# Help needed with DC blocking Caps

Discussion in 'General Electronics Chat' started by MrShhh, Jul 12, 2009.

1. ### MrShhh Thread Starter Member

Jul 7, 2009
24
0
Hi, could someone please point me in the right direction, I don't mind reading articles or using equations:

Suppose I feed the output of an amplifier (audio range, f = 20Hz to 20kHz) to a DC blocking cap, how can I calculate how large the blocking cap must be in order to not affect the bass response adversely?

I have read that a blocking cap should behave close to a short at working frequency i.e. Xc should be minimised where Xc = 1/(2*pi*f*c).

However, I have a valve pre-amp with 4.7uF blocking caps at the output, which at the low end would give Xc = 1/(2*pi*20*0.0000047 = 1693 ohms ! This is nowhere near a short.

It's also confusing as I commonly see very small blocking caps at the input of audio equipment (e.g. 220nF) but much larger caps at the output (10uF and above). Why does this not wreck the bass response at the input?

Many thanks

2. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
513
I am going to assume you know some basic circuit theory, without which you cannot understand the answer. In particular I am going to assume you understand how a potential or voltage divider works.

As far as the signal voltage is concerned, a blocking capacitor forms a potential divider, along with the input impedance of the following stage and the output impedance of the preceding stage.

Look at my sketch. Both the preceding and following circuits may be just part of a circuit a single transistor or IC; they may be a whole circuit such as a preamp or they may be a whole device such as a loudspeaker or input cartridge.

It does not matter the principle is the same.

We want the greatest part of the signal voltage to be developed across the input impedance of the following stage.

So we arrange for the output impedance of the preceding stage to be substantially less than the input impedance of the following one.

We also want the capacitor impedance to be substantially less than either, but at least less than Ri. If the impedance of C (which I see you know how to calculate) was say equal to Ri, half the available voltage would be lost across C.

So let us look at some figures

If the preceding stage is a record cartridge supplying a preamp, the input impedance of the preamp is usually around 47k. So we want the impedance of C to be small compared with 47k.

If the following stage is a loudspeaker, the impedance of a loudspeaker is about 8Ω we want the impedance of C to be small compared with 8Ω.

Normally we would look for small to be less than 1/10 of Ri.

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3. ### MrShhh Thread Starter Member

Jul 7, 2009
24
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Thanks studiot , that makes perfect sense.

So my valve preamps blocking cap Xc of 1693 ohms probably doesn't matter because the input impedance of the power amp following the preamp will usually be very high (10x higher ideally).

The same preamp has a 220nF cap at the input. At 20Hz this gives Xc = 1/(2*pi*20*2.2*10^-7) = 36172 ohms

Does that mean the impedance of the grid to the ECC82 input valve must be very high in order for the circuit to perform well? (I guess the 1M resistor must have something to do with it as well).

Schematic:

http://www.gyraf.dk/gy_pd/g9/g9_sch.gif

4. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
513
Yes the input impedance of valve circuitry is inherently very high, tens or even hundreds of MΩ.
The 1MΩ (R23) across the input is to define the input impedance to the second stage and prevent it being too high. If it is too high you loose the signal in other ways and increase the noise pickup.

I note that the preamp input is 47k, set by R7.

You also seem to have a transformer to convert balanced input to single ended.

5. ### millwood Guest

there are many reasons for you NOT to want the input impedance to be that high.

it is also there to bias the plate.

6. ### millwood Guest

more importantly, the output impedance of the device (a microphone?) that feeds into the preamp must be considerably lower than 16k - which usually is true.

7. ### MrShhh Thread Starter Member

Jul 7, 2009
24
0
Cheers folks That's a big help.

Related to the above, in some audio projects I've made, a larger blocking cap has given a better bass response at the expense of using more power and vice versa.

Is there a way to calculate the power a capacitor will consume (in the audio frequency range)? I know they are never 100% efficient and so must incur losses so does the ESR come into play here?

As a guess I'd say a cap with a lower ESR is more efficient and hence would use less power but that's probably oversimplifying things.