Zener Diode higher voltage at other disconnected end?

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Im confused. I was operating on a KOSS DVD player and found a
defective +5V supply. Just before the +5v output pin, there was a
diode (unknown exact type) I measured the voltage at the supply side
of the Diode and it read +5V and on the "output" pin side that was NOT
connected to anything I read +22V. How the heck is that?
At any rate, I removed the diode from the circuit, and now I have the
+5V back. All is not perfect however, the DVD player shuts off every
hour or so now.
My questions are
1) what do you think that diode was for.
2) why is it possible that I got a higher (much) voltage on the other
side of the diode when it was not connected to anything.

thanks!

 

[Charles]

Im confused. I was operating on a KOSS DVD player and found a
defective +5V supply. Just before the +5v output pin, there was a
diode (unknown exact type) I measured the voltage at the supply side
of the Diode and it read +5V and on the "output" pin side that was NOT
connected to anything I read +22V. How the heck is that?
At any rate, I removed the diode from the circuit, and now I have the
+5V back. All is not perfect however, the DVD player shuts off every
hour or so now.
My questions are
1) what do you think that diode was for.
2) why is it possible that I got a higher (much) voltage on the other
side of the diode when it was not connected to anything.

A shunt regulator circuit is a series resistor and shunt zener. A higher
voltage is normal on the input to a shunt regulator (before the resistor).
Removing the zener would normally result in a higher output voltage. I too
am confused.

 

[mike]

Im confused. I was operating on a KOSS DVD player and found a
defective +5V supply. Just before the +5v output pin, there was a
diode (unknown exact type)

What made you think it was a zener?

I measured the voltage at the supply side

of the Diode and it read +5V and on the "output" pin side that was NOT
connected to anything I read +22V. How the heck is that?
At any rate, I removed the diode from the circuit, and now I have the
+5V back. All is not perfect however, the DVD player shuts off every
hour or so now.
My questions are
1) what do you think that diode was for. isolation
2) why is it possible that I got a higher (much) voltage on the other
side of the diode when it was not connected to anything.

put an oscilloscope on the 5V side and see if there are 22V spikes.
Sounds like your filter cap isn't doing its job.

It's normally a bad idea to take parts out of a circuit you don't
understand. The vendor wouldn't normally waste 2cents on a diode they
didn't really, really need.

 

[Eeyore]

on the "output" pin side that was NOT
connected to anything I read +22V. How the heck is that?

It seems it was conected to something but you must have missed it. Explain how a
diode has an 'input side' and an 'output side' too will you ?

Graham