Circuit of BC817 as a switching transistor

[sankavit]

I have a circuit wherein the supply is from a regulated 5V.

circuit is as in the attachment.

the mystery for me is how there is a voltage of 2.23V at the transistor collector, when the transistor is turned off? How would the circuit close? what am i missing out in my selection?

BMP image replaced with gif format which is smaller and can be viewed directly in browser, [alexan_e]

 

[keith1200rs]

I would guess you are pulling the transistor collector down towards ground with the internal resistance of the multimeter. Try connecting your multimeter between the collector and +5V and see what the voltage is.

Keith.

 

[WimRFP]

How much current do you supply to the base? Measure the voltage drop across the resistor that sets the base current. Maybe the base current is too low, you have just 47 Ohms in series with the LEDs.

You are sure you didn't change the emitter with the collector (as is has low hfe in that case).

 

[sankavit]

he Base has a 2.2K resistor. So, the working current is cut to it.
the collector was isolated from the circuit and found to have no voltage.
similarly, the output from the LED side was measured and found to have the same 2.4V.
I think it is somesort of a LED drop voltage that is visible at the other end. but am not sure of the logic.
can you throw some light on it?

 

[WimRFP]

When the LED's should light, how much voltage is across the 2.2k base resistance, then you know the base current? The base-emitter voltage should be around 0.7V.

It is normal that when there is no supply to the collector, the voltage across collector and emitter is small (<0.5V).

 

[keith1200rs]

I assume you didn't try my suggestion?

Keith.

 

[vimalkhanna]

There are two things which will clear the doubts.
Put a series 1N4148 with the 2.2K at base .No current shall leak thru .
Put a diode at collector of BC817 with cathode to collector .Does the 2.4VDC rise ?.
You shall have an answer to where your currents are flowing from the LEDs ..
Most probably your multimreter is drawing a few microamps which
causes the leakage current on the amber LEDs

 

[FvM]

I think it is somesort of a LED drop voltage

Yes, you have proven that the voltage drop of the LEDs is 1.3V each at a current of 0.24 uA (assuming 10 Mohm multimeter input resistance). This sounds reasonable, although you won't find a specification for ultralow currents in a LED datasheet. Forward voltage at mA operation currents will be around 1.5 to 2V.

As Keith suggested, you'll see no voltage drop, if you measure the respective node voltage referred to +5V.

 

[emresel]

Hi;
I had seen similiar phenomenon with an IR led and a mosfet (instead of bjt), i was measuring via scope (not multimeter) at that time, i couldn't do any comment except that the leakeage current through the transistor which are around a few hundreds nAmps causing it...
Is it possible also because of leakeage curent? What you think friends?

Dear sankavit please do the Keith's suggested measurements and tell us. It will clear my mind also, i am wondering the result:roll:
Thanks

 

[FvM]

The effect hasn't to do with leakage as sankavit already reported, it's simple diode (LED) voltage drop. You don't need a measurement to clarify, just sketch the circuit and think.

 

[ALERTLINKS]

The effect hasn't to do with leakage as sankavit already reported, it's simple diode (LED) voltage drop. You don't need a measurement to clarify, just sketch the circuit and think.

I agree.This can be confirmed by reducing supply voltage to 3V.

Yes, you have proven that the voltage drop of the LEDs is 1.3V each at a current of 0.24 uA

I'm sure now it 'll be near zero. If it had been due to leakage it would be arround 1.3V.