Continuing questions about the S.E.X. amp design

anthony · 9486

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Offline JC

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Reply #15 on: December 06, 2009, 05:17:39 AM
Just a WAG, but I expect this refers to the grid resistor on the following stage, as CB points out.

I note that Morgan Jones identifies this as the "grid leak" resistor, which is a common phrase from some time ago; it is not much in current usage, and neither are grid resistors of that large a value from what I can tell.  But, the use of the term "grid leak" is evocative of an interesting design point: Search up some info on "grid leak bias", which is yet another method by which the negative bias on the grid can be established without the use of a cathode resistor or a negative bias supply.

This method is also not much in use in current designs, but you may possibly run into it in older gear.  I used to see it on the input stages of old guitar amps all the time, and it was a real puzzler until I learned more about it!


Jim C.


Offline JC

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Reply #16 on: December 06, 2009, 05:32:02 AM
Oh, and excuse me, I neglected to respond to your power supply question.  CB offered good advice, and I will only add that it is important to remember that the secondary Voltages specified for the transformer are generally given as RMS Voltages.  In the power supply, after rectification, the caps charge to the peak Voltage of the AC, which is, of course, much higher.  A good rule-of-thumb calculation for converting RMS to Peak is to multiply by a factor of 1.414.

Jim C.


Offline anthony

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Reply #17 on: December 06, 2009, 07:35:40 AM
@caucasian blackplate:  I see.  that makes a lot more sense than the way I was looking at it (the filtering effect)

as for the power supply, i guess i'm just ill-equipped to do my own measurements.  I brought my multimeter to work and it was stolen :(  also I do not own an o-scope and i can't run useful software since my PC won't turn on (i'm on my macbook, which i typically use for audio recording and editing).  So again, I apologize for my inability to do some of this work myself.  I was going to steal someone else's meter at work, but I decided that would be uncool, so i just gotta suck it up and buy a new one i guess, and not bring it to work!

What you and JC said, however is very helpful.  Sometimes I don't know when to use RMS and p-to-p, i guess, but thanks for pointing that out, JC.  One thing JC, what is a WAG? hahaha

I have one more question that I've been meaning to ask, that I haven't been able to find the answer to elsewhere.  In the filament heater circuit, why is the resistor only 0.1 ohms?  What is the purpose of it?

Again, I can't thank you guys enough for all your help.  The reason I'm so curious here is cause after I built the kit, I wanted to try to design an amplifier myself.  I started (back when I was in college and had the necessary equipment) and soon realized that I have a significant amount still to learn before I can build something that will work as intended haha.  I think another kit build will be a good idea (more $ for bottlehead haha), so i've been saving up for that and looking forward to my next project.



Offline JC

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Reply #18 on: December 06, 2009, 08:11:13 AM
Wild Ass Guess! 

I can only imagine that the .1 resistor is in there to drop a tiny bit of voltage to keep the result at the tube pins at or somewhat below the 6.3 Volts specified for the 6DN7 tubes.

As you can see, after rectification and filtering, the DC ends up closer to the peak of the   waveform coming out of the secondary of the transformer.  When the load (tube filaments) is applied, it no doubt lowers the voltage due to the fact that the secondary Voltage on the transformer drops according to how much load is applied to it and how well it maintains its Voltage under load.

I suspect in this case that even the load was not quite enough to lower the Voltage to the safe level, so a small resistance was needed to lower it a bit further.

This will be a good exercise in the use of Ohm's Law.  I am away from home, so I don't have the data readily available, but if you do you can use the heater amperage specification x 2 for the current flowing through the resistor.  That multiplied by the .1 value of the resistor will tell you how much Voltage it is dropping and, therefor, about how much the Voltage would probably be too high if it were not in there.

Of course, .1 Ohms doesn't seem like much resistance, but since the amount of Voltage it drops is also a factor of how much current is flowing through it, it can be significant.

I have often had to use fractional resistors like this in old guitar amps to lower the heater supplies, mostly AC ones.  See, the amps were designed and built at a time when the typical AC Line Voltage in the US was 117 VAC or less.  So, the transformers were specified to operate at that input Voltage, not the 120 - 125 VAC typical today.  That increase on the input translates to a corresponding increase on the secondaries of the transformer, so often an older amp was applying heater Voltages that were higher than +10% of nominal.  This not only shortened tube life, as you can imagine, but the extra heat also caused the cathodes to emit more which could alter the operating points enough for audible differences.


Jim C.