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I have a device which works with 7V. I have a 7812 and 7805 already running in the circuit. Can I use the outputs of this voltage regulators to feed the 7V device?

I know this is possible in theory. But most practical voltage sources can supply current in only one direction. In other words, the device will drain arbitrary amounts of current (\$I_D\$) from the 12V grid, and give it back to the 5V grid. What will happen if the other circuit elements which are connected to the 5V grid drain current less than \$I_D\$? Will it cause the voltage on the 5V grid to rise up and the voltage on the device drop down?

Is this configuration OK in practice?

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3 Answers 3

up vote 7 down vote accepted

NO you cannot!

What most people don't appreciate, apparently even some people with a lot of experience is that a voltage regulator is not the same as an ideal voltage source. The 78XX series can only supply current. In the topology you are suggesting the lower 7805 must be able to sink current into it's output.

What will happen is that once you device starts to pull current the 7805 will see it's output pulled high, it will attempt to control this condition by reducing the control signal on it's pass transistor, shutting it down a little bit, this shows less of a load on the bottom side so the voltage bumps up more etc. etc. until the 7805 output is at the 12 V rail and your 7V device is shut off.

You have to have a power supply that can both source and sink current into it's out put. So a Op-amp style with a complementary output structure.

Another alternative is to use a 7905 with it's bottom rail being your implied ground (18 V negative ). 7905's are design to sink current into a negative rail.

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Your concern is justified. If the current drawn from the 5V line is less than the 7V device is drawing (even momentarily) the 5V line will rise above 5V and possibly damage whatever is connected to the 5V power. Most regulators cannot sink current, only source it.

You could add a dummy load to the 5V supply (but if your 7V device has a capacitor inside then it could cause a transient that would damage things connected to the 5V supply).

It's much better to add a 7V regulator to the 12V supply (or the 18V supply). An LM317 with a couple of resistors will do it, or if you're not too fussy about the exact voltage, a 7805 with a red LED in the GND pin will get you close.

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If you add a load from 5V to ground that always consumes more current than the 7v device, then this will work.

So if your 7v device is rated at 750mA peak current, then using an 800mA load on the five volt output would allow the regulators to function correctly.

However, this is pretty inefficient anyway. If this is all you have available, then you can make it work with a load, but you are essentially using 18v to power a 7v device, so you're dropping 11v, and losing a lot of power in these linear regulators. You would be better off with an appropriate 7v switching regulator.

As currently designed with no 5v load, though, it will fail.

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