Formula for calculating value of capacitor

[klitgt]

Hi

Do any of you know a formula for calculating the minimum value of a coupling capacitor at the output of a preamp if e.g. I want the low frequency response to go as low as 10Hz, the output impedance of the preamp is X ohms and the input impedance of the poweramp is Y ohms.

As far as I know the minimum value of a cap in this situation depends on the lowest frequency at which there is a 3dB drop, the output impedance of the "source" and the input impedance of the "receiver", in this case "preamp" and "poweramp".

Thank's to those of you who can help

[DonoMan]

Just use an RC network where the formula is f = 1 / (2(pi)RC)

The thing to remember is that the impedance of your filter is equal to the impedance of your input at resonant frequency.

That is to say that the formula above comes from this
General reactance formula for a capacitor (which is close enought o the impedance of a capacitor as well) - X = 1 / (2(pi)fc)
Impedance of a resistor = R (excluding negligible inductance and capacitance in wires and such)


X = R = 1 / (2(pi)fc)
solve from there.


You can use that concept in other ways as well. Do with it as you will.

You may be able to pick a cap where X = the impedance of the amps at 10Hz...

[klitgt]

Originally posted by DonoMan
Just use an RC network where the formula is f = 1 / (2(pi)RC)

The thing to remember is that the impedance of your filter is equal to the impedance of your input at resonant frequency.

That is to say that the formula above comes from this
General reactance formula for a capacitor (which is close enought o the impedance of a capacitor as well) - X = 1 / (2(pi)fc)
Impedance of a resistor = R (excluding negligible inductance and capacitance in wires and such)


X = R = 1 / (2(pi)fc)
solve from there.


You can use that concept in other ways as well. Do with it as you will.

You may be able to pick a cap where X = the impedance of the amps at 10Hz...

Thank's but I am not working with a filter. The situation is that I have a preamp with an output impedance of 600 ohms and a poweramp with an input impedance of 10Kohms.
The preamp has DC offset that must be blocked by a "coupling" cap. What value should it have to be sure to keep a frequency response flat down to 20Hz.
I have tried your calculator asuming that the output impedance of the pre is 10K ("equal to the impedance of your input at resonant frequency") and I find something like 8uF. Wrong or right?
I am not knowledgeable in electronics so I may misunderstand much of what you tell me

[DonoMan]

Erm, yes, I messed up my wording there, I meant load, not input... Sorry 'bout that. My actual examples were right, though (as far as I can tell, anyway). Hopefully the output impedance of the preamp is low enough to disregard.

[klitgt]

Now it's me saying Erm...
I mean 0,8uF, not 8uF according to the calculation.
It seems right that a coupling cap of approx. 1uF is enough to make sure that the roll off downwards starts at approx. 20Hz

[klitgt]

And here is what I mean, please look at attachment

[audiopro]

Assuming the impedance is fairly constant (and correct), you will be 3 db down at 20hz with the 0.8uf cap. Only your ears can tell if that is sufficient.

[sam9]

Try this:

X(in ohms) = 1/(2*pi*f*C)

For 20Hz set f=20 then insert the possible values of C from 1uF to 20uF. This is one leg of a voltage divider. The impedance of the load is the other leg. Use the voltage devider formular to figure the attenuation for each value of C.

For nearly all audio purposes except energy storage in power supplies or delay/timer circuits, capacitors can be demystified by thinking of them a a frequency dependent resistor. Expanding on this, RC networks or filters generally, can be looked at a frequency defendent voltage dividers.


One wrinkle to all this: Onve you have figured out the capacitor valuer you need fort flat response at 20Hz, you may want to double that. The objective is not further bass extension but distortion/noise. I don't know if there is an exact rule for this but in many uses an in-series DC blocking cap contributes to the noise floor even at higher frequencies where it has no frequency response impact. The effect is usually gone by the time upper bass or lower midrange is reached, but it is certainly there in the mid-bass. I stumbled on thismyself a couple of years ago and wondered if I was mis interpreting things. But subsequently, I noticed when opening up a subwoofer plate amp that unusually large value caps were used at the input so I conclude that the people who design these recognize the phenomona; also I subsequently re-read D.Self's amp book and see that he mention's it as well.

The fact that you are dealing with a preamp doesn't change this I think because it just a question of which box the DC blocking cap is in. Circuitwize it's the same whether it is preamp output or amp input. This brings up one more thing to think of - your output cap will be in-series with an amplifiers input cap (unless it is direct coupled). Two caps in series have a combined vale of (c1*c2)/(c1+c2). You might want to take that into consideration as well.

[jackinnj]

Originally posted by klitgt



Thank's but I am not working with a filter. The situation is that I have a preamp with an output impedance of 600 ohms and a poweramp with an input impedance of 10Kohms.
The preamp has DC offset that must be blocked by a "coupling" cap. What value should it have to be sure to keep a frequency response flat down to 20Hz.
I have tried your calculator asuming that the output impedance of the pre is 10K ("equal to the impedance of your input at resonant frequency") and I find something like 8uF. Wrong or right?
I am not knowledgeable in electronics so I may misunderstand much of what you tell me

Assuming that you can get inside the preamp -- if you have a DC offset problem on a preamp you should deal with this first -- DC offset doesn't materialize for mysterious reasons.

DC offset is problematic in that any ripple from the power supply is probably making its way into the input of the preamp -- and when you push hard on the preamp the ripple is going to show up as spurs about the fundamental frequency. You can see this if you badly overdrive an amplifier.

There may be a faulty capacitor in the preamp, some crud and grease on one of the traces, a DC imbalance on an operational amplifier from input bias current, or an opamp which has gone out of specification. If the opamps are in sockets in the preamp one of them may have become a little unseated.

At any rate, 4.7uF will work fine:

[AndrewT]

Hi,
a few pointers in selecting your capacitor;
1. The flat response to 20Hz will require an Fc about 3 or 4 octaves below this. Butterworth single pole has a fairly slow roll off that extends many octaves above Fc. Phase angle extends even further but may not be audible.
2. Biamping from the pre will reduce the effective load impedance and in turn raise Fc if you ever go that route.
3. The DC blocking cap in the power amp will be in series with the preamp cap and you have to calculate the effective capacitance for the pair, then recheck Fc.
4. Pay for the biggest film cap you can afford/worth the improvement, 8uF to 10 uF if possible. Some listeners suggest a high voltage rating improves the sound but dielectric is more important imo. teflon, polypropylene, polystyrene.
Hope this helps.