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?
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\$\begingroup\$ Not enough ground current. Take a closer look at the datasheets. \$\endgroup\$– Ignacio Vazquez-AbramsJun 24, 2014 at 3:34
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\$\begingroup\$ Is there a way that I could increase the ground current or would that require a larger input? \$\endgroup\$– BowesAndArrowsJun 24, 2014 at 3:39
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1\$\begingroup\$ What's the input voltage? If it's 12V that would explain it, you need at least 18-19V in. \$\endgroup\$– Spehro PefhanyJun 24, 2014 at 3:55
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\$\begingroup\$ Yes it's 12V in, sorry for not mentioning that. So if I need to boost the voltage, I'll have to look at other options then such as a boost converter and not use this method? \$\endgroup\$– BowesAndArrowsJun 24, 2014 at 4:00
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1\$\begingroup\$ Yes, that's correct, though you could use the 7805 as a post-regulator since boost converters tend to be noisy. \$\endgroup\$– Spehro PefhanyJun 24, 2014 at 4:16
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1 Answer
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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.
answered Jun 24, 2014 at 4:28
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\$\begingroup\$ Nice answer. Is it actually advisable to use this Zenner based aproach? I haven't seen it before. \$\endgroup\$– WalyKuJun 25, 2014 at 14:04
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\$\begingroup\$ There's no problem with it IME. Handy for +/- supplies in some situations. The GND current is close to constant. \$\endgroup\$ Jun 25, 2014 at 16:02