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schematic

simulate this circuit – Schematic created using CircuitLab

To start I am very new to working with electronics, I've read a little theory and thats about it. My goal was to make a circuit with a BC337 transistor in it acting like a switch. My goal was that a logic cell, which outputs 3.3V would be able to control the flow of power for a 12V LED chain of lights.

I hoped to use the 3.3V to control the base of the transistor which would enable the 12V to flow(or not) from the collector to the emitter so I wired the positive end of the 12v to the collector, the 12v ground to the emitter, and was about to connect the positive wire of the 3.3V to the base when I realized I had not figured out a place for the negative end of the 3.3V.

So my question is this: Where do I place the negative end of the 3.3V base on my transistor, so that the transistor can still be controlled by the 3.3V and that the voltage of the emitter will still be 12V(or at least very close)?

I tested several configurations of where the negative end could go but all of them either didn't work or resulted in reduced emitter voltage: 9V or less. I do not know if it is relevant but my 12v comes from a 1.5A wall voltage converter and my 3.3V comes from a coin battery. Any help would be appreciated thank you.

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    \$\begingroup\$ Can you share a schematic? The grounds for the 3.3 and 12V should be connected together. \$\endgroup\$
    – vir
    Dec 29 '20 at 16:48
  • \$\begingroup\$ It sounds like you're using an NPN BJT here, in which case your emitter voltage should probably be 0 V, not 12 V. Your collector is what should be close to 12 V (and even then, it should only be near 12 V when off; when on it'll drop to about 0.1 volts, which is what you want). Show us a schematic of what you're doing. \$\endgroup\$
    – Hearth
    Dec 29 '20 at 16:49
  • \$\begingroup\$ I am very new to this how do I make a schematic? \$\endgroup\$ Dec 29 '20 at 17:05
  • \$\begingroup\$ @KidWithComputer There's a built-in schematic editor on the site, but even one made in LTspice or drawn by hand and photographed is better than nothing. \$\endgroup\$
    – Hearth
    Dec 29 '20 at 17:07
  • \$\begingroup\$ Ok thank you I drew my best schematic I hope it helps \$\endgroup\$ Dec 29 '20 at 17:12
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Here is your circuit, redrawn a bit, with the LED string in the correct polarity and showing how you'd hook up a 3.3V independent source so that you can correctly power the base of the transistor.

Note that circuit ground is arbitrary, at least assuming that the wall-wart you're powering this off of is isolated. You can choose any node on this circuit, and it'll work (well, not-work, that's coming) the same.

I'm going to assume that the strip needs less than the 800mA that a BC337 can switch. If this isn't true, then the best you can hope for is something that kinda-sorta works, possibly while burning up the transistor.

But you have worse problems. To really switch on hard, the transistor needs to saturate. The rule of thumb for saturating a transistor of this age (it's from the 1980's) is that you need about 1/10th the collector current going into the base. So you need to put 80mA into the base of the thing. A watch battery simply isn't going to supply that much current -- and if it could, it wouldn't do so for very long.

The easiest answer to the problem of base current is to use an N-channel FET that's designed for logic-level operation and can carry the full current of your LED strip. You need to find a part that's rated for your desired LED current, at 3.3V gate to source. You could pretty much drop that part into this circuit, replace R1 with a wire, and it would work. In fact, it should work for a very long time, because FET's don't pull any gate current.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ @Hearth's schematic is a more practical solution to just making the @$&% thing work -- I wanted to stick as closely to your schematic to explain what's going on with your proposal. \$\endgroup\$
    – TimWescott
    Dec 29 '20 at 17:37
  • \$\begingroup\$ Thank you Tim I think I may try your suggestion of just replacing the transistor with an with an N type mosfet. Aside from making sure it is rated for 1.5 amps should I make sure about anything else when buying on of these. Is the 3.3V gate to source important? \$\endgroup\$ Dec 29 '20 at 18:08
  • \$\begingroup\$ @KidWithComputer Yes, you need to make sure that it has a low RDSon at 3.3 volts. (this is not the same as having a threshold voltage of 3.3 volts, your FET will need a threshold voltage well below 3.3 V.) Many MOSFETs aren't fully on until the gate-source voltage reaches about 8-10 volts. It doesn't need to be rated for 1.5 A though; it just needs to be rated for what your LED strip actually draws. \$\endgroup\$
    – Hearth
    Dec 29 '20 at 18:11
  • \$\begingroup\$ 1.5 amps is what the Led strip actually draws, or at least this is what I would guess, given that the wall plug came with the led strip in question. Are there any that don't require such high voltage for the gate source, closer to maybe 3 or 4.5V? \$\endgroup\$ Dec 29 '20 at 18:16
  • \$\begingroup\$ @KidWithComputer Yes, they're called logic-level FETs. Please use the @ feature so that people get notified when you ask them questions. \$\endgroup\$
    – Hearth
    Dec 29 '20 at 18:42
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Now that you've added your schematic, I can immediately see a number of problems with it.

Firstly, you must have a current-limiting resistor on the base of your transistor, or your transistor will go up in smoke.

Secondly, there's no reason here to use an npn transistor for high-side switching; you should put it on the low side.

Thirdly, you've got your LED connected backwards. It's not going to light, and might be fried if you hook it up to power.

I'm assuming your LED strip has current limiting resistors built in, so leaving that off the LED is presumably fine.

Here's a corrected schematic for it:

schematic

simulate this circuit – Schematic created using CircuitLab

I didn't bother to work out what value resistor you should use, because I don't know how much current your LED strip needs. Be aware that the BC377 can't handle more than 0.8 amps though.

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  • \$\begingroup\$ I appreciate your quick answer thank you. Could you explain what is meant by high/low side switching and why my current must be limited? Also I previously was able to turn the transistor on and off by plugging in and out the 12v to the base, which has 1.5 amps, this worked fine for a long time with no transistor deterioration, even though like you said the BC337 is rated for 0.8 amps is there an explanation for this? \$\endgroup\$ Dec 29 '20 at 17:26
  • \$\begingroup\$ You need a bigger transistor that has lower resistance and more power dissipation to run cool as a switch for >0.5A \$\endgroup\$ Dec 29 '20 at 17:33
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    \$\begingroup\$ If you connected the transistor collector and base to +12V, and ran the lights off of the emitter, then you would have made an emitter follower, AKA common-collector amplifier (you can search those terms). It would have put about 11.3V on the emitter, and naturally limited the base current. Hearth's circuit is a common emitter amplifier (search that one, too); it doesn't naturally limit the base current (the base-emitter voltage "wants" to limit at 0.7V, and the transistor will fry if you try to force it). \$\endgroup\$
    – TimWescott
    Dec 29 '20 at 17:40
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    \$\begingroup\$ High side switching is what you have, where the transistor is connected to +vcc and the load. Low side switching is when it's between the load and ground. \$\endgroup\$
    – Hearth
    Dec 29 '20 at 17:40
  • \$\begingroup\$ When you were running 1.5A through the transistor, did you check its temperature? Normally I recommend feeling for heat, but in this case I'd suggest you start by feeling the air close by, or tapping the thing the way you'd check a pan suspected of being hot. You're probably running the thing at a way-elevated temperature -- semiconductors can stand that for hours, but not thousands of hours. \$\endgroup\$
    – TimWescott
    Dec 29 '20 at 17:42

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