I thought I’d explore the phenomena around the audio voltage charging of a capacitor used to provide a DC control voltage (CV) used in many designs of tube compressors. I shall be using the formula used by Morgan Jones in his book Valve Amplifiers, chapter on Rectifiers Page >300. He uses it for mains rectification but it also applies to audio voltages too.
The subject is important because it shows how it’s possible to reduce the effects of the feedback oscillation that sometimes happens in these designs. So, why does feedback even occur in a compressor when it’s supposed to be just a DC signal? The answer lies in the ripple.
Morgan Jones’s formula for ripple is :-
Vp-p = t*I/C
Where t is in seconds, I is in Amps and C is in Farads. V is peak to peak ripple voltage.
For 50Hz mains rectification, t is 0.01seconds, which is 1/100Hz as the negative cycle is inverted, so two peaks instead of one.
To cut down on the variables I have chosen 10VDC as an arbitrary CV. I have also assumed a 1M grid resistor so we can arrive at a current of 10/1M which equals 10uA. I am not going to argue over exact figures, you can work those out for yourselves.
If you look through a lot of schematics, you will notice that nearly all the designs use CV caps between 100nF and 1uF, but I will just use those two values to illustrate the effects.
Let’s see what we get with 1kHz, that will have two peaks when rectified,
So t = 1/2000 = 0.0005, putting that into our formula we get:
0.0005*0.00001/0.000001 = 5mVp-p ripple with a 1uF cap and 50mV with a 100nF cap. If the phase is wrong then these ripple signals could be positive feedback.
If we look at the bass frequencies it gets worse, 40Hz has two peaks so
t = 0.0125 seconds and in our formula that gives 125mVp-p ripple with 1uF and 1.25Vp-p with 100nF.
Now I know these figures are wrong and we don’t compress sine waves, we compress music, but they serve to illustrate why we can sometimes have problems. If you have a regular beat of say 120bpm, then that translates to a very low frequency of 2Hz and t = 0.5 seconds. Maybe that's enough to trigger motor-boating. Check the number of caps in the signal path. Check their values, you can see what happens with 40Hz, maybe you don’t want 40Hz reaching your CV rectifier? I have found that going to a 1uF cap reduces audio ripple on the CV so that it's no longer a problem in some designs.
Best
DaveP
The subject is important because it shows how it’s possible to reduce the effects of the feedback oscillation that sometimes happens in these designs. So, why does feedback even occur in a compressor when it’s supposed to be just a DC signal? The answer lies in the ripple.
Morgan Jones’s formula for ripple is :-
Vp-p = t*I/C
Where t is in seconds, I is in Amps and C is in Farads. V is peak to peak ripple voltage.
For 50Hz mains rectification, t is 0.01seconds, which is 1/100Hz as the negative cycle is inverted, so two peaks instead of one.
To cut down on the variables I have chosen 10VDC as an arbitrary CV. I have also assumed a 1M grid resistor so we can arrive at a current of 10/1M which equals 10uA. I am not going to argue over exact figures, you can work those out for yourselves.
If you look through a lot of schematics, you will notice that nearly all the designs use CV caps between 100nF and 1uF, but I will just use those two values to illustrate the effects.
Let’s see what we get with 1kHz, that will have two peaks when rectified,
So t = 1/2000 = 0.0005, putting that into our formula we get:
0.0005*0.00001/0.000001 = 5mVp-p ripple with a 1uF cap and 50mV with a 100nF cap. If the phase is wrong then these ripple signals could be positive feedback.
If we look at the bass frequencies it gets worse, 40Hz has two peaks so
t = 0.0125 seconds and in our formula that gives 125mVp-p ripple with 1uF and 1.25Vp-p with 100nF.
Now I know these figures are wrong and we don’t compress sine waves, we compress music, but they serve to illustrate why we can sometimes have problems. If you have a regular beat of say 120bpm, then that translates to a very low frequency of 2Hz and t = 0.5 seconds. Maybe that's enough to trigger motor-boating. Check the number of caps in the signal path. Check their values, you can see what happens with 40Hz, maybe you don’t want 40Hz reaching your CV rectifier? I have found that going to a 1uF cap reduces audio ripple on the CV so that it's no longer a problem in some designs.
Best
DaveP