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Old 30th July 2009, 12:46 AM   #1
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Default Power supply resistors power ratings

After looking at some RCRC networks, I wonder how the resistors used don't burn up. For example :
http://diyaudioprojects.com/Tubes/KT...Pull-Tube-Amp/

Scroll down and look at the PS resistor , a 100 ohm 5W unit.

How is this NOT burning to a crisp ? It is providing ALL the power for a pair of Class A KT88's ! Thats gotta be close to 50W combined on the plates .

Can someone enlighten me , PLEASE ?



As long as we're looking at this schematic, what is the benefit of loading the V2 tubes' screen with a resistor rather than directly tying it to ground as per the Melvin Lebowitz original design ?


Thank you................................Blake
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Old 30th July 2009, 12:56 AM   #2
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P=I^2 * R
For 100 mA, R=100, P=1W
For 200 mA, R=100, P=4W

The screen of V2 is not tied to ground, the grid is. The tube is drawn upside down.
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Old 30th July 2009, 01:04 AM   #3
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"The screen of V2 is not tied to ground, the grid is."

My fault. Any idea why he used a resistor inline there ?

"P=I^2 * R
For 100 mA, R=100, P=1W
For 200 mA, R=100, P=4W"


Never saw that formula before.


Sweet shortcut !



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Old 30th July 2009, 01:57 AM   #4
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The design uses grid stopper resistors, while the one in the original article does not.
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Old 30th July 2009, 10:55 AM   #5
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Default grid stopper resistors

"The design uses grid stopper resistors, while the one in the original article does not. "

Yeah, I noticed that it is different than the original. I was wondering WHY he did it.

I thought a grid stopper went from the grid to ground , not in series with the grid.


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Old 30th July 2009, 09:46 PM   #6
chrish is offline chrish  Australia
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Quote:
Originally posted by Nihilist
"The screen of V2 is not tied to ground, the grid is."

My fault. Any idea why he used a resistor inline there ?

"P=I^2 * R
For 100 mA, R=100, P=1W
For 200 mA, R=100, P=4W"


Never saw that formula before.


Sweet shortcut !



Thanks...................Blake
Yeah, it is that esoteric, rarely used, difficult to find thing called Ohm's Law


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Old 30th July 2009, 10:48 PM   #7
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Nobody likes a smarta** .


P=I^2 *R is NOT Ohms Law. It's not even Thevenin's Power Law.

Ohms law is V/I=R or I * R = V or V/R=I .

Amps squared , multiplied by Resistance equals Power is not taught in ANY school or books that I know of.

How did you come to know it ?


Any takers on the Grid Stopper question? A series resistance inline with the grid constitutes a grid stopper ? As I already said, I thought a parallel resistance from the grid to ground was considered a grid stopper. Can anyone clarify ?


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Old 30th July 2009, 11:07 PM   #8
Gordy is offline Gordy  United Kingdom
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Quote:
Originally posted by Nihilist

Any takers on the Grid Stopper question? A series resistance inline with the grid constitutes a grid stopper ? As I already said, I thought a parallel resistance from the grid to ground was considered a grid stopper. Can anyone clarify ?
A series resistance in line with the grid forms a low pass (i.e. high cut) filter with the internal capacitance of the tube, and hence helps 'stop' it oscillating at high frequency.

Going back to the resistor / power question; remember that it is the power dissipated in the component / device that is to be considered and it is not going to be the same as that dissipated in the target load.

Power = Current x Voltage
Power = Voltage(squared) / Resistance
Power = Current(squared) x Resistance

The amount of power a component can take varies with ambient temperature (and sometimes with physical orientation).
Always choose a component that can handle higher power than the calculated figure. For example if you calculate a dissipation of 0.8W in a resistor it would be wise to choose a 2W resistor.
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Old 30th July 2009, 11:11 PM   #9
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Quote:
A series resistance inline with the grid constitutes a grid stopper
Yes, used to stop oscillations. Hence the name stopper.

Quote:
I thought a parallel resistance from the grid to ground
that is just a grid resistor, sometimes called a grid-leak resistor ( although I believe that term is not really applicable unless it is being used to provide bias, I may be nitpicking. )
used to drain away charge from the grid.

Quote:
Amps squared , multiplied by Resistance equals Power is not taught in ANY school or books that I know of.
You can't be serious. I can't think of any elementary book that doesn't give it, and any advanced book doesn't have to.


Quote:
How did you come to know it ?
It is simple algebra. If power is voltage times current, and if voltage is current times resistance, then power must be current times current times resistance, current squared times resistance

P = I * V
V = I * R
P = I * I * R
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Old 30th July 2009, 11:53 PM   #10
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Thanks for the info. I had "Basic Electricity" (a 1 hour a day course) and "Electronics" ( a 2 hour a day course) in high school , and the variants of the power law using the square of voltage or current were never taught.

I never thought my teacher really knew what he was doing.

That was some time ago, about 15 years, but seriously this is the first time I have seen it used this way.

Thank you.


Here's a schematic of my current setup. Do I need stoppers or "leak" resistors ?

I plan on using CCS later.


http://img353.imageshack.us/img353/1...rrentsetup.png


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