todd
New Member
Posts: 15
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Post by todd on Jun 5, 2020 22:30:38 GMT
Need replacements for the output transistors. Q5 & Q6 L&R. Rockola part # 43506. It says on the transistor delco 7305626.
I know they are probably obsolete, I read that i could use 2N3791 and replace R55 & R585 L&R with 4.7 ohm resistors?
Amplifier is a 47160-A in a model 448.
Thanks Todd
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Post by jukenorman on Jun 6, 2020 9:46:11 GMT
Yes, 2N3791 would be a suitable silicon replacement for the germanium output transistors. I might be inclined to think that 4R7 is a bit on the low side for the replacement resistor in the base circuit and would veer towards at least 6R8 or even 8R2. Norman.
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todd
New Member
Posts: 15
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Post by todd on Jun 6, 2020 12:06:01 GMT
Thanks for the advice Norman. Will post results when parts arrive. Todd
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Post by ajmills522 on Mar 24, 2024 17:21:52 GMT
Hello,
I know this is a super old post but I am working on a Rock-ola 437 Deluxe with a 43500-A Amplifier that uses the same output transistors but appears to be set up a little differently that the 47160-A.
"I read that i could use 2N3791 and replace R55 & R585 L&R with 4.7 ohm resistors?"
If I was to use a 2N3791 to replace the Delco Outputs what resistors would I need to change. R55 in my amp is a 47ohm resistor. R58 is a 565-285 Standee Resistor so it doesn't appear that this modification can be accomplished with the same procedure.
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Post by jukenorman on Mar 24, 2024 21:26:56 GMT
In your amplifier, it's R61 and R64 that would be changed - they are in the equivalent positions to R55 and R58 in the 448 amp.. 1 watt resistors.
Norman.
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Post by ajmills522 on Mar 26, 2024 22:41:08 GMT
In your amplifier, it's R61 and R64 that would be changed - they are in the equivalent positions to R55 and R58 in the 448 amp.. 1 watt resistors. Norman. Thanks a million! I REALLY REALLY appreciate it!
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Post by jukenorman on Mar 27, 2024 17:26:46 GMT
I take it that you are aware that if you go this route (and there's no reason not to!), you will have to adjust the bias pots as instructed in the manual?
Norman.
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