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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.

KidWithComputer
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    Can you share a schematic? The grounds for the 3.3 and 12V should be connected together. – vir Dec 29 '20 at 16:48
  • 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. – Hearth Dec 29 '20 at 16:49
  • I am very new to this how do I make a schematic? – KidWithComputer Dec 29 '20 at 17:05
  • @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. – Hearth Dec 29 '20 at 17:07
  • Ok thank you I drew my best schematic I hope it helps – KidWithComputer Dec 29 '20 at 17:12
  • [This](https://electronics.stackexchange.com/questions/151326/bjt-driving-an-led-above-or-below/151328) SE EE Q&A shows you what to do plus provides several good explanations. The left hand circuit is most flexible and easiest to implement. Read all answers and adjust component values to suit. The LED is on when the input is high. The right hand circuit is valid for some applications but not for yours. || For a REALLY good discussion of the issues see [this](https://electronics.stackexchange.com/questions/57845/why-would-one-drive-leds-with-a-common-emitter/57849#57849) SE EE Q&A – Russell McMahon Dec 29 '20 at 17:23
  • For a 12V LED supply and 12V LED strip you do not need R5. For a strip that draws substantial current (say more than 50-100 mA) using an N Channel MOSFET with on resistance (Rdson) under 100 milliOhms (preferably much less) will provide a much better overall experience. – Russell McMahon Dec 29 '20 at 17:24

2 Answers2

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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.

Hearth
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  • 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? – KidWithComputer Dec 29 '20 at 17:26
  • You need a bigger transistor that has lower resistance and more power dissipation to run cool as a switch for >0.5A – Tony Stewart EE75 Dec 29 '20 at 17:33
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    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). – TimWescott Dec 29 '20 at 17:40
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    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. – Hearth Dec 29 '20 at 17:40
  • 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. – TimWescott Dec 29 '20 at 17:42
  • "High side switching" means that the switching device (transistor, relay, switch, etc) is connected between the voltage supply and the load. "Low side switching" means that the switching device is connected between the load and ground. Because the current through the transistor (between the collector and emitter) depends on the current into the base, which depends on the voltage between the base and emitter, an NPN transistor isn't ideal for high side switching since the emitter voltage will change drastically when the load is on, sometimes enough to turn the transistor back off. – vir Dec 29 '20 at 17:42
  • @KidWithComputer The transistor needs to have the current limited because the base-emitter junction is a diode, so its current increases rapidly with voltage above about 0.7 volts. And that current means increasing power dissipation, which means an unhappy transistor. – Hearth Dec 29 '20 at 17:43
  • It's a pretty good rule of thumb with silicon devices that if you can put your finger on it and hold it there, it'll do OK even at elevated ambient temperatures. That's about 50C. For power devices that will be running in a room-temperature environment, being able to touch them and hold for a second before you have to snatch your hand away that's probably OK -- but it'd be too hot for industrial, military, or automotive temperature ranges. Don't burn yourself in the testing (ask how I know!) – TimWescott Dec 29 '20 at 17:44
  • If the object of the exercise is just to switch on/off the 12V rated LED strip then put the 12V across the series combination of the LED strip and a switch. –  Dec 29 '20 at 17:51
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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

TimWescott
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  • @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. – TimWescott Dec 29 '20 at 17:37
  • 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? – KidWithComputer Dec 29 '20 at 18:08
  • @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. – Hearth Dec 29 '20 at 18:11
  • 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? – KidWithComputer Dec 29 '20 at 18:16
  • @KidWithComputer Yes, they're called logic-level FETs. Please use the @ feature so that people get notified when you ask them questions. – Hearth Dec 29 '20 at 18:42
  • Search for "logic-level FET", and be careful about reading the datasheets. Logic-level FETs are specifically made to work at a Vgs of 5V or 3.3V -- but to know *which one*, and *what the specs are at that voltage* you need to read the datasheet carefully. It's not uncommon to see "10A Logic-level Switch!!" and then find out that it'll only switch 10A with 12V on the gate and a heat sink the size of a brick. – TimWescott Dec 29 '20 at 18:43
  • @Hearth thanks for the tip – KidWithComputer Dec 29 '20 at 18:46
  • @TimWescott I've been searching for a long time looking for the right mosfet and I think I finally found a good one: https://datasheet.octopart.com/IRF3205PBF-Infineon-datasheet-163446.pdf however the last graph concerns me: stating that at 1.5A the case temperature would be around 175 degrees Celsius. Is that right? I feel I must be reading the document wrong – KidWithComputer Dec 29 '20 at 20:54
  • I'm not sure what graph you're talking about -- figure 9 is the maximum *allowable* drain current vs. temperature -- not the temperature you'll **get** for a given drain current. – TimWescott Dec 29 '20 at 21:04