Kevin Weddle Posted April 18, 2004 Report Posted April 18, 2004 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. Quote
ante Posted April 18, 2004 Report Posted April 18, 2004 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 ::) Quote
Kevin Weddle Posted April 20, 2004 Author Report Posted April 20, 2004 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. Quote
ante Posted April 20, 2004 Report Posted April 20, 2004 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 ::) Quote
Kevin Weddle Posted April 20, 2004 Author Report Posted April 20, 2004 What you are saying is that a large capacitor on the output will allow more current when there is a sudden current burst. Quote
Kevin Weddle Posted April 21, 2004 Author Report Posted April 21, 2004 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? Quote
ante Posted April 21, 2004 Report Posted April 21, 2004 Kevin,Can you make a drawing of the circuit so I can explore is? It is possible that it is a kind of floating ground circuit. Ante ::) Quote
Kevin Weddle Posted April 27, 2004 Author Report Posted April 27, 2004 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. Quote
ante Posted April 28, 2004 Report Posted April 28, 2004 You said earlier it was a transistor / Zener circuit; witch regulator are you trying to find the data sheet for? Ante ::) Quote
Kevin Weddle Posted April 30, 2004 Author Report Posted April 30, 2004 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. Quote
ante Posted April 30, 2004 Report Posted April 30, 2004 Good question, there are a lot of questions here I guess :-\Ante ::) Quote
GPG Posted May 9, 2004 Report Posted May 9, 2004 The zener is the reference for the regulator.MJE800 is a darlington transistorhttp://www.fairchildsemi.com/ds/MJ/MJE803.pdfDoes that make it look like the att. The unlabelled resistors are part of the darlington. Quote
GPG Posted May 9, 2004 Report Posted May 9, 2004 Attachment did not attach so its herehttp://homepages.slingshot.co.nz/~peg/ Quote
Kevin Weddle Posted May 10, 2004 Author Report Posted May 10, 2004 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. Quote
GPG Posted May 11, 2004 Report Posted May 11, 2004 The zener is the reference for the regulator.When the voltage on the base of the transistor exceedsthe zener voltage plus Vbe, the transistor conductstending to remove base drive from the pass transistor.So the regulated output is(Vz + Vbe)*(1+R1/R2)What are the original values? Quote
audioguru Posted May 11, 2004 Report Posted May 11, 2004 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. Quote
audioguru Posted May 11, 2004 Report Posted May 11, 2004 Sorry for my repeat of GPG's formula. I didn't see page 2 when I replied. Quote
GPG Posted May 11, 2004 Report Posted May 11, 2004 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 Quote
audioguru Posted May 12, 2004 Report Posted May 12, 2004 Kevin,I think that you "roasted" the little zener if you put 1.5A through it.Next time use a current-limiting resistor in series. Quote
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