Author Topic: About those heckin J201s  (Read 5733 times)

thermionix

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Re: About those heckin J201s
« Reply #20 on: February 15, 2018, 04:12:58 AM »
What exactly does the bypass cap do?

Boosts the gain of frequencies over xxx.

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CE-2 (in progress)

Me too.  Mine's actually done, just waiting on MN3007 and MN3101 to get here, pop 'em in the sockets, set the trimmer, and (hopefully) that's it.  I already have the knobs on, and the battery installed.

jellyjams

Re: About those heckin J201s
« Reply #21 on: February 15, 2018, 04:18:06 AM »
If the bypass cap is big then the preamp will have a higher gain than without it.

However, if the cap isn't that big  then the gain at low frequencies will be small low and the gain at high frequencies will be high.  If the frequencies are in the bass-range then it will act kind of like a bass cut.  If the frequencies are in the mid-range it acts like a presence boost, and if the frequencies are high it acts like a treble boost.   The fetzer page has some comments about this.

I understand that's what it does, but I don't understand how/why it does it.
"Tagawasak ng pekpek" (I'm smarter than you think)

Boxed builds: TS9, Rat, Blue Box, Pitch Pirate, Deep Blue Delay, Equinox II, CE-2 (in progress)

reddesert

Re: About those heckin J201s
« Reply #22 on: February 15, 2018, 04:32:41 AM »
It's often useful to think about the reactance of a capacitor. This is basically the equivalent resistance to signals of a given frequency: reactance = 1/(2 * pi * f * C), where f is the frequency, C is the capacitance, and the units of reactance are ohms. So the cap behaves like a frequency dependent resistor, low resistance at high frequency.

Putting the bypass cap in parallel with the source resistor gives high frequency signals an easier path to ground. Effectively, it lowers the source resistance at high frequencies. Since the gain is ~ Rdrain/Rsource, that makes the gain higher at higher frequencies.

The bypass cap can often be found in the tube amp gain stages that the JFET amps are modeled after.

Rob Strand

Re: About those heckin J201s
« Reply #23 on: February 15, 2018, 04:39:39 AM »
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I understand that's what it does, but I don't understand how/why it does it.

The simplest explanation for the unbypassed case is the gain is the ratio of the drain resistor to the source resistor,

Av   ~ - Rd   / Rs

However the JFET acts like there is a resistor in series with the source  with value rs  = 1 / gm.  Where gm is called the transconductance.    With this included you get a more accurate expression,

Av  =  - Rd / (Rs + rs)

When you bypass the source resistor this becomes

Av = - Rd / rs

Since the denominator is smaller it means the gain is higher.

When you add the bypass cap is creates a transition between the two gains.   The frequency where the transition occurs depends in the value of the cap and the values of Rs and rs.

It's possible to explain why the gains are what they are a different way.  The input voltage effectively appears across the total source resistance; (rs + Rs) for unbypassed and rs for bypassed.   That means a current  vin / (rs + Rs) flows in the drain resistor.   For a JFET the drain current and the source current are the same value so that means the  source current is also vin/(rs + Rs).   Now observe that the source current flows through the resistor Rd.   The output voltage = Rd * source current.   So that means the output voltage is Vout =  Rd * (Vin / (Rs + rs) ).   The gain is Vout/Vin   = Rd /(Rs + rs).   So that shows where the results above come from.

For more formal reasoning you need to find some PDF's on the gain of common-emitter amplifier with an without bypassed emitter resistors.  Most of this stuff is explained regarding transistors (BJT's)  not JFETs but the ideas are exactly the same.

Most formal stuff will have more maths.   In the above I've started with the least mathematical way to "explain" it.


« Last Edit: February 15, 2018, 03:57:15 PM by Rob Strand »
Send:     . .- .-. - .... / - --- / --. --- .-. -

Rob Strand

Re: About those heckin J201s
« Reply #24 on: February 15, 2018, 04:42:23 AM »
Quote
t's often useful to think about the reactance of a capacitor. This is basically the equivalent resistance to signals of a given frequency: reactance = 1/(2 * pi * f * C), where f is the frequency, C is the capacitance, and the units of reactance are ohms. So the cap behaves like a frequency dependent resistor, low resistance at high frequency.

The point where the gain starts to rise is
    f_low  = 1/(2* pi * Rs * C)
and the point where it starts to level off again is,
     f_hi    = 1/ (2* pi * Rt * C)
where Rt  = Rs in parallel with rs (=1/gm).
Send:     . .- .-. - .... / - --- / --. --- .-. -

fryingpan

Re: About those heckin J201s
« Reply #25 on: February 15, 2018, 06:50:41 AM »
If the bypass cap is big then the preamp will have a higher gain than without it.

However, if the cap isn't that big  then the gain at low frequencies will be small low and the gain at high frequencies will be high.  If the frequencies are in the bass-range then it will act kind of like a bass cut.  If the frequencies are in the mid-range it acts like a presence boost, and if the frequencies are high it acts like a treble boost.   The fetzer page has some comments about this.

I understand that's what it does, but I don't understand how/why it does it.
You know how emitter degeneration stabilises BJTs, but lowers voltage gain? The idea behind the bypass cap is that DC doesn't see it (so it's an emitter degeneration stage for biasing purposes) but AC, according to the size of the cap, might have a direct path to ground (so it's closer to a regular common emitter amp for AC signals, so higher gain). If the cap isn't large enough though, you'll have varying gain within the audio range. My undergrad EE studies basically skip JFETs as they are niche devices (maybe obsolescent?) but I guess the underlying idea is the same.

thermionix

  • Guest
Re: About those heckin J201s
« Reply #26 on: February 15, 2018, 03:32:56 PM »
An unbypassed (cathode/emitter/source) resistor is a form of NFB.