I'm working on a project for which I need to use a 7805 voltage regulator. I feed 9V to its input pin and power a 5V board with the output. However, when the board is connected to a computer, the board outputs 5V through the same cable: even if I turned off the 9V power supply before connecting the board to a computer, would the voltage regulator be hurt by the 5V input to its output pin? Can I fix this with a diode?
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3 Answers
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The TI data sheet says this: -
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\$\begingroup\$ Thank you! I'll have that in mind, but I was talking more about a circuit like the one in here: imgur.com/rvjnI24. Would that diode prevent the 7805 from damage once the board began to output 5V? \$\endgroup\$ Commented Oct 25, 2020 at 10:03
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1\$\begingroup\$ That won't be an effective circuit at producing 5 volts on the target circuit because of the 0.5 volts to 0.7 volts dropped across the diode when forward conducting. Not recommended if you target board needs a stable 5 volt supply from the 7805 @EightBlueLemons. With my circuit, you can leave turn off the 9 volts without worry but, of course if you leave it connected then there may be a "fight" in that the 7805 might want to regulate at 5.05 volts and the external (board fed) 5 volts might be 4.95 volts. \$\endgroup\$– Andy akaCommented Oct 25, 2020 at 10:15
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\$\begingroup\$ Oh don't worry, when I connect the board to the computer I'll disconnect the 9V power supply, I just wanted to know if once i disconnected the 9V and the board started to output 5V it would harm the 7805. \$\endgroup\$ Commented Oct 25, 2020 at 10:22
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1\$\begingroup\$ @EightBlueLemons The circuit in my answer is the way to avoid that happening. \$\endgroup\$– Andy akaCommented Oct 25, 2020 at 10:30
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\$\begingroup\$ I'll heed your advice, thank you! \$\endgroup\$ Commented Oct 25, 2020 at 23:52
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for which I need to use a 7805 voltage regulator
And right there is your mistake. No you don't. Unless you're doing a historical project on antique voltage regulators, what you actually need is a 5V regulator. You don't need a specific one, you need one which meets your technical requirements.. The 7805 is not it.
If you're connecting to a computer, chances are you're using digital logic. In that case a switch-mode regulator is usually the best choice. But let's assume there's a reason you want a linear regulator. Regulators such as the LM2940 have protection built in against the various ways you can damage it and are a far better choice.
Basically, instead of kludging around with a near-obsolete device, design your circuit with something that isn't 40 years old. It's not 1980 any more. We have alternatives.
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\$\begingroup\$ So, if I were to use the LM2940, i wouldn't have to worry about the 5V backfeeding that i'm talking about? And if it still were necessary to use the 7805,would the 5V damage it significantly? \$\endgroup\$ Commented Oct 25, 2020 at 23:50
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\$\begingroup\$ @EightBlueLemons The L2940 is protected against reverse driving that way. As for the 7805, yes you can damage it like that \$\endgroup\$– GrahamCommented Oct 26, 2020 at 0:07
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1\$\begingroup\$ But the L2940 is more expensive. The 7805 is old but not that obsolete, It does its job even today. \$\endgroup\$– FredledCommented Oct 26, 2020 at 0:26
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\$\begingroup\$ It's not really a matter of price, I just want to know if the 7805 will be damaged once i plug the board to a USB and how i could avoid that. I've already added a diode to the circuit, but I want to be sure. \$\endgroup\$ Commented Oct 26, 2020 at 7:07
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1\$\begingroup\$ The caveat with the LM2940 being, not just price, but also more quiescent (off-state) current leakage (10-15mA vs 6), and less stability in the output voltage (a power supply rejection ratio (PSRR) of 64-72dB vs 80). These differences may be small, and thus unlikely to matter in many applications, but it's not impossible for these factors to be important in some cases, so they're still worth mentioning. No device still sold today is better than another in every way, there are always some trade-offs. \$\endgroup\$– Dee S.Commented Oct 26, 2020 at 10:01
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If you don't short or heavily load the regulator input to ground the backfeeding is unlikely to damage the 7805.
If the +5 isn't too critical you could do something like this to prevent backfeeding:
simulate this circuit – Schematic created using CircuitLab
R1 establishes a minimum output current of ~5mA to match Iq.
answered Oct 25, 2020 at 17:35
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1\$\begingroup\$ A single diode at the input only may be better. 78xx will not be damaged if there is no load at the input. \$\endgroup\$– fraxinusCommented Oct 25, 2020 at 19:07
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\$\begingroup\$ So with your circuit, output pin ic will be at about 5v+(0.5-0.7v). And circuit output(after diode) will remain to 5v, right? \$\endgroup\$– HidaCommented Apr 19, 2022 at 20:35