What is this circuit doing?

Thread Starter

Ylli

Joined Nov 13, 2015
965
A little circuit that generates +18V from the power transformer in a NAD C375bee Amplifier. Why the complicated circuit just to generate the +18 volts? What is it doing different than an simple rectifier/filter? Does it have a faster/slower rise time? Faster/slower decay? Sure doesn't seem to regulate at +18 too well, but the designer must have done this for some reason. Attached schematic snippet from the NAD manual and my .asc.
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Papabravo

Joined Feb 24, 2006
13,969
A little circuit that generates +18V from the power transformer in a NAD C375bee Amplifier. Why the complicated circuit just to generate the +18 volts? What is it doing different than an simple rectifier/filter? Does it have a faster/slower rise time? Faster/slower decay? Sure doesn't seem to regulate at +18 too well, but the designer must have done this for some reason. Attached schematic snippet from the NAD manual and my .asc.
View attachment 187413
Depends on what is on the other side of the connector -- THAT YOU NEGLECTED TO SHOW US!
PS I don't usually download files.
 

Thread Starter

Ylli

Joined Nov 13, 2015
965
The connector that is labeled "From Main Transformer" is from the main transformer. Pins 1 & 2 are the ends of the winding, and pin 3 is the CT. The "+18out" from the circuit eventually feeds the input to a +5.2V regulator. The manual for the NAD is available here (see page 58): https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=2ahUKEwitq-Pnp4jlAhVHF6wKHYq9BTUQFjAAegQIABAB&url=https://www.manualslib.com/manual/1397194/Nad-375bee-C.html&usg=AOvVaw01dTajUliHr98reZ66Zseo .
 

Jony130

Joined Feb 17, 2009
5,176
They do it probably to reduce the power losses in the MOSFET. Add the load resistance to your simulation and plot to see it by yourself.
 

Thread Starter

Ylli

Joined Nov 13, 2015
965
They do it probably to reduce the power losses in the MOSFET.
That wouldn't explain why a simple rectifier/filter with simple regulator would not be used.
Add the load resistance to your simulation and plot to see it by yourself.
I did try with a load, and the output voltage is very load dependent. That's why I mentioned it did not regulate at +18 too well.
 

Jony130

Joined Feb 17, 2009
5,176
That wouldn't explain why a simple rectifier/filter with a simple regulator would not be used.
Becouse a simple rectifier/filter will produce a high "DC voltage", from 36V AC voltage we will get around 50V DC voltage that we need to drop across series pass transistor. And this "pre-regulator" will help reduce the power losses.
 

Jony130

Joined Feb 17, 2009
5,176
It looks like this +18V is going into Darlington Q206/Q207 (series pass transistor) which creates the +12VD (input selector relay supply voltage).
 

Thread Starter

Ylli

Joined Nov 13, 2015
965
It looks like this +18V is going into Darlington Q206/Q207 (series pass transistor) which creates the +12VD (input selector relay supply voltage).
Yes. During standby, the switching power supply is operating and providing +4.65 volts to the MCU. When operating, +12 is feed to the 7806 regulator and that feeds the 5.2V bus. What was happening is that the processor was resetting during that switchover.
 
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