7805 Voltage regulator - have I broken it?

[dugeen]

Hello all

I put together a primitive power supply with a 12V DC output straight from the rectifier, and a 5V DC output which was achieved using a 7805.

It worked fine until I shorted the 5V output. Now there's only 1.4V between the output and ground pins on the 7805, where there was 5V before.

Is that how voltage regulators fail, when they fail?

 

[flat5]

What voltage do you have on the input to the regulator?
Perhaps the problem is there.

The reg. should tolerate a short.
It does not like a backwards voltage on it's input or output.

 

[Hero999]

A short circuit shouldn't kill it - the LM7805 is short circuit proof. There must be something else wrong.

 

[dugeen]

Thanks both - the input voltage is about 12V, just like it was before. The symptoms are the same whether or not there's any load on the other output from my supply, the 12V one.

flat5, the input pin is definitely at +12V compared to the ground pin now. But suppose a negative voltage had been applied to the regulator before, would that cause it to behave like it is doing?

 

[flat5]

You are not allowed to send reverse current to either the input or the output of the 78 series of regulators.
It will cause damage to them.
If it may happen, use a series or shunt diode to protect the device.

I'm not an expert but I believe this is correct.

 

[Boncuk]

The chip was probably "Made in China" and didn't have short circuit protection.

Boncuk

 

[audioguru]

The chip was probably "Made in China" and didn't have short circuit protection.

Boncuk

Then it is a counterfeit one.

 

[killivolt]

Something like Vanbodia.

Then it is a counterfeit one.

It probably came from Canada.

Somewhere in that province of Chi-wan-eez.


Ha ha.


kv

 

[audioguru]

It probably came from Canada.

Somewhere in that province of Chi-wan-eez.


Ha ha.

I moved away from Vancouver, Canada when it became Hongcouver.
I don't think voltage regulator ICs are made there.

 

[98formulaLS1]

I'll bet you probably shorted some other IC on your board, and that is why you're measuring 1.4V. I would take the 7805 off of the board, place it on another breadboard and retest.

 

[dugeen]

I replaced the regulator with a new one and the situation is just the same.

I've attached a circuit diagram to show what I'm trying to do. The 12V terminals have an external switching module connected to them, and it switches the 5V supply, which goes to a motor. The idea is to replace the two separate 5V and 12V supplies I'm having to use at the moment.

 

[flat5]

You say:

"I put together a primitive power supply with a 12V DC output straight from the rectifier, and a 5V DC output which was achieved using a 7805."

So you have AC going to this bridge?
Your bridge is wired wrong.
You need to reverse two of the diodes.

 

[audioguru]

18VAC makes a peak voltage of 25.4V which is reduced to 23.4V by the bridge rectifier.
The 5.6uF capacitor has a value that is way too low for a filter so the "12V" is pulses of 0V then +25.4V then 0V then +25.4V over and over.

The motor starts on one +25.4V pulse then its inductance causes a very high voltage as the supply voltage drops to 0V and the high voltage spikes destroy the 5V regulator.

 

[Roff]

Your bridge is wired wrong.

Really?

 

[98formulaLS1]

Follow these steps!!!

1. Take everything off the board except your 7805.
2. Measure the output voltage from the 7805.
3. If you measure 5v, then you have miswired your circuit or some other component is blown. 7805's are a dime a dozen, you buy some more and retest.

 

[kchriste]

Change the 5.6uF cap to something a lot bigger like a 2200uF or 4700uF. Observe correct polarity and voltage ratings. As noted by AudioGuru, 18VAC will become 24V when rectified and properly filtered. Change the transformer to a 9Vac or 10Vac one so that you get apx 12Vdc unregulated.
Also, add a 0.1uF cap on the 5V side of the 7805.

 

[dugeen]

Will try with a bigger filter capacitor, and an extra one on the other side of the regulator.

I specified the transformer wrongly in my diagram, it's 18V in that it's +9/0/-9 centre tapped, and I'm just using the +9 and 0 terminals. So I should have drawn it as a 9V one. There is definitely 12V DC at most between the two sides of the rectifier.

The rectifier is correctly shown, though, isn't it? (Circuit Simulator doesn't have a symbol for the IC rectifier that I'm actually using. It's got +, - and ~ ~ clearly marked, enabling me to connect it properly).

 

[Mike_2545]

Rectifier looks OK to me

 

[kchriste]

it's 18V in that it's +9/0/-9 centre tapped, and I'm just using the +9 and 0 terminals.

While that will work, the way to get the most power out of your transformer, and still have 12Vdc output, is to connect it like this:
**broken link removed**

The rectifier is correctly shown, though, isn't it?

Yes.
Flat5 was in error on that point.

 

[flat5]

Doh! I was seeing the 12v as the source instead of the 18v.
Very sorry for the dumb mistake.