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Kevin Weddle

70 volt regulator

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I have this power supply that I was given that takes the 120vac and converts it to 50vac. The current of the bridge is 1.5A max. The regulator has a control line, input, and output. The output is fed to the base of a darlington and the collector goes to the control. The emitter has a zener diode that gets it's voltage from the ouput. The input to the regulator is 50vac filtered to 70v dc. I suppose the output is approx. 70v dc. What is the voltage of the zener? Is it 70v? I have already taken the circuit apart so that I can reboard it. The circuit has a large capacitor on the output, which in my opinion is negligent design.

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Hi Kevin,

Do you have a multimeter? If so why not measure the voltage over the Zener? Why do you think a large capacitor on the output is negligent design?

Ante ::)

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I already took apart the circuit. The large capacitor on the output is open for debate. I have seen smaller capacitors on the output. I think this is correct because the voltage regulator regulates not only the dc, but the fluctuation as well. If you design the voltage regulator to have a small impedance output, then it would be alright. I tend to think it was designed for an impedance equal to the voltage divided by resistance.

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Quite often the things you connect to a PSU are equipped with big capacitors on their input. If you have a big capacitor on the PSU output it reduces the rushin current from the powerstage.
The zener can easily be checked by connecting it in series with a resistor and measure the voltage across.

Ante ::)


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I found the zener to be a 8.5v approx 1.5 A . This would make the control voltage approx. 12v because of the transistor. The input voltage is approx. 70v though. This is very high, but there are large divider resistors which means there is not much current from input to the control. The current comes through the collector of the transistor through the control line. Can a regulator accept an input voltage of 70v down to 12v on the output?

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I can't find the data sheet on the regulator. I know the circuit was operating at 25vdc with the input, the ouput, and the control all at this voltage. It might be a variable voltage regulator. I must have the data sheet to do anything with it.

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The regulator is a motorola 513 je 800. I think the je800 is the part. The zener is on the emitter of the NPN control transistor. The collector goes to the control of the regulator. The base is connected to the ouput through resistors. The zener gets its voltage this way. I wonder about the zener. Why is it there? Why is it better than a resistor.

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The zener is the reference for the regulator.
MJE800 is a darlington transistor
http://www.fairchildsemi.com/ds/MJ/MJE803.pdf
Does that make it look like the att. The unlabelled resistors are part of the darlington.

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That's the circuit. What does the zener help in that position? Why not use a resistor? Does the zener present a higher impedance to the signal? Also, I am unable to measure the diode of the device, it's just low resistance all the way around.

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The zener is the reference for the regulator.
When the voltage on the base of the transistor exceeds
the zener voltage plus Vbe, the transistor conducts
tending to remove base drive from the pass transistor.
So the regulated output is
(Vz + Vbe)*(1+R1/R2)
What are the original values?

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That circuit provides a regulated output voltage by this calculation:
Zener voltage plus 0.65V (base-emitter) multiplied by (R1 divided by R2) plus 1.
The zener is the voltage reference that the divided output voltage is compared with. Since a zener is low impedance, changes in current through it does not produce much voltage change across it. If a resistor was used instead, then there would be no reference voltage (it would change with any output change) and the circuit would not regulate.

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If the zener reads low both ways it is :'(
You say "I found the zener to be a 8.5v approx 1.5 A "
How did you test it?. If it has some writing on it tell us, and the values of the resistors

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