67V output with Bo Hansen Phantom PSU ?

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bernbrue

Well-known member
Joined
Mar 13, 2006
Messages
1,516
Location
Wolfenbüttel - Germany
Hi,
I made Bo Hansens 48V Phantom PSU and the output voltage is 67V. Is that allright? I did the first PSU shown here:

http://web.telia.com/~u31617586/#48%20volt%20phantom%20power%20supply

Does'nt the 51V Zener prevent to get more than 51V? Could anyone give me a hint, what I made wrong?
best regards
Bernd
 
67V will likely cook some microphones, most mic manufacturers use 63V electrolytics in the mics, which will be somewhat stressed depending on the current requirements of the mic and the resulting voltage drop.

Is the zener good? Was it possibly installed backwards? If so, it would have drawn heavy current until it burnt up. In that particular circuit, the zener will clamp the output voltage for all it's worth. Are the two 1K resistors in place? This is what limits the current, keeping the zener from burning up.
 
Bernd,

You shall have around 67 volt with out load if the zener diode is not connected, or if the zener have a interruption inside.

(48 volt ac multiply with 1,4 is approx. 67 volt dc after rectifying and smoothing with out any shunt/zener regulation connected)

So, check if you have done the circuit connection right and also the zenerdiode condition.
The mark/circle shall be on the last 1 kohm resistor side.

Don´t worry, I not think you have damage any microphone, this is a "passive" PSU, so when the condenser microphone get to high supply, it will draw more current, and with more current the passive supply will drop more voltage on the output.

--Bo
 
If the zener is connected backwards (all other parts being connected corectly), the output voltage will be around 2V (not 67V). This can't do any damage to the zener because of the high value (10k) of the forward resistor. I suspect that the reg. transistor (or its implementation) is faulty.

Regards,
Milan
 
If it's the first circuit, there isn't a series pass transistor; the second circuit doesn't have a 51V zener that I can see.

If the zener is backwards, the output will be around .5V - .6V , whatever the forward voltage drop of the zener happens to be. As the zener gets warm, this voltage will drop further. If the zener is backwards and it cooked itelf until the junction went open the voltage would be high as you describe. Typically though, they short and just draw whatever current is available and you would have very little voltage on the output in this case.

The series resistance is shown as 2K, two 1K resistors in series as part of the PI filter, not 10K. With a dead short, I calculate 33.5mA, not enough to kill the zener.


I think you have an open zener, or it isn't connected correctly in the circuit.
 
Of course, if the zener is backwards, you have the forward voltage drop of the zener, and the output voltage is drop to around a volt.

As usual, it is a middle of the night when I´m on line here, so probably my head think backwards diode :oops:
(I will correct my last post)

As BYacey says, you have an open zener, or it isn't connected correctly in the circuit.

So, check that you have connect all components righy, and then change to a new 51 volt, 1 watt zenerdiode, and be sure that you connect the zener with the mark/circle to the last 1 k resistor, and the other end to ground.

--Bo
 
Hi mates,
thx a lot for your responses. I checked the connections once again and indeed the zener was open. I replaced it and everything was fine. I used a 52V 2W Zenerdiode. It works now as it should. Thanks a lot for your input, Bo.
best regards
Bernd
 
Hi,
just wanted to mention, that this Phantom PSU indeed gives 67V Phantompower. I strapped a 6K8 resistor across +/- to set it under load (simulates the load of two microphones, thx Olaf for this advice) and then exactly 48V are achieved. A very simple and straight forward Phantom PSU I can highly recomend. Thx a lot, Bo!
best regards
Bernd
 

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