# Spurious voltage readings for open transistor

Discussion in 'General Electronics Discussion' started by seanspotatobusiness, Nov 20, 2017.

193
4
Sep 11, 2012
I have a circuit I made for an experiment with four LEDs in series behind a transistor with a floating base so the circuit is open and LEDs are unlit. When I measured the voltage across the LEDs I got a stable value of about 2.4 V but when I measure the voltage across the collector and emittter I get 2.1 V most of the time with a brief increase to about 12.4 V every second; I don't have access to an oscilloscope right now so I can't tell exactly what is happening. When I measure the voltage across the LEDs and the transistor, I get a constant 8 V which is as expected (I have my bench power supply set to provide 8 V). What is happening?

The reason for me doing this is because I want to use a transistor to turn on and off 50-60V LED filaments (which are a bunch of about 20 LEDs in series) and I was hoping I could use the BC337 transistor to do this. The maximum rated voltage across the collector and emitter of the BC337 is 45 V but I was thinking that even when the transistor is off, maybe the LED filament would drop some of the voltage so the transistor would never experience too much but this weird behaviour with brief pulses of at least 12.4 V makes me question that.

The floating base was at -0.6 V so maybe grounding it will change the behaviour?

2. ### Harald KappModeratorModerator

11,524
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Nov 17, 2011
• You have the transistor backwards. The emitter of an NPN transistor needs to go to the lower potential, 0V in youre circuit. Swap emitter and collector.
• How's that possible? Your circuit shows an operating voltage of 8 V. Is your meter operating correctly?

• To ensure the transistor is on or off, have the base controlled (via a series resistor) by either a positive voltage (on) or 0 V (off). A floating base can (and wil) give unstable results.
• To ensure the LEDs wil not burn out you need to limit the current through the LEDs.

See our ressource section for more detailed info:
Cheers,
Harald

By the way: There's no such thing as a filament in an LED. An LED is a solid state (semiconductor) light emitter.

3. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,497
2,838
Jan 21, 2010
You are measuring voltages where there is a high impedance in series with some non-linear resistances. The voltage you're reading from a moderate impedance multimeter is pretty much undefined. The act of measuring the voltage changes the circuit state.

4. ### Harald KappModeratorModerator

11,524
2,655
Nov 17, 2011
One possibility for a reading higher than 8 V is when the LEDs are lit by an externallight source. Then an LED acts as a photodiode, generating a few µA. As the transistor is open/non-conducting, the only way for the current to flow is back through the multimeter. Your multimeter has 10 MΩ input impedance when measuring voltages > 4 V. Even a very small photocurrent (e.g. 1 µA) generates a measurable voltage drop:
1 µA * 10 MΩ = 10 V

Check for this cause by shielding your LED string from any ambient light.

5. ### BobK

7,682
1,688
Jan 5, 2010
Turn the transistor around, place a 10K resistor from base to emitter. This will eliminate the floating base.

And don't use a transistor rated for 45V with a 60V supply, you are just asking for trouble, get one rated for 80V or more.

Bob