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Support Forum » Question about 7805 Voltage Regulator

August 10, 2011
by Alexp
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I connected up my circuit for the first time and received no victory display. So I started debugging. I put the voltage regulator in a separate part of the breadboard and I was only measuring like 1.3 volts from the output of the regulator to ground. I measured the voltage out of the battery and was only getting 5.3 volts. I replaced the 9V with a new one that was over 9V and the regulator operated as expected, and when connecting back into the circuit I received the victory display.

So my question is, looking at the datasheet for the 7805, how do you tell what the minimum input voltage is?

August 10, 2011
by Rick_S
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The Drop Out Voltage is the amount of voltage above the output that is required for the regulator to work. In this case the datasheet states the dropout is typically 2v max of 2.5 so you would need at least 7 - 7.5V to ensure the 5v version (7805) works.

At least that's the way I understand it Wink

Rick

August 10, 2011
by rajabalu21
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Please see this. Page # 3. V Drop = 2 V so the minimum is 7 V (5 V + 2V) for the regulator to work correctly.

-Raja

August 10, 2011
by rajabalu21
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Sorry Rick. I think we posted almost at the same time.

-Raja

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