Drop voltage on transistor base

[Bigz1]

Hi

I am using a NPN transistor as a switch. I am taking the base from a LED. My problem is their is a constant voltage of 1.87v when the LED is off(1.97v when it is on). I was thinking of a zener diode of 1.9v to drop the voltage but am unable to source one. Is their an alternative way? My knowledge of electronics is very limited and appreciate any help.

 

[Bluejets]

Show the circuit.

 

[Bigz1]

My problem is the base always has voltage.

 

[(*steve*)]

Show us the circuit (the whole circuit)

 

[Bluejets]

Bit shown is all backwards...

 

[dorke]

The obvious problem is the 12v supply is connected in the opposite direction.
Change it and recheck ,like so:

 

[Bigz1]

Sorry my mistake drawing (12 v supply is right way in circuit). 12v LED is constantly on. I can't draw all the circuit. Its a UPS that's not accessible. Its remotely turned on but I want an indicator to show its status. The D1 Led is on the UPS(shows when battery is on).

 

[dorke]

Try measuring the on/off voltages on the junction of R1 and D1.
What led is 12V?
Another problem is ground reference between the 12V battery and the UPS.
Please show it on your drawing.

 

[(*steve*)]

If you completely remove the connection to the base of the transistor, does the LED stay illuminated?

If so, your first problem is one of:

1. The transistor is connected incorrectly
2. The transistor is PNP
3. The transistor is damaged
4. It is not a transistor

Do you have a part number for Q1? Values for your resistors might be nice to see also.

With the circuit you've shown the base of the transistor is effectively not connected because the rest of the circuit does nothing. However I'm assuming there is more to the circuit that you've shown. However we have no idea what it is, or even what reference you used in reading your voltages. As @dorke has said, there needs to be a ground reference, and we also need to know what the floating connections are connected to. If drawn conventionally, that diode on the far left is reverse biased, but I can't be sure that you're following any conventions in drawing your diagram.

As you can see, we're really lost without lots more information.

 

[BobK]

Are you saying when D1 is on the voltage is 1.97 and when it is off it is 1.87? I find that hard to believe.

Bob

 

[CDRIVE]

If your merely desirous of switching Q1on and off why are there two paths biasing the b,e junction?

Are these components that you're adding or are they OEM?

Chris