Calculating Bypass Cap Gain - How?!

[Paul Marossy]

I need to calculate the effect of adding a cap in parallel with the source resistor on an FET. Just how do you do that?! I can't seem to round up any info on how to do that on an FET circuit...

[R.G.]

The fact that it's a FET makes no difference. What matters is when the capacitor's impedance begins being lower than the source resistor value.

That frequency is when Rs = Xc = 1/(2*pi*F*C).

Rearranging, F = 1/(2*pi*Rs*C)

or the cap for a specific rolloff frequency is
C = 1/(2*pi*Rs*F)

FET-ness has nothing to do with it.

[Paul Marossy]

So does that mean that only frequencies at a certain threshold get a boost in gain? I was always thought adding a bypass just generally added a certain amount of dB/gain to the circuit. I never really thought about it in terms of specific frequencies being boosted, but that makes sense.

Thanks RG.  

[R.G.]

The gain of any active device - tube, transistor, FET, is approximately the plate/collector/drain resistor divided by the cathode/emitter/source resistance.  The reason this is true is that the cathode/emitter/source resistor is a series feedback element, and the feedback converges the gain to that as closely as the gain of the active device will let it.

That feedback gain is less than the gain of the raw active device. If we shunt the cathode/emitter/source resistor with a cap, the impedance of the resistor and cap in parallel become the gain determining element.

At low frequencies, the resistor dominates, and the gain is just the ratio of the two resistors. As the cap's impedance gets lower with frequency, the cap eventually takes over and the gain goes up until it gets to the gain of the raw active device when the cap impedance is low enough that the plate/collector/drain resistor over the cap impedance is comparable to the gain of the active device itself.

So - a source/cathode/emitter cap keeps the gain low til some frequency where it increases at +6db per octave until it approaches the gain of the device with no cathode/source/emitter impedance, and it then flattens off at that higher gain.

[Paul Marossy]

The gain of any active device - tube, transistor, FET, is approximately the plate/collector/drain resistor divided by the cathode/emitter/source resistance.

This is my understanding. This whole thing came up when I looked at a FET guitar preamp that Donald Tillman designed. He claims that the preamp only has a gain of ~3dB, but when I calc. it out, it's more like 9.5dB. So, I'm thinking that possibly Tillman's claim to only 3dB of boost is incorrect...

In two of my own guitars, I have one with the Tillman FET preamp circuit with a 4.7uF cap in parallel with the source resistor and another with a 10uF cap in parallel with the source resistor. The 10uF cap provides a helluva boost! So I started to wonder just what kind of a dB boost that is. That's what prompted the question in the first place.

Thanks for your explanation RG, that really helps.  

[Ucho]

Bringing up again this discossion.

So, if I understand right, to have a similar frequency behaviour, halfing the resistor at source, I should double the parallel cap?

Example: I'm "playing" with my new Peppermill (http://www.diystompboxes.com/smfforum/index.php?topic=49273.0).
If I want to have a roughly similar frequency given by the 1k mosfet source res with added 4.7uF cap parallel to that res, using a 470R i could use a 10uF cap in place of 4.7uF?

[George Giblet]

I have to disagree with RG here the JFET does affect the frequency response.  Also, there's more to it.

The frequency response with a bypassed cap has *two* frequency points.

At low frequencies the gain is if the cap was not there.  At high frequencies the cap shorts the source resistor Rs.  In other words the frequency response isn't like a high pass filter where the gain keeps dropping off at low frequencies.  What happends is it shelves off, like a shelving bass-cut equalizer.  A shelf response has two frequency points.  The first, at lower frequencies is where the gain starts to rise about the low frequency gain.  As the frequency the gain rises further until it gets to the second higher frequency point where the gain flattens off again.

The first (lower) frequency point is when the cap impedance is the same as the source resistance Rs, ie.

     f_low   = 1/(2*pi*Rs*C);   that's the equation RG gave.

The second (upper) frequency point is when the cap impedance is same as the source resistance, Rs,  in parallel with the JFET's internal source impedance (which is equal to 1/gm, where gm is the transconductance of the JFET) ie.

   f_hi       = 1/(2*pi*Rx*C)

where Rx  = 1/(gm + 1/Rs) = Rs / (1+Rs*gm)

The second frequency is often more important because it's sort of like the -3dB cut-off of the high-pass filter (provided f_low and F_hi aren't too close together).  Note also f_hi depends on the JFET.

As for the gain,

Without bypass cap:  gain = -Rs' / (1/gm + Rd)
With bypass cap:      gain = -gm*Rs'
In this equation Rs' is Rs in parallel with any load resistance.

Assuming the 2N5457 version of the tillman preamp the gm for the JFET is about gm=1.3e-3, Rs' = 6.8k in parallel with 51k = 6k, Rd = 2.2k

Unbypassed gain = 2   or about 6dB
Bypassed gain     = 7.8 or about 18dB.

If you have a 2.2uF bypass cap,

f_low = 33Hz
f_hi   = 127Hz  (here Rx = 570ohm using Rd = 2.2k and gm=1.3e-3)

Notice the ratio of Bypassed gain / Unbypassed gain = 7.8/2 = 3.9, and the ratio of f_hi/f_low = 127/33 = 3.8 which is about the same.  This is always the case with these circuits.

The J201 version of tillman will have a higher Bypassed gain (say 20dB) , this will increase f_hi.

The equations work for MOSFET and transistor circuits all you need to do is use the correct gm value.

[Ucho]

Thanks George. Your post have been really very informative. Now I'm going experimentig...