I have a switch mode DC-DC converter assignment which is supposed to convert 36-72V input DC voltage to 9V DC voltage. The circuit I'm working on contains a control circuit (implemented with a custom and multipurpose microcontroller) which requires 5V DC supply (and drains 55mA current from it) to run.

I feel myself in a paradox; a DC-DC converter is needed to run a DC-DC converter. How do I simply obtain a 5V supply for the control circuit? How do other practical and commercial circuits do this?

Since the input voltage can be up to 72V, I can't use a 7805 directly. Is there any way of using it by making a puzzle-like connection out of it, or cascading two 7805 in a smart way? Also, I can't convert the 9V DC at the output to 5V, because in my design, without the control circuit creating PWM signal, the output voltage will be 0V.

I'm working on a scientific assignment in which I can only use the most simplistic semiconductor devices, like (BJTs, MOSFETs, 78xx, etc). So, please don't offer me using a high level IC for this purpose.

  • \$\begingroup\$ Easy way: the 72V-9V converter is always on. A 7805 generates 5 V from 9 V to run the uC. The uC controls a pass transistor that can disconnected the other loads from 9 V when you want them turned off. \$\endgroup\$
    – The Photon
    Jan 23, 2012 at 23:16
  • 1
    \$\begingroup\$ Is this a switching or linear supply? \$\endgroup\$
    – tyblu
    Jan 23, 2012 at 23:17
  • \$\begingroup\$ @ThePhoton Sorry for not specifying it in my question. I edited it to add this sentence: "Also, I can't convert the 9V DC at the output to 5V, because in my design, without the control circuit creating PWM signal, the output voltage will be 0V." \$\endgroup\$ Jan 23, 2012 at 23:23
  • \$\begingroup\$ @tyblu Sorry for not specifying it in my question. I edited it to specify that my DC-DC converter is a switch mode DC-DC convertor. \$\endgroup\$ Jan 23, 2012 at 23:24
  • \$\begingroup\$ I recommend to use a switching converter controller that doesn't require the uC to be active in order to operate. There are lots of choices out there. \$\endgroup\$
    – The Photon
    Jan 23, 2012 at 23:26

2 Answers 2


A common approach is to make a low power supply which will start the microcontroller and which is then either

  • disabled when the main circuit starts. Or,

  • the low power circuit may be low enough power to be left running.

The arrangement below shows how this can be done using a very minor modification to @Pentium100 's circuit. The load in his circuit is R2. I've left R2 in place but it now serves as a very light nominal load when no power is being drawn by the microcontroller.
If you supply the microcontroller via a diode from the emitter of Q, then, if you subsequently supply the load from a slightly higher voltage, the low power circuit will supply no power as the diode will be reversed bias.
eg Say you microcontroller will operate with 3V < Vdd < 5V. If you arrange for the emitter of Q to be at 4V then the output end of the diode will be a diode voltage drop below Q1 emitter.
A diode drop is typically about 0.6V so the microcontroller will see about 4V-0.6V = about 3.4V. So the microcontroller has Vdd =~3.4V Vdd to start up with.
BUT if you then feed 5V to Vdd the diode will be reverse biased and supply no power.
The only power then drawn by the low power circuit is the power fed to DZ1 by R1. This could be say 0.1 mA for a power dissipation of V x I = 72 V x 0.0001A = 7 mW. Even at 1 mA its only 72 mW.

Super rough also borrowed diagram ...

enter image description here

Upper uode is part of the buck regulator feed - may or may not actually use a diode in the location shown but this gives the general idea.

The above diagram is modified from one found here. An interesting and useful page but only somewhat related to this topic.

  • \$\begingroup\$ I'm not sure about if there are components that can do this (e.g. a transistor that can have about 70V Vce) but it's a smart idea! \$\endgroup\$
    – clabacchio
    Jan 24, 2012 at 9:16
  • \$\begingroup\$ I think I've found one for 0.13$/1pc \$\endgroup\$ Jan 24, 2012 at 14:19

You can make a linear voltage regulator out of a transistor, resistor and a Zener diode. Linear voltage regulator

You can calculate R1 with this formula:

R1 calculation

Where: Vs - power supply voltage, Vz - Zener diode voltage, Iz - Zener diode current (a few mA), Ib - transistor base current (load current / hFE) K - 1.2 to 2 (so the resistance is low enough)

In practice a 10K resistor would most likely work, assuming the control circuit does not use a lot of current or the transistor has high hFE.

The output voltage will be lower by about 0.6V than the Zener voltage, so use a 5.6V Zener diode to get around 5V output.

If you really want to, you can use this regulator to drop the voltage to one that is acceptable for 7805 and then use the 7805 to get 5V.

Schematic and formula taken from Wikipedia, I was too lazy to draw my own.

  • \$\begingroup\$ Note 68 V * 55 mA = 3.6 W burned by the transistor. Whether this is a problem or not depends on OP's total power budget. \$\endgroup\$
    – The Photon
    Jan 23, 2012 at 23:41
  • 1
    \$\begingroup\$ @ThePhoton - Presumably, this would only be drawn for a few milliseconds while the buck regulator starts up. After that, it could be powered by a 7805 or similar regulator running off the 9V supply. \$\endgroup\$ Jan 24, 2012 at 0:17
  • \$\begingroup\$ This is a good answer along with @KevinVermeer's comment. \$\endgroup\$ Jan 24, 2012 at 0:30
  • 1
    \$\begingroup\$ @KevinVermeer; agreed, but then the answer should probably show how to make that happen. And how to deal with what will happen during debugging if the uC code hangs, or if the board is powered up with an un-programmed uC, etc. \$\endgroup\$
    – The Photon
    Jan 24, 2012 at 0:47
  • \$\begingroup\$ Or if not explain how to fix those problems, at least point out that they exist. \$\endgroup\$
    – The Photon
    Jan 24, 2012 at 0:49

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