Hoffman Amplifiers Tube Amplifier Forum
Amp Stuff => Tube Amp Building - Tweaks - Repairs => Topic started by: kagliostro on May 29, 2013, 11:12:05 am
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Reading the tread about the resistor for the cathode biased EL34 I give a look to the DD-30 on ddawgamps.com
and I get my attention captured by the 1ML pot that is between V1a and V1b
at the same time I remembered something of unusual I've seen yesterday in the schematic of a JCM2000
there is a diode in parallel with the 1R bias resistor
Please can someone explain this two (difficult for me) circuits ?
Many Thanks
K
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First question: Looks like the pot is blending in V1b. (best guess)
#2: those are current sense resistors and they help measure current for the bias supply.
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To the first question:
It looks like the second stage is a inverting feedback amplifier that has variable feedback resistor Rf.
More info here:
http://www.aikenamps.com/FeedbackAmp.htm (http://www.aikenamps.com/FeedbackAmp.htm)
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About the preamp question I'll read the material on the link, Thanks
the pot in the circuit, for sure, isn't the usual gain as often seen
About the 1R resistor I know it is as to read the bias current (read as mV with an 1R resistor)
what is odd to me is D2, the 1N4007 paralleled with the resistor
I don't remember to have seen one in that position before
K
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> diode in parallel with the 1R bias resistor
"Protects" the resistor.
IMHO, a dumb idea.
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Thanks for the answers
About the inverting feedback amplifier I've read the explanation on the link (a bit too much math for me)
and what I understand is that, as VMS say, a variable feedback was added via the 1ML pot
but I'm really not able to go deeper on the reason for the choice of such a circuit, as I've understand the use of that kind of circuit is to achieve a low output impedance as to be able to drive ToneStack without loss, but the amp has the TS positioned on the further gain stage
which will be the purpose of an inverting feedback amplifier with variable feedback in the amp ?
(I'm not able to follow that)
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About the resistor with the paralleled diode, my poor knowledge told me that a diode in parallel with a resistor will act as a shunt, so I'll not be able to measure a voltage across the resistor (of course I'm wrong, or that circuit will not be in an amp)
K
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> I'll not be able to measure a voltage
Typical tube-amp bias in a 1 ohm resistor is about 50mV.
The Silicon diode does not start to conduct well until 500mV.
So normal operation is not affected.
The only way the diode does anything is if the tubes conduct *500mA*.
The diode won't let voltage go much over 0.75V maybe 1.0V. 1.0V across 1 ohm is 1 Watt. The resistor is rated 1 Watt. It is protected against long-term abuse. The $0.50 part will not burn-up.
However...
If we assume 400V across the tubes, and 500mA= 0.5 Amps through tubes and resistor, we have 200 Watts in two 30 Watt tubes. $30 tubes *will* burn-up.
If the resistor is not "protected", there is a *chance* the $0.50 resistor will burn-open before the two $30 tubes. And that the charred resistor will give a clue what happened.
Now, it is *possible* that some fire-code discourages letting resistors burn-up, and the diode is "required" under a legal rule that does not really cover this situation. Or it may be a young designer who is in the habit of putting diodes everywhere (they are very cheap in bulk). They can be good for protecting transistors and chips.
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Many Thanks PRR
Your evaluation of things and disquisition is always illuminating !
K
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which will be the purpose of an inverting feedback amplifier with variable feedback in the amp ?
it just varies the gain of the second gain stage, so it's a gain/volume control