DC Blocking cap value

[TheWinterSnow]

I feel embarrassed for asking this as it seems so elementary but I did everything I could to try to figure this out but keep drawing blanks.  So I have a circuit that has a blocking cap for biasing purposes but the resistance network makes it kind of hard to figure out what the cap value should be without being too big to cause blocking distortion.  Problem is the biasing resistor has to be calculated to be low enough as it seems it would be a high pass filter which means the biasing resistor has to be low enough to allow for low frequency response.

So here is the Thevenin equivalent of the circuit:



RS = Voltage Source Resistance
RB = Bias Resistance
RL = Load of the circuit

How would I go about determining C1 and RB to have a 20Hz-20KHz response with as little phase lag and blocking distortion in the low end.  R1 and R2 have to be the same value but can't be made any lower than 1K even though I prefer it to be 10K.  They can be somewhere in between if need be.

[daverdave]

Rb and C1 do not form a high pass filter. Rb would act as a decharge path for the capacitor, it also forms a potential divider with the source resistance so ideally it would be quite large. You're going to get a high pass effect from C1 and R2, with R1 and R2 forming another divider, dividing the input signal by a half. If you ignore R1, then you'll have a passive high-pass filter, if C1 was 470nF then your cutoff frequency would be 33.87Hz approximately, since fc = 1/(2*pi*R*C). Increasing R2 would DECREASE the cutoff frequency.

I don't know why you need R1 in there as I don't know the application. Your phase lag will decrease as the incoming signal frequency is increased above the cutoff frequency, I don't know what you mean to blocking distortion.

[merlinb]

So here is the Thevenin equivalent of the circuit:

You haven't really drawn the Thevenin equivalent- you need to do some further simplification; then it should become obvious to you.

RS and RB are effectively in parallel, so can be represented as a single resistor (the source resistance).
R2 and RL are in parallel so can be represented as a single resistor (then see next step)
R1 and R2||RL are in series, so can be represented by a single resistor (the load resistance).

You will then see a more familiar high-pass filter, and work out the cut off frequency for yourself.

[GibsonGM]

Hmm, it's ok to be new, and this is the way to get information to learn.  

So - what is this FOR?  What kind of signal are you trying to pass, from where to where ("voltage source" is what?) and do you want full low-end response?  What would happen if you had a cutoff at 100Hz or something?

What dave said - your rolloff frequency, or "half power point", is given by 1/2pi * R'     where R' = your DC resistance PLUS the output impedance of the device your filter follows.   <in many cases the impedance, Z, may be safely ignored>  

Now - where your C is in that pic, you will not likely have blocking distortion, since you have a resistance in series after the C!

What I'd do is model an R/C filter with LT Spice after doing 1/2pi R C...then add another section, model them together....see what the overall response is, and go with it.... Experiment....it's ok to lower the value of C if you are experiencing "fartiness"  ;o)  

BTW: 20Hz is too low for most applications, and 20KHz is above what we need. Unless you're designing radar, ha ha.   Try 80 Hz to 10kHz, for guitar, see what you get....unneeded freq's lead to instability and oscillation...

>> I see Merlin got here first...AWESOME job on the book, Merlin!  Extremely understandable yet 'human' enough to be interesting and captivating.   Thanks for doing it ) 

[R.G.]

As noted, the circuit you've shown is not reduced down to what is needed to properly estimate a rolloff. A Thevenin equivalent circuit consists of a voltage source and a series resistance, so there's some work to do. As MB said, the load the cap works into for rolloff purposes is RL paralleled by R2, and then that is in series with R1. RL is so much larger than R2 that the load works out to 19975 ohms instead of simply 20K. This is all Ohm's law stuff.

On the other side, you get to do a real Thevenin equivalent. A Thevenin equivalent circuit is calculated from the open circuit voltage and the short circuit current. If C1 is not there, just momentarily for the purposes of calculating this equivalent, then the open circuit voltage is whatever the voltage input is, divided by the resistor net of Rs and Rb. So the Thevenin equivalent voltage is Vsource* Rb/(Rs+Rb); the Thevenin equivalent resistance is Rs and Rb in parallel.

Setting C1 for a specific rolloff point isn't that hard, it just takes some juggling and some introspection. Let's look at juggling first. Thevenin's theorm says that we can replace the voltage source, Rs and Rb with a single voltage source of Vs*RB/(Rs+Rb); Ohm's law says that we can replace R1, R2, and RL with R1+R2||Rl. There's one more step.

Since what we are looking for in terms of frequency response is the frequency where C1's impedance causes the current in R1+R2||Rl to diminish to half the power of the midband value, we have to look at the current through C1. For that, we can do a final simplification. The Thevenin equivalent resistance of Rs||Rb is in series with C1, and that's in series with R1+R2||Rl; The current in a string of series devices does not change if you rearrange their order.  So we can rearrange Rs||Rb to be in series with R1+R2||Rl to get a series resistance of (Rs||Rb)+R1+(R2||Rl) on the load side, and C1 in series with that from the signal source. Now you can calculate C = 1/ (2*pi*R*F) for your low frequency rolloff.

But I'm left wondering about this:

How would I go about determining C1 and RB to have a 20Hz-20KHz response with as little phase lag and blocking distortion in the low end.  R1 and R2 have to be the same value but can't be made any lower than 1K even though I prefer it to be 10K.  They can be somewhere in between if need be.

For minimal phase lag on the bottom end, you need to make the low frequency rolloff as low as possible, ideally making C1 have a half-power point of no higher than 2 Hz. Unless the device being driven is specifically a vacuum tube with a grid that goes into conduction at Vgk>0, or a JFET where you get gate conduction at Vgs>(a diode drop) then blocking distortion really doesn't exist as such.  Blocking distortion, if I understand how you mean that, is caused by a signal big enough to drive the following load into one-sided conduction, which loads up charge on the coupling cap and shifts the bias point.

The simplest way to avoid blocking distortion is to (1) don't use a tube or a JFET, or (2) if you must use a tube or JFET, don't push so much signal into it that it's driven into control-node conduction, or (3) arrange that the control node conduction is forced to not be asymmetrical by providing a separate conduction path for the opposite polarity of signal. 

Blocking distortion, again if I understand how you're using the term, is a characteristic of the signal size and driven device much more than the size of the coupling cap. Reduction of the coupling caps to minimize blocking distortion is a stock practice for tube amps where they're being overdriven for some reason and the charge buildup on the grid can't be controlled some other way.

[TheWinterSnow]

In my schematic I was over thinking the role of R1 in the divider instead of having a simple 1 cap 1 resistor as a HPF.  It was after R.G.'s assessment and final Thevinin circuit in which I realized it doesn't matter where the resistance is, the total series resistance is all that matters.

Something really really simple that I was overthinking.  Figured it was better safe than sorry to ask.

Also I guess I was tired or just fried from a day of working but I realized that the circuit would not be producing distortion AND is using an opamp balanced to unbalanced converter...meaning I never needed to worry about blocking distortion, now I feel even more dumb.  And yes R.G. you were right on what I meant by blocking distortion.