It should not. The drive is dependent on the 5v and the negative rail. You have drive. Now you have to determine if it's of sufficient amplitude.
If you believe that the un-powered IC is loading the drive circuit more than it should, lift pin 1 to remove the IC from the drive circuit.
If you believe that the un-powered IC is loading the drive circuit more than it should, lift pin 1 to remove the IC from the drive circuit.
With your meter, measure the AC voltage on the output of the LM211. Black probe on the audio ground pin of the driver board. This is for reference.
Then measure the voltage on pin 1 of the 21844 with the black probe on pin 3 or 5 of the 21844.
Post the voltage you read for both.
Then measure the voltage on pin 1 of the 21844 with the black probe on pin 3 or 5 of the 21844.
Post the voltage you read for both.
The problem is the ir218442.
U1 & U2
With pin#1 connected output is appx 1.5v.
Without pin#1 connected output is close to 10v.
U1 & U2
With pin#1 connected output is appx 1.5v.
Without pin#1 connected output is close to 10v.
Attachments
I haven't seen anything that shows them to be defective.
Have you seen the posts about low-voltage testing by adding windings to the main transformer?
You could replace the regulators and power up normally or use the low-voltage testing which gives the amp a better chance of surviving if there is a problem in the output stage.
Have you seen the posts about low-voltage testing by adding windings to the main transformer?
You could replace the regulators and power up normally or use the low-voltage testing which gives the amp a better chance of surviving if there is a problem in the output stage.
Start reading here and the next couple of pages:
4000.1
The B+ limiter doesn't help, many times, because the caps get charged and then the circuit is enabled. If there is a problem, the outputs may have to try to survive discharging all of that stored energy. Using the low-voltage significantly reduces the energy stored.
4000.1
The B+ limiter doesn't help, many times, because the caps get charged and then the circuit is enabled. If there is a problem, the outputs may have to try to survive discharging all of that stored energy. Using the low-voltage significantly reduces the energy stored.
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