So. I'm assembling a power supply kit from one of the DIY kit retailers. It works well, but the TL783 regulator providing 48V gets really hot with no load attached. I think that it might be the diode which is for some reason connected across the Vi and Vo pins of TL783. I cannot find it on any schematic in the application note and it looks like it shorts the in and out (Vi to ground should be 15V but now I measure it at about 100V). Is that it?Schematic

  • 1
    \$\begingroup\$ How are you heatsinking the regulator? \$\endgroup\$ – Brian Drummond Mar 25 '18 at 9:59
  • 1
    \$\begingroup\$ Looks like you have a Cockcroft–Walton (or some variation thereof) voltage multiplier producing the ~90VDC from an AC input. The diode is useful- you will find it shown in figure 21 of the data sheet \$\endgroup\$ – Spehro Pefhany Mar 25 '18 at 14:47

First of all, you can't get Vo of 48V from a Vi of 15V, that's not how linear regulators work. Linear regulators work by varying the resistance of a pass transistor, turning any excess voltage in to heat. That means the output voltage will always be lower than the input voltage.

Because Vi is always higher than Vo, the diode doesn't conduct because it is reverse biased (notice that the cathode is to Vi). Instead this diode will only conduct when you turn the input voltage off so as to discharge the output capacitor, protecting the regulator from reverse conduction.

The diode/capacitor arrangement on the input is a diode voltage multiplier which takes an AC voltage input, and produces DC voltage of approximately five times the AC peak.

If you are feeding 15VAC RMS into the circuit, that means a peak AC input voltage of 21.2V. Accounting for diode losses, that means you should get ~100V at the regulator input, which is what you are measuring.

As to why it would get warm without any load, you have to remember there is a load, R1+R2. With a 48V output, R2+R1 will give you about 16mA current draw.

With a 100V input, you get 52V across the regulator @16mA, which gives 0.8W dissipation in the pass transistor. So even with no off-board load connected, your regulator will be dissipating a chunk of energy simply because that is how it regulates. If the board has minimal cooling, 0.8W will produce a reasonable amount of heat in the regulator.

If the regulator is so hot that it burns your hand instantly, there is likely something wrong with your implementation.

  • Is the output voltage stable at 48V?
  • Have you checked all of the solder joints for shorts?
  • Have you checked the resistor values of R1 and R2 are correct?

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.