# NPN transistor as switch for LED problem

Discussion in 'LEDs and Optoelectronics' started by David Lacroix, Sep 22, 2016.

1. ### David Lacroix

13
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Sep 15, 2016
Hi,

I used an NPN transistor as a switch to light up a small LED. The circuit works great but when I cut the base current for the transistor the LED stays lit just a tiny bit. I'm using a 220 ohm resistor for the LED and a 100 ohm for the base of the transistor.

What causes this?

Thx

2. ### davennModerator

13,813
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Sep 5, 2009
and what voltage supply ?

you may also need a ~ 1k resistor from the base to 0V to ensure the transistor is properly cut off

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

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Jan 21, 2010
And/or maybe a moderate value resistor across the LED so the leakage current does not flow through the LED. The resistance must be such that the leakage current causes a voltage drop significantly less than Vf for the LED

davenn likes this.
4. ### CDRIVEHauling 10' pipe on a Trek Shift3

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May 8, 2012
Steve, I've never had to shunt a LED. Have you experienced that? Are you referring to situations with very high VCC or when using Germanium Transistors?

Chris

Arouse1973 likes this.
5. ### David Lacroix

13
2
Sep 15, 2016
12V from 8 AA batt.

Circuit is 12V to 220ohm to LED to collector then emitter to ground and then 12V to 100ohm to base.

6. ### David Lacroix

13
2
Sep 15, 2016
What is a moderate value for you?

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

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Jan 21, 2010
It depends on the leakage current. If it is 0.1 mA and you don't want more than 1V across the LED, then 10k would be the value I'd start with.

If you are pulling the base low with a similar value resistor then a leakage of 0.1mA is pretty huge.

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

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Jan 21, 2010
Oh, and a base resistor of 100 ohms! If taken to 12V then that's over 100mA of base current. I hope your transistor can handle that. Its also more than that used by the LED!

HellasTechn likes this.
9. ### hevans1944Hop - AC8NS

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Jun 21, 2012
That unspecified NPN transistor is probably toast by now with a 100 Ω resistor connected from base to +12 V... unless it is a 2N3055 power transistor, which would be a bit of overkill for lighting up your typical LED. IIRC, the 2N3055 needs a huge amount of base current to turn on properly and the emitter-collector leakage current approaches one milliampere when the base-emitter voltage is zero, Whatever, I would raise the base resistor value to 10 kΩ, shunt the base-emitter junction with 1 kΩ, and replace the transistor.

HellasTechn likes this.
10. ### HellasTechn

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Apr 14, 2013
Changing the LED resistor will also affect the brightness of the led when fully on.

11. ### CDRIVEHauling 10' pipe on a Trek Shift3

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May 8, 2012
Try this. R1 can be as shown and will provide about 18mA of LED current. Raising it to 1K will reduce LED current to about 11mA. This assumes a RED LED.

Chris

12. ### David Lacroix

13
2
Sep 15, 2016
I guess I've probably toasted those transistors. I've just started playing around with electronics, I'm at step 1 haha...damn

ps: when a transistor is toasted, could it conduct current straight from collector to emitter without anything from base, because that would answer some questions about my last experiment...

13. ### David Lacroix

13
2
Sep 15, 2016
I'm going to try this next time and see what happens.

14. ### CDRIVEHauling 10' pipe on a Trek Shift3

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May 8, 2012
Yes, a shorted Collector - Emitter junction will conduct all the time. Are you sure you have the pinouts correct?

Chris

15. ### hevans1944Hop - AC8NS

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Jun 21, 2012
Absolutely! A transistor is a piece of doped silicon with precisely implanted ions that define where the emitter- base, and base-collector junctions are. If you "toast it" all that stuff melts and fuses together into a single conductive chunk of silicon. Usually. You can also vaporize the internal wire bond connections from the package terminals to the silicon and create an open circuit, often at the same time letting the "magic smoke" out when a plastic encapsulation opens up to atmosphere.

16. ### hevans1944Hop - AC8NS

4,608
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Jun 21, 2012
Sorry, Chris... there is no collector-emitter junction. There is the emitter-base junction and base-collector junction between the emitter and the collector. Melt those junctions to create an emitter-collector short. You can tell you have done this "right" by measuring the resistance from emitter to collector with an ohmmeter. Should get the same (low) reading with either polarity of the two probes. Don't ask me how I know this.

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

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Jan 21, 2010
I believe it is possible to produce doped channels across the base. These produce a resistive connection between emitter and collector without significantly affecting the base-emitter or base-collector junction.

Apparently this results in increased leakage, but with transistor action still taking place.

It's not a typical failure mode, especially after smoke-releasing incidents.

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18. ### CDRIVEHauling 10' pipe on a Trek Shift3

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May 8, 2012
A mis-speak. I should have said "a short between the emitter and collector pins". I've seen it happen many times when max Vce is exceeded or from HV kickback. I didn't include any statements regarding the base because the transistor would be useless anyway.

Chris