UCD180 questions

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Jan-Peter said:
Steff,

Because of the internal boost trap circuit to drive the positive fet you have an output voltage of 12VDC when the amp is OFF and without a load. You can remove this high impedance DC voltage when you connect a 47k resistor from the output to -V.

Or the best way would be to use a 12V zener and 33k from the output to -V. The kathode of the zener should go to the output of the amp, anode to the resistors and other side of the resistor to -V.

Regards,

Jan-Peter

Thanks JP for the answer.

But where come from the -20.6V with no charge and MUTE ON?

And why 180mV with 100 Ohms charge?

I can see the 12V when MUTE is OFF with no charge.
 
Jan-Peter said:
??? I don't know....

Does the amps works ok?

Perhaps you can post the full schematic of your setup?

JP

Not done yet but the test circuit is really simple.

PSU (MBR10100 + BHC ALS10S = 42V) -> module (mute directly soldered on GND)

Input shorted. Output tested with charge and without charge. For the test, the GND is not connected to the box. I test the speaker output directly. The DC protection circuit have been completely removed.

The module have been tweaked as this : new op-amp OPA2134 and coupling caps replaced with 47UF BG NX HiQ (no 22UF in stock sorry).

The module works well connected to a 4 Ohms speaker.

What's the dark side of the zener+resitor tweak?
 
Hi all,

I am considering a pair of UcD180 modules to play with... but before I know what I'm dealing with, I would prefer to use some 24V transformers I have at hand (to keep costs down).

Is this possible, or will I get into trouble?
(I saw the data sheet for the UcD400, hich would indicate problems with my supply voltage being too low.)

Jennice
 
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