I'm playing around with a BD179 transistor, trying to get my LED bulb to light up but without luck. My mistake is probably something very simple but I'm learning so bear with me.
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2\$\begingroup\$ Please label the connections for the LM2596 \$\endgroup\$– DKNguyenAug 2, 2019 at 20:52
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2\$\begingroup\$ Your schematic is extremely difficult to understand as you're not following standard circuit conventions as: ground at the bottom, supply/battery at left or right etc. Have a look at how similar schematics are drawn and follow that. I agree with DKNguyen that you need to add the names of the pins of the DCDC module. Adding a text like you just did is not sufficient. We're here to help but we do expect that you make things as clear as possible. \$\endgroup\$– BimpelrekkieAug 2, 2019 at 20:54
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1\$\begingroup\$ Also your LED is in series with the emitter of the NPN, that means the NPN will not work as a switch and you will not get the voltage you (probably) want across the LED. Also why step down to 3.4 V, that makes no sense. Yes you're learning, good. Here's a tip: go search how others switch on/off LEDs using a transistor. Then do the same. Trying to "figure it out on your own" is a recipe for disaster because: it will extremely likely not work and/or might also damage components. \$\endgroup\$– BimpelrekkieAug 2, 2019 at 20:58
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1\$\begingroup\$ @Bimpelrekkie There, I hope it's better now. Essentially I'm trying to switch a higher load from a microcontroller, I'm simulating it here with this ~3V. It's an LED Bulb, not a simple LED. \$\endgroup\$– php_nub_qqAug 2, 2019 at 21:02
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2\$\begingroup\$ @Bimpelrekkie problem is this is just a test setup. I'll be switching the transistor from a microcontroller which has only 3.3V, so I've either got the wrong transistor or I wired it incorrectly. I suppose it's likely the latter. \$\endgroup\$– php_nub_qqAug 2, 2019 at 21:10
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2 Answers
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There are two potential problems in your circuit.
1. The 2kOhm base resistor is too high.
By applying 3.4V through a 2K resistor, and accounting for the base-emitter voltage drop of the BJT (in the datasheet) you get a base current of:
\$ I_{collector} = \frac{V_{supply} - V_{be}} {R_{base}} = \frac{3.4V - 1.3V} {2kOhm} = 1.05mA\$
In the datasheet, the BJT's DC current gain ranges anywhere from 15 to 160 which means your collector current will be anywhere between 15 to 160 times the base current which is 16mA to 168mA.
But your bulb is a 12V, 6W bulb which means it runs at: \$ I = \frac{P}{V} = \frac{6W}{12V} = 500mA\$
2. The LED should be on the collector side of the transistor.
Put simply, the current flowing between the base and emitter terminals of an BJT determine how much it turns on by. The BJT can ONLY see the voltage difference between its terminals. It does not and cannot know about voltages anywhere else.
Your power supply is applying to the base resistor relative to ground. But the voltage and current parameters that the BJT actually cares about are those between the base and emitter terminals. If your source pin is not connected to ground then what you BJT cares about is not the same as what the supply is providing. Things get distorted
As the transistor turns on and conducts current through your bulb, the voltage across the bulb rises pushing the source terminal voltage away from ground which reduces the base-emitter voltage difference (and the voltage across the base resistor which reduces the base current). This acts as negative feedback and fights the transistor turning on more.
This negative feedback has its uses, but not when using the transistor as a plain old switch. It's mostly for amplifiers and analog circuits.
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\$\begingroup\$ So if I understand this right, I need to move the bulb before the transistor on the 12V line and I need to increase the current on the base. \$\endgroup\$ Aug 2, 2019 at 21:14
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\$\begingroup\$ I'm sorry if it's a dumb question but why does it make a difference whether the load is before or after the transistor on the collector-emitter path? \$\endgroup\$ Aug 2, 2019 at 21:16
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3\$\begingroup\$ @php_nub_qq Read the part about the transistor only being able to care about the base and source. Read it carefully. It's the most important thing in there. \$\endgroup\$– DKNguyenAug 2, 2019 at 21:18
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\$\begingroup\$ Huge gratitude for the time you took to explain this to me like to a complete moron. Little feedback - I just switched the bulb to the collector side and it immediately lit up. I then switched to a 370 ohm resistor since that's the lowest I have laying around and it seems to be as bright as if I connect it to the supply directly. \$\endgroup\$ Aug 2, 2019 at 21:41
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As designed, your circuit is an emitter follower. You're applying 3.4V to the base of an NPN, and taking power off of the emitter. The transistor will try to hold the emitter voltage at roughly \$V_{be} - 0.7\mathrm{V}\$, or about 2.7V. That's not nearly enough for your LED.
You want something like the following.
You need to choose a transistor that can pass 500mA (because it's a 6W, 12V "bulb" -- that works out to half an amp). Then you need to choose a resistor that'll cause the transistor to turn on hard. If you used a 2N4401, you'd need about 50mA into the base -- that would require a resistance of \$R_1 = \mathrm{(3.4V - 0.7V) / (50mA)} = 54\Omega\$.
However, you have a problem, because you mentioned that you're driving this from a microcontroller, and there aren't any microcontrollers out there that can drive \$50\mathrm{mA}\$. So you either need to use a Darlington (which has a higher collector-emitter drop than a plain BJT), or you need to search around for a "super beta" transistor (they're out there, and they're nice -- look for high \$H_{FE}\$ in saturation), or you need to use a logic-level FET that's rated for a gate voltage of 3.3V
simulate this circuit – Schematic created using CircuitLab
answered Aug 2, 2019 at 21:17