Zen Output Question

What happens connecting the Zen output this way?
 

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I see at least one potential problem. C1 and C2 will not divide the rail voltage evenly. Assuming that they're normal electrolytics, and rated +10-20% or some such, you should see that the voltage divides unevenly according to the ratio of their actual capacitance. The usual solution is to use a resistive divider to force equal voltage across the two capacitors. If you can manage to get the amp to set an equal DC potential--and hold it--you might have a viable circuit.
In principle you can delete the Source resistor under the MOSFET.

Grey
 
GRollins said:
C1 and C2 will not divide the rail voltage evenly.The usual solution is to use a resistive divider to force equal voltage across the two capacitors.


Grey

Well they do and hold the point and dont seem to give problems . The speaker see just fiew millivolt if the DC output point is at half the rail .I was thinking of the resistive divider too and I am sure it may help in case the dc point of the amplifier isnt half the rail ( Zen V9 for example )


Regarding the possible hum , It is easy to understand that if the amp itself is well filtered ,to add two more capacitors par rail cant hurt . Infact I didnt noticed any additional hum testing that situation .
 
Good. I think wuffwaff is on the right track in that the circuit itself is able to source/sink sufficient current that it holds the node between the capacitors steady.
Don't be surprised if you get a moderate-sized pop through the speaker if you connect the speaker while the amplifier is already on. The caps will have charged to some arbitrary voltage and they'll have to equalize with the circuit through the speaker. Connecting the speaker only while the amp is off will circumvent this problem.
The resistive divider probably won't accomplish much the way things are going, so just leave it off.

Grey