# Simple NPN transistor circuit

Discussion in 'General Electronics Discussion' started by electric, Dec 19, 2013.

1. ### electric

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Dec 19, 2013
I'm new to electronics and circuitry, so bear with me.

Attached is a file of a circuit I have constructed. Basically, what I would like is for the transistor to act like a relay, so that when there is current flowing from the timer (i.e. when the timer hits the zero mark), it forces the transistor to complete the number 2 current loop. What I don't want is for current from the higher voltage battery to flow through the timer. I essentially would like the two loops to be completely separate, so that they are independent of each other, except for one turning the other one on. Do I need a diode?

I guess my confusion stems from my lack of complete understanding of transistors and also how current flows in multiple loops. If you have any light to shed on either of those topics, please feel free to share!

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

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Jan 21, 2010
Almost exactly right except the timer goes between the emitter and base.

Note that you would be wise to have some series resistance (say 100 ohms).

To turn the transistor on the base has to be more positive than the emitter and you'll need about 1/100th of the load current flowing through the base (this differs greatly depending on the gain of the transistor)

edit: I realised looked at the circuit diagram and read the load as a resistor. It is in fact a LED, so some comments further down indicate how to handle that (and then the OP says it's really a light bulb!)

Last edited: Dec 19, 2013
3. ### iimagine

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Oct 23, 2013
Here is one of the most efficient way of driving an LED

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4. ### electric

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Dec 19, 2013
Ok so now I've changed the circuit to the configuration you said and have posted a new picture. Problem is, even when the loop with the timer (loop 1) is not even closed, the bulb lights up (dimly). If I then close loop 1 without even including the timer and its associated battery (by simply bridging the gap with a wire), then the bulb gets extremely bright. How can this be explained?

Also, for the base to be more positive than the emitter, wouldn't current have to flow from the base to the emitter (I thought the current in the low voltage "controlling" loop (loop 1 in this case) had to flow from the emitter to the base)?

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5. ### Harald KappModeratorModerator

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Nov 17, 2011
To turn off the transistor completely put a resistor (e.g. 10kOhm) from the base to the emitter.

The current in loop 2 needs to go in the other direction as it flows into the base and out of the emitter. The timer circuit needs to output a current into the base of the transistor. Depending on how the timer's output is constructed, you may need an additional resistor in series with the base to limit the base current.

Last not least you need to limit the current through the LED by adding a series resistor in the LED current loop 1.

Read more in our tutorial section "got a question about driving LEDs?"

6. ### electric

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Dec 19, 2013
Before I go any further, I should mention that I'm actually working with a light bulb and not an LED. I thought they were somewhat interchangeable before I read that article but that is obviously not the case.

To Harald Kapp-

"To turn off the transistor completely put a resistor (e.g. 10kOhm) from the base to the emitter."

-Then why does the bulb still glow when I put an infinite resistor across it (i.e. when I open the loop)?

"The current in loop 2 needs to go in the other direction as it flows into the base and out of the emitter. The timer circuit needs to output a current into the base of the transistor."

-I thought it was this was for a PNP transistor? According to this article (https://www.electronicspoint.com/introduction-bipolar-junction-transistors-t222878.html), the controlling circuit is supposed to flow from the emitter to the base and the controlled circuit is supposed to flow from the emitter to the collector for an NPN transistor.

"Depending on how the timer's output is constructed, you may need an additional resistor in series with the base to limit the base current. "

-What would happen if I didn't have that resistor? Would it just burn up the transistor because too much current would be flowing through it?

"Last not least you need to limit the current through the LED by adding a series resistor in the LED current loop 1."

-Assuming I was actually using an LED, wouldn't I want to put a resistor in loop 2 rather than loop 1 to limit the current flowing through it?

7. ### BobK

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Jan 5, 2010
Conventional current flows from positive to negative. Some people use electron current, which flows the opposite direction because electrons are negative. The arrow on the bipolar transistor symbol indicates the direction of flow of conventional current.

Bob

8. ### electric

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Dec 19, 2013
So that article was referring to electron current?

9. ### mursal

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Dec 13, 2013
The arrow in the transistor is facing towards the emitter, so its a NPN. To get it to work as a switch (what you want) you will allow 0.6V to drop across the base emitter junction and the rest of the timer output voltage must be dropped across the base resistor. Conventional current flows into the base and out the emitter. This will determine the size of the base resistor. (Vout-0.6V)/Ibase = Rbase
Where Ibase is the base current
This small base current (collector current/10) will allow the larger collector, emitter current to flow. If you dont tie the base to ground )off) using the resistor already explained above, the base will "float" and turn the collector, emitter circuit on, turn the load on a little bit (dim).

10. ### electric

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Dec 19, 2013
Ok I think I understand it a bit better now.

One more question though - will the current from the higher voltage source in the number 2 loop also flow through the timer loop once the current from the timer loop has "opened" the transistor? If so, how do I prevent that?

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

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Jan 21, 2010
No, the base-collector junction is reverse biased and current will not flow

(it can if your transistor can't withstand the voltage of the battery, but this is one of the specifications of the transistor and you choose one with ample margin. At the low voltages you're talking about it's not likely to become an issue)