Bad voltage regulator?

I think I have a bad 7805 (TO-220) voltage regulator, but maybe I'm just not using it properly. If I apply 5V to the input, I get a surge of current (warming up my battery and cutting its voltage from the normal 6V to about 1.5). That's with the output unconnected. With no power applied, I find there's a 4K resistance between ground and output (is that normal??)

Reply to
strnbrg59
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How hot does the 7805 get?

Reply to
ian field

"ian field" wrote in news:40TIj.34443$ snipped-for-privacy@newsfe3-win.ntli.net:

Probably. I get 4.33 K on a known good 7805.

Reply to
Lostgallifreyan

Hot enough to give off an odor of burning plastic, and way too hot to touch.

Reply to
strnbrg59

Look at the output with a scope to see if its oscillating.

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Reply to
Michael A. Terrell

Those type of regulators back feed if the out side of the regulator is higher in voltage than the input side in which cause, your post regulator will be acting like a short and your battery will be trying to over come the 5 volts etc..

Normally this problem shows up in cases where you have a charged capacitor on the output side and then there is no power to the input in which case, causes the regulator to short the output to the input and thus drain the cap via the reg. This can some times damage the regulator and placing a diode on the input and output normally fixes it. In your case, you would need one in series.

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Reply to
Jamie

You need to provide the input of the 7805 with *more* than 5 volts. The regulator needs some "headroom". Give it 8 volts or more on the input. Also, you may have cooked it, so get a new one and give it at least 8 volts input. You are using it to charge a battery, so you need to protect it against current going the wrong way with a diode.

Ed

Reply to
ehsjr

Make that "If you are using it to charge a battery, you need to protect it against current going the wrong way with a diode."

Ed

Reply to
ehsjr

Reply to
Wicaksono

WHY are you applying 5V to the INPUT of a 7805 ? It won't work properly ! It's a linear regulator . That means the input needs to be (several volts) HIGHER than the rated output voltage.

A 7805 WILL NOT charge a 6V battery !!!!! You need around 6.8V to charge a 6V (lead acid) battery.

Uh ?

Measuring the 'resistance' of ICs is pointless.

Learn about how to use these things before trying next time.

Graham

Reply to
Eeyore

You've probably broken it.

Graham

Reply to
Eeyore

You arrogant prick!. Why do you think he came here in the first place? to ask why it was happening! "To learn", you prick!

But then again, should I expect any better from you?

Your manners leave a lot to be desired, you British honker!

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Reply to
Jamie

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