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28volt power supply

Kevin Weddle

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1) As I said before, a high voltage zener diode (yours is 28V) has poor voltage regulation and its voltage increases as it heats up. If the input is exactly 30.0V then the 10 ohm resistor feeding the zener diode has a VERY high current of 2V/10 ohms= 200mA (minus 23mA in the pots). Then the zener must dissipate 177mA x 28V= 5W which is extremely hot so its voltage will increase a lot as it warms up. If the input is a little high at 33V then the zener diode must dissipate 7.8W!

Use a 5.6V zener diode at a low current. It will have pretty good voltage regulation and its voltage does not change when the temperature changes. If you feed it from an active current source made with an opamp then its voltage regulation will be excellent. The output amplifier can easily amplify it to 28V.

2) Why do you use TWO opamps when only one is required?

3) With a 30.0V supply, if you use ordinary 741 opamps then their maximum output voltage is about +26V. The darlington output transistor has a maximum voltage drop of about 2V so the maximum output of your "28V" regulated power supply is only +24V. If the 30V has ripple then the maximum regulated voltage will be even less.

4) It has NOTHING to prevent it from burning up if the output current is too high or if the output is shorted.

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