After looking at some RCRC networks, I wonder how the resistors used don't burn up. For example :
http://diyaudioprojects.com/Tubes/KT88-Push-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
http://diyaudioprojects.com/Tubes/KT88-Push-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
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.
..................Blake
"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.
..................Blake
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 ?
.................Blake
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 ?
.................Blake
Nihilist said:
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.
Yes, used to stop oscillations. Hence the name stopper.
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.
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.
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
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/img353/1543/mycurrentsetup.png
.......................................Blake
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/img353/1543/mycurrentsetup.png
.......................................Blake
Nihilist said:
I don't know about the push-pull part, but on the first triode you need a grid resistor from ground to grid to allow the miniscule grid current somewhere to flow. Typical values seem to be in the region of 250K to 470K.
Then it seems sensible to include a stopper in series immediately before the grid. Typical values seem to be 250R to 1K.
So it will be... input socket > resistor to ground > grid stopper > grid.
Hopefully someone else will be able to give you hints about the push-pull part.
suggestions for study and investigation
use all of the 6AN8. The pentode section can be run at the top of a SRPP design effectively becoming a CCS load for the triode, giving great performance and overcoming miller etc that may otherwise load up the triode on its own.
Watch out for cathode/heater voltages in that small bottle!
use all of the 6AN8. The pentode section can be run at the top of a SRPP design effectively becoming a CCS load for the triode, giving great performance and overcoming miller etc that may otherwise load up the triode on its own.
Watch out for cathode/heater voltages in that small bottle!
maybe my math is wrong but
I just did a quick check on your voltages against hte 6AN8's characteristics. I reckon AT BEST in its current configuration, your circuit will only generate around 30V p-p swing in the triode - not enough by a factor of around 2.
Anyone confirm?
Assumptions:
135V at the anode at -3.5v grid bias means about 10ma idle current; via a 15k anode load @ 10ma means 285V B+
2Vp-p input gives (around) 30V p-p output at the anode in a perfect world.
The 29V grid bias on the output tubes also seems a bit high, as does the anode idle voltage, but I'll defer to others on that...
I just did a quick check on your voltages against hte 6AN8's characteristics. I reckon AT BEST in its current configuration, your circuit will only generate around 30V p-p swing in the triode - not enough by a factor of around 2.
Anyone confirm?
Assumptions:
135V at the anode at -3.5v grid bias means about 10ma idle current; via a 15k anode load @ 10ma means 285V B+
2Vp-p input gives (around) 30V p-p output at the anode in a perfect world.
The 29V grid bias on the output tubes also seems a bit high, as does the anode idle voltage, but I'll defer to others on that...
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