Did I get this right?? Calculating the frequency response for an input capacitor

[Derringer]

This is basically the input stage of the Hot Silicon.
I'm trying to fully comprehend how an input cap can be chosen.



f = 1/(2*Pi*R*C)  where R is the input impedance of the following stage

f is the lowest frequency that will pass un-attenuated. (or will the frequencies below f simply be blocked?)

The input impedance of the following stage is the parallel resistance of the 22K and 150K resistors.

R = 1/(1/22K+1/150K) = 19186 ohms

If C = 0.1 uF then f = 83 Hz

Since a low E string is at about 83 Hz then a 0.22 uF cap would probably be better. It would allow lower (unneeded frequencies) to pass but would no doubt allow a low E as well as a low D.

If C = 0.22 uF then f = 38 Hz


Would the input impedance be more accurate if the 1M resistor is also taken into account as being in parallel with the 22K resistor? I know that it wouldn't change much but ...

If  R = 1/(1/1M + 1/22K + 1/150K) = 18825 ohms

If C = 0.1uF then f = 85 Hz

If C = 0.22 uF then f = 38 Hz


So am I getting this right?


Thanks,

Bill

[ayayay!]

My theory is probably way off here, so I'm actually opening this up for discussion.  Here's goes:

The 150K is setting the voltage of Q1's Base, so it has little to nothing to do with tone shaping.  (Am I wrong here guys?  Or half right?)

Cx and 22k set your frequency cutoff filter.  For this, we use our super duper expensive R-C filter calulator:  http://www.muzique.com/schem/filter.htm

[R.G.]

@ Ayayay!: My personal opinion is that using cased-up calculators will delay, perhaps forever, your understanding what's really going on. Learn it first, and if you still need calculators, fine. Otherwise you're resigning yourself to not understanding, perhaps forever.

As to the 150K, it does primarily help set bias, but everything that connects to the cap is involved, at least a tiny bit, in setting frequency response.

@Derringer: You're almost right. To do it accurately, you need to take into account the loading of everything that connects to the capacitor's output side.

You are correct in replacing the power supply with an AC short circuit to ground and calculating the parallel combination of the 22K and the 150K. However, there is another thing to take into account - the transistor base. The "resistance" of the transistor base enters into it too.

The 1M does not, because the 1M loads the source before the capacitor can start causing frequency-related loss. It has to be taken into account in the source loading, but does not affect frequency response with Cx.

Here's how to estimate the base loading. All bipolars have an internal hidden resistor that's part of the device physics. Worse, it's variable with emmiter current, being about 26mV/Ie in ma. In this case the current is about 6ma, so that becomes 26/6 = 4.3333 ohms. Ugh. But there is a 1K unbypassed emitter resistor. The ACTIVE resistance of the base is approximately the current gain times the sum of internal base resistance and external, so it's hfe times 1004.33. for reasonable gains, say 100 or so, this gets to being 100K to 500K. We do not know until we get the specific transistor to measure what hfe is at the operating point, so we have to estimate. Using the lowest value gives a conservative answer of 100K. Reality will be better than that. This 100K appears in parallel with the 22K and the 150K real resistors.

So the resistance to use is 1/ (1/22K + 1/150K +1/100K) = 16.097K.

Now to picking Cx.

The rolloff point is arbitrarily defined as the frequency where half of the voltage is dropped across the capacitor. The voltage is -6dbv at this point, and if we simply pick 82Hz and do the math, we will have a low E string that is half-power. Especially since some people like to do drop-D tuning, we probably ought to make Cx bigger than the 82Hz number would say. You're very conservative making it 1/10 the frequency, or 8.2Hz. But that may make Cx too big. Let's pick something in the middle - 20Hz.

Now we can finally do Cx = 1/(2*pi*R*F) =1/(2*3.14*16097*20) = 0.494uF. I'd use 0.47.

Notice: the 22K dominates. The 150K and equivalent 100K B-E resistances lower it a bit from 22K to 16K. But picking how conservative you want to be has a big effect.

[Derringer]

ayayay!: thanks for jumping in!

R.G.:

Awesome!

Thanks so much for taking the time!

Bill