PNP transistor help

Thread Starter

vilaemail

Joined Jul 4, 2014
2
I am creating a project and as a start i began building prototype parts of circuit on protoboard. However it didn't work and as I went on and investigated i came up with this circuit that is not behaving as I expected (see attachment).

I am asking you to confirm how this circuit should behave? I am expecting the transistor to be "closed" and voltage on collector to be 0V. As I understand in order for current to flow through transistor there must be difference in voltage between emitter and base - e.g. a current must flow through emitter base in order for current to flow through emitter and collector.

However I observe 3V on collector.

P.S. My first post here, looking forward for your replies and hoping this is not too stupid question.

Thanks,
Filip.
 

Attachments

GS3

Joined Sep 21, 2007
408
Either the transistor is bad or something is not connected right. Double check everything.
 

GopherT

Joined Nov 23, 2012
7,983
Please confirm HOW you are measuring voltage.

If measuring from battery positive to collector, then 3 volts. (Wrong way to measure). If transistor is off (not letting current flow), this method is like measuring the voltage of a battery or the potential across an open switch.

If measuring with black probe on battery negative to collector, you should have zero volts. This is the correct method. Essentially, you need to measure voltage drop across a resistor to determine if current is flowing. Then you can use ohms law to calculate current flow.

If you did measure the correct way (across the resistor in your picture), then you saw 3 volts across an 800 ohm resistor and can calculate 3.75 mA of current. Then you can assume that something is wrong with the pin assignments ( check datasheet*) or that the transistor is damaged. You should, in most cases, have a resistor to limit current flow into the transistor base.

* note that BC type transistors have pin assignments flipped vs 2N-type transistors. Check the datasheet.
 
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Thread Starter

vilaemail

Joined Jul 4, 2014
2
Thank you all for your replies.

I omitted the base resistor in this circuit because I was testing my own sanity here.

Voltage is measured between ground and collector which is effectively voltage on 800 Ohm resistor. And the reading is 3V.

I double checked everything and I found out i miss connected the transistor. When i was connecting I looked up the emitter pin and assumed that the middle pin is base. I was wrong, the middle pin is collector. So effectively in my circuit the current was actually flowing from emitter to base through resistor to the ground.

Thanks for all your help. I indeed feel stupid :)
 

Attachments

GopherT

Joined Nov 23, 2012
7,983
Thank you all for your replies.

I omitted the base resistor in this circuit because I was testing my own sanity here.

Voltage is measured between ground and collector which is effectively voltage on 800 Ohm resistor. And the reading is 3V.

I double checked everything and I found out i miss connected the transistor. When i was connecting I looked up the emitter pin and assumed that the middle pin is base. I was wrong, the middle pin is collector. So effectively in my circuit the current was actually flowing from emitter to base through resistor to the ground.

Thanks for all your help. I indeed feel stupid :)
We learn most from our mistakes. Make them early, make them on simple projects. Then you can eventually build more complex projects with fewer mistakes.
 

ScottWang

Joined Aug 23, 2012
6,819
Make mistake in the experiment is quite normal, many mistake can be speculate from the theory, as your circuit I think there is no way to have any current flows through from (+) to (-) of battery, and I used the same value of resistor to do the testing, yes, that was match the theory, so what I thought about the 0.7V that was easily to think about the diode(Vbe=Vbc=0.7V) of bjt, it must be you connected the wrong pins of bjt, the current flows through the Vbe(0.7V) to resistor and to the negative of battery, V_800 = 3.7V-0.7V = 3.0V.
 
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