# 7805 Regulated DC Power for two devices

I need to create a circuit specifically for powering two devices. These devices include a device, which has to be powered via 5V USB and another device which has to be powered with at least 6V (all DC).

Now, 12V power supplies are quite cheap.

Can i simply use a 7805 regulator to power the 5V device? Does it support 12V as an input? What about the other device? I thought about using a simple resistor for this one, as current drain (should) remain constant over time. Is that feasible? How would you implement such circuit?

• Question "Does it [7805 regulator] support 12V as an input?" is best answered by reading the datasheet. – Nick Alexeev Nov 7 '12 at 18:36
• One thing others haven't mentioned is that if you in the end have to go with linear regulators, you can chain several together so that heat dissipation is spread out. For example you could put a 7809 before the 7805. This will cause lower heat dissipation in 7805. – AndrejaKo Nov 7 '12 at 21:03

You can use your 7805 regulator to provide a 6V output. How? Take a look at the datasheet:

The way the 7805 works makes it possible to get any voltage between Vin-2V and 5V on the output. The 7805 works in such a way, to ALWAYS achieve 5V between its ADJ and VOUT pins. So the output voltage is as follows: $$\frac{5V}{Vout}=\frac{R1}{R1+R2}$$

Pick R1 and R2 to get 6V on the output. Afterwards, you can use diodes to drop that voltage a bit. A single diode provides a forward frop of approx. 0.7V. Depending on your actual requirements, you might get away with using two diodes and a 1.4V drop. You can get 6V and 4.6V. Depending on your requirements maybe you can get away with 6.2V and 4.8V. You can also try using diodes with different voltage drops to get a better output.

OR, you can use a separate regulator, since they're so cheap.

• Wow. That's what I call a comprehensive answer :) I didnt think of that at all. Maybe I will go with the diodes. Thank you, dear sir. – s1ck Nov 7 '12 at 18:58
• Did you notice that he does not need 6V, he needs at least 6V. Probably 12V will do fine (but this needs to be checked!) – Wouter van Ooijen Nov 7 '12 at 21:57
• @WoutervanOoijen You're right, I missed that! – Christoph B Nov 8 '12 at 6:15
• Can you please tell me whats the value of R1 & R2 used in the above circuit and will it be necessary to add heatsink with transistor? Thank you. – user54656 Oct 6 '14 at 12:16
• @EhsanulMomin There is a formula to calculate the values of R1 and R2, and it is in the answer. For more information refer to the datasheet – clabacchio Oct 6 '14 at 12:25

For the 5V device you can probably use an 7805 (don't forget the decoupling C's, two 100nF's will do fine), but you must check the (maximum) current your device consumes. An 7805 can deliver 1A, but with 12V input it would dissipate 7W, which requires a good heat-sink. A reverse diode over the 7805 is also a good idea (because your device probably has a large elco at its input). And when you are at it, include a series resistor at the input too.

For the at-least-6V device: find out the maximum input voltage. There has to be a maximum, 120V will definitely let out the magic smoke. If 12V is below the maximum you can indeed use as 12V supply. Provided of course that it can deliver the total maximum current.

• Thank you very much! Indeed, 12V is below the maximum input voltage. However, the USB device drains 700mA so the 7805 will have to crunch 4.9 watts. Will I need cooling (the environment is pretty warm)? Could I use some sort of LM2576 just as easy instead, without cooling? – s1ck Nov 7 '12 at 18:55
• The ballpark power-without-heatsink for a TO220 is 1W, so yes, you will definitely need some heatsink at 5W. A switching regulator like an LM2576 will dissipate much less power, and will probably not need heatsinking. But the circuit will be a little more complex. – Wouter van Ooijen Nov 7 '12 at 21:59

Yes, the 7805 does work with Vin of 12V. And the other device can be powered from the same 12V with its own regulator for whatever voltage it needs, e.g. 7809 for 9V.

The only thing to note is the current consumption of your device(s): Those regulators produce heat proportional to the current drawn. Making 5V out of 12V @ 1A generates 7W of excess heat which is too much for the 7805 without massive heat sinking. At 100mA however, there's only 0.7W, which may be quite tolerable. (Still gets hot though! You have been warned :)). Look at the datasheet for your regulators to find out what maximum current they can provide in your installation.

If you find that you need more current than the regulator(s) will withstand, or that you don't want the massive waste of energy dissipated as heat, you can have a look at some switching regulators, like for instance those KIM-... (KIM-055 for 5V) modules they sell off everywhere these days.

• Thank you! Please read my comment on Wouter's answer. Are the KIM-XXX modules cheaper than some step-down regulator like the LM2576 ? – s1ck Nov 7 '12 at 18:56