Tube Amps / Music Electronics
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|9/9/1997 6:41 PM|
||Preamp-Plate & Cathode Resistors|
Take the typical 100k plate resistor and 1.5K cathode resistor. Double both values. What affect on tone? Would it be basically the same or a very definite change. I've tried a variable pot in the cathode and that does make a difference. Thanks, Terry
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|9/9/1997 8:53 PM|
These 2 resistors will affect the tone, gain, headroom and harmonic dist. of the tube. One tone change comes from the change in the "RC" of the res/cap combination. As the res value goes down the cap has less effect on the mids and lows. As the value goes up it has more. RxC = T 1.5(k) x22(mf) = 33 If you want the same (up to a point ) Freq. responce change the cap value to keep "T" the same - 1(k) x 33(mf) = 33 3.3k x 10mf = 33 etc... Of course, changing "T" may give you a nice tone change. The lower value res will bias the tube hotter so more current flows and the tube will be able to take hotter input signals before clipping. When it distorts it will have slightly more 2nd order harmonics. A "rounder" tone. 820 ohms is about as low as you can go to sustain bias to the tube. Higher value cathode resistors will lower current flow and the tube will not take as hot of an input before clipping, and will have more 3rd ordered harmonics when distorting. A "sharper" tone that you can distort easier. The higher you go the less gain out of the tube untill you reach a point of unity gain- 1 in 1 out. The plate res is a little harder to get. Increasing the value actually lowers the plate voltage, But this increases the gain because its the P. res & the tube itself that develops the wave. Gain doesn't really come from the voltage, it comes from the voltage divider of res and tube. To high of a res (470K + ) and there wont be enough voltage to "swing" the signal - it'll clip. Also as P.res increases current decreases (remember - less plate voltage). Both resistors work together to set the operating point, so "dink" with both at the same time - pots work great, but remember the high DC on the plate! It's best to set the range with fixed resistors in series with the pot. Try a 820 ohm resistor with a 25k pot on the cathode, and a 50k resistor with a 250k pot on the plate. For a good clean 1st stage try to change the plate res to put 1/2 of the B+ on the plate while changing the cathode resistor to get 1ma of current. My fav in fender style front ends is a 150k plate resistor with a 1k (33mf) cathode resistor. One benifit of this is that since the tube can handle a hotter signal I can get rid of the 68K grid stopper resistors for a better tone. Whew! PAUL C
|9/10/1997 9:00 AM|
Great reply Paul!
Keep it up!
|9/10/1997 4:20 PM|
Hi Paul, First, I want to thank you very much for your reply. You did a great job. Matter of fact, I printed your reply so I would have it handy. I have experimented in the past with a set plate resistor of 100k and also 220k and then varied the cathode resistor only. Of course the plate voltage goes up and done accordingly. What I believe your're saying is to also vary the plate resistor to balance everything. I'll have to give that a try soon. I'm limited on time lately. A reply like yours makes this board and others very worthwhile and I for one really appreciate it. Thanks You, Terry
|9/24/1997 9:18 AM|
In the second to last sentence you mentioned getting rid of the 68k grid stopper resistors to improve tone because of your tube being able to handle hotter signals. The grid stopper will not actually decrease signal input because of the extremely high input impedance of the grid. The grid does have capacitance to the cathode, which combined with the 68k resistor creates a high frequency roll off at around 10khz (I think). Anyway sometimes I can't stand the brightness of my home-made amps so I use these... Yeah, that was a great post. Take it easy!-eth
|9/25/1997 8:44 AM|
Can someone explain to me the function of the 68k grid stopper resistors on the input circuit of my 50w super lead Marshall?
When I study the schematic it seems to me that, depending on which input is selected, no current can flows: therefore there can be no signal volt drop across it?
|9/25/1997 10:23 AM|
You are correct that no signal current can flow through the grid, but not that this resistor causes no loss.
There is a parasitic capacitor from the grid of a tube to both plate and cathode. A grid stopper forms a voltage divider with these capacitances so that at frequencies where the capacitive impedance is about the same as the resistance, the net signal to the grid starts to drop as the capacitance shunts away signal. The input signal current flows through the grid stopper and then through the grid capacitance.
The effect of the grid-cathode capacitance is obvious, but the grid plate capacitance is not. The signal voltage across the grid-plate capacitance is larger than the actual input signal by the amount of gain in the tube, and inverted in phase. If you go through the math, this makes the grid-plate capacitor look like an equivalent capacitor to ground equal to the real grid-plate capacitor times the voltage gain from grid to plate. This is called the Miller Effect, and the capacitance is called the Miller Capacitance.
The Miller capacitor actually comes in handy in guitar amps. By using a grid stopper, you can tailor the high frequency cutoff of the stage to limit the highs in a distortion preamp. I've used up to 100K as a grid stopper in a distortion channel. It's also very useful in cutting out RF pickup, and in stopping parasitic oscillations in some cases.
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