Emitter Resistor Value Changes.

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#1
I have a ton of .33 Ohm emitter resistors and I don't want to buy any more. What should I expect if I used them in an amp that currently has .05 Ohm emitter resistors? I tried to use my Navy Math but since I partied this weekend, my 2 brain cells are still arguing about it. I wouldn't think there would be a huge difference. Maybe eat up a couple watts.
 

mlucitt

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#2
My understanding is that the emitter resisters act as current load-sharing devices. Increasing the value could help with stability while slightly lowering the maximum power output. If you consider that some of the .33 resistors in our amps were 10% tolerance, that takes you down to .03 Ohms. I would try it and see if you can get good bias readings. Probably will not affect the sound quality at all.
 

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My understanding is that the emitter resisters act as current load-sharing devices. Increasing the value could help with stability while slightly lowering the maximum power output. If you consider that some of the .33 resistors in our amps were 10% tolerance, that takes you down to .03 Ohms. I would try it and see if you can get good bias readings. Probably will not affect the sound quality at all.

I think you used my navy math.... .33 ohm resistor times 10% =.03

.33-.03=.29
 
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#4
I think I'll change them and do some testing. At $2 plus, each for good resistors, it's worth a try with over a dozen in each amp
 

nakdoc

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You'll need to do some math to get the bias current right. If bias is 1 mv for 0.05 ohm resistors, you need to set 6.6mv for .33 ohm. larger value emitter resistors add quite a bit of thermal stability by producing local feedback. other consequences are a tiny drop in output power, and a lessening of damping factor (feedback ratios interact with emitter resistors) You see modern amps, mostly MOSFET, with 0.1 or less because they really aren't needed at all except for safety. Since you are building large analog anps, I see no reason for 0.05 ohm. If you want to test something, try comparing distortion at 100 watts between 0.1 ohm and .47 ohm emitter resistors. I doubt you'll measure a difference. My advice is use the .33 ohm.
 

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#6
two .33 ohm resistors in parallel are 0.165 ohm. If you want 10 watt emitter resistors, this is one way to achieve that. many modern Denon amps have parallel 0.22 ohm 2w resistors in the emitters. The sad fact is I know this because they blow up. How many blown amps do you see with 0.47 ohm emitter resistors?
 

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You'll need to do some math to get the bias current right. If bias is 1 mv for 0.05 ohm resistors, you need to set 6.6mv for .33 ohm. larger value emitter resistors add quite a bit of thermal stability by producing local feedback. other consequences are a tiny drop in output power, and a lessening of damping factor (feedback ratios interact with emitter resistors) You see modern amps, mostly MOSFET, with 0.1 or less because they really aren't needed at all except for safety. Since you are building large analog anps, I see no reason for 0.05 ohm. If you want to test something, try comparing distortion at 100 watts between 0.1 ohm and .47 ohm emitter resistors. I doubt you'll measure a difference. My advice is use the .33 ohm.

Interesting........ The book calls for 5mv at idle with the .05 ohm resistors. You don't want me to use my navy math do you?
 

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two .33 ohm resistors in parallel are 0.165 ohm. If you want 10 watt emitter resistors, this is one way to achieve that. many modern Denon amps have parallel 0.22 ohm 2w resistors in the emitters. The sad fact is I know this because they blow up. How many blown amps do you see with 0.47 ohm emitter resistors?
The resistors I have are 5 watt which is what the originals are. Double them up??? I like that idea.
 

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#9
Just hijacking this tread a little. Using the .33 ohm emitter resistor in the Phase Linear needs .350-.380 volts to bias properly

Is there a calculation for this for different emitter resistors and transistor types? your requirements are so much lower for the Carver?

I am sorting out making bias level correct on a dual rail setup. I may run separate supply resistors thought the rail voltage switch to give correct bias voltage on high and low voltage. Any information is welcome
 

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Just hijacking this tread a little. Using the .33 ohm emitter resistor in the Phase Linear needs .350 plus volts yo bias properly

Is there a calculation for this as your requirements are so much lower for the Carver?

I am sorting out making bias level correct on a dual rail setup. I may run separate supply resistors thought the rail voltage switch to give correct bias voltage on high and low voltage. Any information is welcome

Hijack away bro..... We'll have to wait for NAKDOC to show us the formula. I need it for doubled up .33 ohm (at original .05 ohm, 5 mv) and you need to find out your set up...
 

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#11
I think you used my navy math.... .33 ohm resistor times 10% =.03

.33-.03=.29
I used my own Navy math... I should have pointed out that the .03 ohms would be the tolerance change and not the total resistor change. But, you already knew that.

300 Watts into 8 Ohms is 6.12 Amps at full power. 6.12 Amps shared across 8 devices is 765 mA per device. 765 mA across a .33 Ohm emitter resistor drops .252 volts. Does this make any sense? I never tried to figure this out with Navy math...
 

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I used my own Navy math... I should have pointed out that the .03 ohms would be the tolerance change and not the total resistor change. But, you already knew that.

300 Watts into 8 Ohms is 6.12 Amps at full power. 6.12 Amps shared across 8 devices is 765 mA per device. 765 mA across a .33 Ohm emitter resistor drops .252 volts. Does this make any sense? I never tried to figure this out with Navy math...
where did the 300 Watts come from, and where did you get 8 devices? I'm doing a Carver M-1.0 with 6 emitter resistors per channel or was that directed at Glen's question?
 

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#14
Bias is a current (I) that the output transistors conduct with no signal present. Use ohms' law to solve for current I=V/R I = current, V= voltage, and R is resistance. It is very important to keep track of powers of 10. So the factory spec I = 0.005 V / .05 ohm I= 0.1 amp (100mA)
So if we want .22 ohm emitter resistors and 0.1 a bias, 0.1a= V / 0.22 V = 0.1 * 0.22 = .022 V or 22mV
We don't gain much by computing watts dissipated at idle in the emitter resistors, because idle is the smallest signal condition. Generally 5watt resistors will work for any amp below 200 watt. A 350 watt amp would need 10 watt resistors to be safe.
 
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