog).  In #225 he 
discusses using a PWM signal to control the voltage of a power supply. 
It needs to be DC to do this, so he uses a filter. He uses a first-order 
RC filter, with the output going to the "+" input of an op amp.  With a 
10% duty cycle on the input he shows the output at 10% of the input 
peak, with some ripple.

What I don't get is why the filter capacitor doesn't (eventually) charge 
to the input peak voltage.  'Cause the input never goes negative, and 
the op amp input impedance is too high to have any affect (or is it?). 
Why doesn't it just keep charging?

If it matters, the input is 5v peak, R is 10k, & C is 100nf.

It's embarrassing that I don't understand something as simple as this, 
but I was a software engineer, so that might explain it <G>.

Bob



"Bob Engelhardt"
> > I've been watching some of Dave Jones videos (EEVblog). In #225 he > discusses using a PWM signal to control the voltage of a power supply. It > needs to be DC to do this, so he uses a filter. He uses a first-order RC > filter, with the output going to the "+" input of an op amp. With a 10% > duty cycle on the input he shows the output at 10% of the input peak, with > some ripple. > > What I don't get is why the filter capacitor doesn't (eventually) charge > to the input peak voltage. 'Cause the input never goes negative, and the > op amp input impedance is too high to have any affect (or is it?). Why > doesn't it just keep charging? > > If it matters, the input is 5v peak, R is 10k, & C is 100nf. >
** The cap charges via R during the "on" time and also discharges the same way during the "off " time. As the off time is 9 times longer than the on time - the average value is 10% of the peak and this is the voltage seen on the cap. ... Phil
Phil Allison wrote:

> ** The cap charges via R during the "on" time and also discharges the same > way during the "off " time. > ...
Ooooh, of course. Discharges when input is 0v ... it doesn't have to go negative! Oh, damn - I'm going to have to get an alias for posting here if I want to avoid the shame of such stupid questions. Sorry for the noise, Bob
Bob Engelhardt wrote:

> I've been watching some of Dave Jones videos (EEVblog). In #225 he > discusses using a PWM signal to control the voltage of a power supply. > It needs to be DC to do this, so he uses a filter. He uses a first-order > RC filter, with the output going to the "+" input of an op amp. With a > 10% duty cycle on the input he shows the output at 10% of the input > peak, with some ripple. > > What I don't get is why the filter capacitor doesn't (eventually) charge > to the input peak voltage. 'Cause the input never goes negative, and > the op amp input impedance is too high to have any affect (or is it?). > Why doesn't it just keep charging? > > If it matters, the input is 5v peak, R is 10k, & C is 100nf. > > It's embarrassing that I don't understand something as simple as this, > but I was a software engineer, so that might explain it <G>. > > Bob > > >
I didn't see the blog so I am only picturing here. But the input does not have to go negative (-), The source of where the PWM is coming from goes to common, that is shorting the cap.. If you were to put a diode in the path, then yes, it would just charge it up. Like I said, I didn't see the blog but, if he was using a function generator, that generator does pull to common in the off duty cycle which simple shorts that network to common, back drains the cap. So some charge is being removed from it in the 90% off cycle and because the on duty is endless current supply, it's going to dominate, which is why even at 10%, it will make it there. :) Jamie
On Apr 2, 11:17=A0pm, Bob Engelhardt <bobengelha...@comcast.net> wrote:
> Phil Allison wrote: > > ** The cap charges via R during the "on" time and also discharges the s=
ame
> > way during the "off " time. > > ... > > Ooooh, of course. =A0Discharges when input is 0v ... it doesn't have to g=
o
> negative! =A0Oh, damn - I'm going to have to get an alias for posting her=
e
> =A0 if I want to avoid the shame of such stupid questions. > > Sorry for the noise, > Bob
Not to worry Bob, if I was worried about looking stupid, I'd never post anything! I still don't really see why you need R5 to be 1 k ohm and not zero ohms in your previous thread. (This may be becasue I don't have a good model for the FET.) I've built some opamp to FET current drivers with R5 =3D 0... but I do have the gate resistor R4 and the 'roll-off' cap. George H.