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I am looking at the 7805 specs sheet and there is one chart I find missing. It is the relationship between output voltage with input voltage, specifically when input voltage is lower that 5 volts, the "nominal" output voltage.

I think it would be desirable to have 0 volts on output until 7805 can provide 5 volts, but I think this is not the real situation.

So, my question is : what voltage will 7805 deliver as output if the input voltage is 3 volts ?

Any pointer is welcome. Thanks.

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    \$\begingroup\$ It's not specified for this region of operation. It could output anything, and will depend on the exact implementation; it varies between manufacturers. \$\endgroup\$
    – Hearth
    Nov 7, 2018 at 17:43
  • \$\begingroup\$ I don't think there is an answer what the voltage will be or how to calculate it. \$\endgroup\$
    – Justme
    Nov 7, 2018 at 17:46
  • \$\begingroup\$ If you want a regulator that shuts down when the input is too low, then you don't want a 7805. \$\endgroup\$
    – JRE
    Nov 7, 2018 at 18:36

3 Answers 3

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You can see typical behavior shown in the datasheet:

enter image description here

Around 3V in it will start to turn on so the output voltage will be fairly unpredictable with that input voltage (probably will vary from unit-to-unit and with temperature).

It sounds like you want some kind of supervisor functionality. There are many such chips and there are a few regulators that provide an output "power good" signal when the output voltage is fairly close to regulation.

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    \$\begingroup\$ Thank, Sphero - this is the datasheet I was missing \$\endgroup\$
    – Sebastia.Net
    Nov 8, 2018 at 18:28
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Figure 2 in the datasheet seems to imply that the output is shut down when the voltage difference between the input and output is less than ~1.5V.

It is best to run a test.

enter image description here

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  • \$\begingroup\$ They don't shutdown. The good old 7805 didn't waste transistors dealing with a situation you weren't supposed to put it in. \$\endgroup\$
    – JRE
    Nov 7, 2018 at 18:19
  • \$\begingroup\$ @JRE Can you explain what the figure from the datasheet means, because it certainly suggests to me that the output current goes to zero if the input voltage is too low. Is it the word "shutdown" you don't like? \$\endgroup\$
    – Elliot Alderson
    Nov 7, 2018 at 19:12
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    \$\begingroup\$ That chart doesn't show the output voltage. It shows the difference between input and output voltage. Even if the voltage is too low, current will still flow. Imagine you have a resistor from output to ground. There will be a voltage difference across the regulator - current will flow, even if the input is below the usual drop out voltage. The 7805 simply has no provision to shut off the output. \$\endgroup\$
    – JRE
    Nov 7, 2018 at 19:25
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below about 7V in the 7805 behaves kind of like a resistor. they don't guarantee an output voltage

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    \$\begingroup\$ Based oh Spehro's chart that's not correct. \$\endgroup\$
    – Drew
    Nov 7, 2018 at 20:53
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    \$\begingroup\$ @Drew: Actually, the chart shows exactly that until the input voltage gets too low. The output voltage linearly follows the input down to 3V. The chart also shows different lines for different loads - so it behaves like a resistor from about 7V to around 3V. \$\endgroup\$
    – JRE
    Nov 7, 2018 at 21:19
  • \$\begingroup\$ I wouldn't exactly call that resistive, but I'll remove my downvote. Edit: sorry I can't \$\endgroup\$
    – Drew
    Nov 8, 2018 at 3:42
  • \$\begingroup\$ The graph shows a fixed voltage drop (which is like a diode) and current dependance (which is like a resistor) \$\endgroup\$
    – Jasen Слава Україні
    Nov 8, 2018 at 7:42

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