I'm boosting an LM7805 to 16V using Zener diodes in series at the GND pin (pin 2) of the 7805. The Zener Diodes I'm using are 1 of each of the following: 3.3V, 3.0V, and a 4.7V. Now theoretically I should be getting 16.0V out (3.3+3.0+4.7+5.0) but for some reason the most I've gotten is only 10V. All these zeners are brand new and I have them oriented properly with the cathode connected to pin 2 of the 7805 and the anode to ground of my input. The reason why I have so many diodes is to power other parts of the circuit like how you'd tap off of resistors in a voltage divider for its divided power. So is there something else that I should've added and that's why I don't have a full 16.0V out or am I doing something wrong?
Electronic – LM7805 and Zener Diodes
boostdiodesvoltage-regulatorzener
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Best Answer
The problem is that with the LM7805 (or any linear regulator) the output voltage must be less than the input voltage by a certain amount, which varies from a fraction of a volt (for so-called LDO = Low Drop Out regulators) to several volts. LDO regulators bring with them some disadvantages (stability can be an issue) so they're not a panacea.
In the case of the 7805 it's at least a couple of volts. You can find the number in the "dropout voltage" specification in the datasheet.
Since you're only giving it 12 volts, it can't output higher than about 10V (and if truth be told, it's best to keep it a bit less than that, maybe 9V).
Options if you need more output voltage than input voltage include boost converters (which store pulses of energy in an inductor then spit them out at higher voltage and lower current) or a DC-DC converter (which chops up the input voltage into AC and passes it through a transformer or does something similar to the boost converter, then converts it back to DC.
One of the cheapest boost converters is the (old) MC34063- it requires a fairly large inductor but it's cheap. Newer ones from Linear Technology and TI can work with smaller inductors due to high operating frequency.