Princeton Clone

DavebF

Member
2018-08-11 11:40 pm
I built an amp very similar to a Princeton with the main difference being I used a 6AU6 in a pentode mode as the second gain stage. The pentode stage has local feedback in order for it to have the same stage gain (about 60) as the 12AX7 stage it replaces. The only issue I see in the resulting amp is that the 1 meg audio/algorithmic pot volume control that feeds the second stage doesn't seems as "linear" in the volume control as on a "standard" Princeton clone I have. All the volume seems bunched at the top end of the pot as if it wasn't a logarithmic pot at all. Can't figure out why? Could this be due to the difference in input impedance between a pentode and a 12AX7 triode???
 

DavebF

Member
2018-08-11 11:40 pm
Sure, attached is the Princeton version I went from, again except for the second gain stage coming off the volume control is a 6AU6 pentode . I only have a schematic of that circuit in my note book but, I'll try to get a photo and upload it. It is a pretty basic pentode circuit:
R load is 180K ohms, Plate voltage 145 v
R screen 1 meg ohms (bypassed), Screen voltage 50 v
R cathode 2,200 ohms (bypassed), 1.5 v bias
Feedback resistor of 1 Meg from plate to grid (through .022 cap)
Input grid resistor 100k (connected to volume pot center tap.)

If I look at a schematic of the Princeton Reverb with signal values at test points, all my values are at the prober levels except I need to turn the volume knob past the "5.5" they say in the notes to get 56mV (.158 v p-p) from the center tap of the volume pot
 

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PRR

Member
Paid Member
2003-06-12 7:04 pm
Maine USA
> 1 meg audio/algorithmic pot volume control ...All the volume seems bunched at the top end of the pot....
Feedback resistor of 1 Meg from plate to grid (through .022 cap)
Input grid resistor 100k (connected to volume pot center tap.)


It's not really "the pentode". It is the NFB connection. Taken simplistically it is 100k input instead of >100Meg. Due to finite gain it may be more like 200k. This is still a *heavy* load on a 1Meg pot. Yes, you have to turn it up further.

And why target gain of 10? The 12AX7 gives gain of 25-50. This also will need you to turn it up more.
 

PRR

Member
Paid Member
2003-06-12 7:04 pm
Maine USA
My sim does not have a 6AU6 but with the 10Meg NFB I'm getting order-of 280k input impedance. So at nominal half-gain (about 7 on the knob) it will be about 6dB lower than the same gain pot driving a naked grid.

Simple experiment. Take off the NFB and also the cathode bypass cap. Gain of 6AU6 in that condition may be near 60-70?
 

DavebF

Member
2018-08-11 11:40 pm
Interesting enough, I've been down that road before. When I first tested the circuit I had no feedback circuitry and both the screen resistor and cathode resistor bypassed, that's when I measured the open circuit gain of 164. That was way to much, the 6AU6 turned out to be very microphonic, laying a screw drive on the bench rang through the speaker! So what you suggested was the first/easiest thing I tried but, it dropped the gain down to somewhere in the 30's (I'll have to check my notes). That's why I when the more complicated method of feedback, so I could get the gain right at 60. OBTW I also tried un-bypassing the screen resistor and that really reduced the gain, down to about 10 as I recall.
 

DavebF

Member
2018-08-11 11:40 pm
I was wondering if just a different pot would help the situation? I see pots on schematics that imply there are different logarithmic tapers you can buy for audio pots, i.e. 15% and 30% but, I don't see any place that advertises pots like that for sale.
 

Gnobuddy

Member
2016-03-01 4:10 pm
...get the gain right at 60...un-bypassing the screen resistor...really reduced the gain, down to about 10...
It doesn't have to be all or nothing. You could partially bypass the screen resistor.

To do this, instead of wiring the screen bypass cap direct from screen grid to ground, put a resistor in series. Twiddle the resistor to set the gain to 60.

Calculating the required resistance value isn't super easy or accurate, so this is one case where I'd suggest using a potentiometer and twiddling it to find the right resistance by experiment, then replacing the pot with a same-value fixed resistor.


-Gnobuddy