BJT PNP went ZAP

[e44-72]

Hello

I understand bipolar transistors (NPN and PNP) are reliant on current. With PNP I belive current flows out or away from the base and the rule is if the base is positively charged then its as if the switch is open and if grounded then it is as if its closed.

So I put a 12v led after the transistor from the collector and my supply voltage of 9v in the emitter of the transistor. I grounded the base of the transistor and when i put my meter on the base to test the volts there the transistor gave out a spark and got fried.

Why is it this happened, diagram too explain below

I a'm an ametur at electronics and learning lots from this forum (my first time playing with transistors)

Thank you for your replies.

 

[atferrari]

No surprise! No resistors to limitate currents!

Look for them in a basic explanation of how transistors are used.

 

[e44-72]

Thank you for your reply

What I don't understand is when the base was attached to ground without a current limiting resistor it was ok, it only burnt out when I put the voltmeter on it?

 

[Reloadron]

It may have "Looked" OK but I guarantee that was only going to be for a very short period of time. You need current limiting for the LED as well as the transistors base. What was the transistor part number?

Ron

 

[jpanhalt]

Did the voltmeter give a reading?

If so, it was not wired as you have shown.

John

 

[audioguru]

your voltmeter is measuring no voltage across a piece of wire.
No resistor limits the base current so the maximum available current from the battery blew up the base of the transistor since the base-emitter is actually a junction diode with a maximum allowed base voltage of about 0.7V, not 9V.

No resistor limits the LED current so it would have blown up if the transistor was connected properly.

This is how it should be connected:

 

[e44-72]

Thank you for your posts

The part number for the transistor is as follows layed out as follows:

K 6F
A733
P

I thought a resistor at the base may have been neaded as it was getting a bit warm, Thank you for explain why this was. The LED however already has a integrated resistor so no resistor needed there and why I put the voltmeter as I said i did i can't even understand myself, it was a late night

 

[RayMillTN1]

No worries on blowing parts up - When I was in basic electronics class, we were all gun shy from all the electrolytic caps exploding from people plugging them in backwards... If you ain't making mistakes you are not learning.

 

[audioguru]

Your "A733" transistor is an Oriental 2SA733 that is also available in India and Russia.
It is very old.
I found an old datasheet from NEC but it is written mostly in Japanese.
Without the series base resistor in your circuit its base current is as much current as the little 9V battery can supply, maybe 2A (!) if it is a brand new alkaline battery. The base current needs to be only 0.002A in your circuit so the resistor can be 3.9k ohms.

In your circuit the transistor will be warm because it dissipates about 2A x 1.0V= 2W! Its maximum allowed continuous heating is only 0.675W.
If the battery is a brand new alkaline battery then it dissipates about 8W x 2A= 16W! Only 1W makes it very hot.
The transistor and/or battery is burnt out.

 

[e44-72]

Thank you audioguru for your reply.

The fact its very old does not surprise me as it is from a basic 200 in 1 electronics set I was givern years ago which also has very old TTL series IC's. Lots of stuff on it has burn't out, ever since I first used it most the leds were already dead.
Can I also ask how you got 3.9K. If volts = amps x ohms then the answer I get is 9 / 0.002 = 4500 ohms or 4.5k

Thanks for the information, I know were I am giong wrong now.

 

[audioguru]

No.
If the battery is brand new at 9.0V then the base-emitter of the transistor has 0.7V and the base resistor has 9.0V - 0.7V= 8.3V.
Then 8.3V/2mA= 4.15k ohms.

But 4.15k ohms is not a standard value and nobody makes it.
3.9k ohms is the next lowest standard value.