How to choose audio coupling electrolytics?

[Steve Jones]

Now it's my turn to sound like a newbie, but hey, I'm a repairer, not a designer... :grin:

How do designers choose the value of interstage audio coupling caps? I have some gear that uses 4.7 microFarad caps, such as some ADR stuff, yet I notice with Joe Malone's circuits that he supplies for his new preamp boards some of them are in the thousands of microFarads. Is it a function of impedences of the stages? How is HF and LF rolloff effected by such divergent values?

 

[mcs]

It's a function of the impedances in the circuit. It's also important to size them correctly in multi-stage amps with loop feedback. But I guess some "designers" would just use caps that are "big enough"...

Best regards,

Mikkel C. SImonsen

 

[squib]

hey Steve...crikey i just sent you an email and then jumped into the Lab and here you are:
Coupling caps....the old formula F ( -3dB ) = 1/ 2*pi*R*C, or
C=1/ 2*pi*F*R is the starting point. Find out the loading resistance, choose the -3dB frequency and determine C. Lets say there is a 10K load and you want the -3dB point at 10Hz, C= 1.6uF. The problem with this simple approach is it pays no regard to the phase shift that will occur within the audio band above 10Hz, thats why designers such as Neve etc generally used values of at least 10uF for stages with light loading ( 10K up ) and 100uF elsewhere.
My general rule of thumb is 22uF within circuits where the loading is known and generally around 5K to 20K and 220uF on outputs where loading is from an external and unknown device and could be as low as 300R. This gives the factor of 10 margin, whereby all components that due to loading wether R or C can cause a change in amplitude are 10 times larger than the absolute minimum.

 

[Steve Jones]

Thanks guys, I will look at all that math tomorrow morning after I've had a big cup of TEA :razz: and not before...

 

[Flatpicker]

I just (re)found this thread and was checking the formula - plugging in 10Hz and 10K doesn't yield 1.6uF (.0000016F) and neither does plugging in 1.6uF and 10K yield 10Hz. Also, F= 1/ 2*pi*R*C is not the algebraic equivalent of C=1/ 2*pi*F*R.

squib or anyone have the answer?

 

[mcs]

[quote author="Flatpicker"]F= 1/ 2*pi*R*C[/quote]
Try f = 1 / (2*pi*R*C). Then you get 10Hz with R=10k and C=1µ6.

Best regards,

Mikkel C. Simonsen

 

[squib]

yeah sorry.... i left out the brackets

 

[BYacey]

One thing to keep in mind is as capacitance of an electrolytic goes up in value, so does it's internal resistance and inductance and leakage. This can have undesirable effects depending on how it is implemented in the circuit.

 

[AMZ-FX]

I was thumbing through some old issues of Elektor recently and came upon an article about capacitors... an interesting finding of the cap tests was that, if you parallel a big electro with a smaller value film cap, the film cap should be no less than 1/10 the value of the electrolytic... so if the electro is 10uF you should use a 1uF film.

regards, Jack

 

[clintrubber]

That ratio of 10 sounds familiar. Hmm, I thought the SSL uses 100nF // 100uF (ratio of 1000...) but it definitely does help as I understood :? :roll:
(my own SSL not ready yet)

 

[Flatpicker]

Yes, it works both ways with the brackets there. Thanks! :thumb:

 

[Flatpicker]

[quote author="AMZ-FX"]...if you parallel a big electro with a smaller value film cap, the film cap should be no less than 1/10 the value of the electrolytic...[/quote]Interesting. I remember someone asking once how to choose the film cap's value, but I didn't have a good answer other than it didn't have to be very large since it only had to pass the high audio frequencies.

Did they give an explanation for using 1/10th the value?