I need a circuit to turn on a LED with 0.2 volts. I want to do it with a transistor as a switch. What kind of transistor will turn on at this voltage?
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6\$\begingroup\$ Pretty much none of them. I would recommend amplifying the signal before trying to use it as a switch, perhaps with a comparator. \$\endgroup\$– HearthCommented Apr 8, 2019 at 12:04
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7\$\begingroup\$ Instead of asking about a transistor which needs to turn on with only 0.2 V (there aren't any but please prove me wrong!) and assume that you can do this with only one transistor (you can't but I'd like to be proven wrong again), describe what your requirements are and ask what solutions exist. A LED will not light up with 0.2 V so there must be some other voltage present. Why 0.2 V? Do research on similar setups (switching LEDs) and learn how that's done. \$\endgroup\$– BimpelrekkieCommented Apr 8, 2019 at 12:11
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3\$\begingroup\$ Highly related if not duplicate: electronics.stackexchange.com/questions/129183/… \$\endgroup\$– winnyCommented Apr 8, 2019 at 13:00
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1\$\begingroup\$ 0.2V isn't enough to run an LED; if you have a rail that'll drive an LED then you can use a comparator. \$\endgroup\$– TimWescottCommented Apr 8, 2019 at 15:54
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1\$\begingroup\$ What's the LED and what's the current and voltage it requires? What power supply rails do you have? \$\endgroup\$– jonkCommented Apr 8, 2019 at 16:21
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4 Answers
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No single transistor can do what you want. Instead, you need to use multiple transistors. A simple voltage comparator will do what you want:
simulate this circuit – Schematic created using CircuitLab
R2 and R3 set the reference voltage for the base of Q2. As long as the voltage at the base of Q1 is less than Vref, all of the current through R1 will flow through Q2. As soon as the input voltage rises above 0.2V, Q2 will cut off and all of the current will flow through Q1 and the LED instead.
Note that since the emitters of the transistors need to be at a negative voltage, this circuit requires a negative supply. An alternative would be to use an IC comparator — you can find single-supply units that can work with input voltages close to ground.
answered Apr 8, 2019 at 12:23
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1\$\begingroup\$ You can approach the lower rail by using PNPs 'upside down' and a single supply. \$\endgroup\$– Russell McMahon ♦Commented Apr 8, 2019 at 17:34
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\$\begingroup\$ @RussellMcMahon: But then there's no overhead left for the load (LED), at least, not without adding additional transistors. At that point, just get the IC comparator, which is essentially the same thing. \$\endgroup\$ Commented Apr 8, 2019 at 17:36
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\$\begingroup\$ This circuit really needs some positive feedback to achieve a clean switch. \$\endgroup\$– EinarACommented Apr 9, 2019 at 7:17
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Whilst I'm inclined to agree with @Bimpelrekkie that this is probably an X-Y problem, for those who might actually need such a device there are some options. Most MOSFETs are sold as enhancement or depletion mode (less commonly) devices, however it's possible to tune the threshold voltage to approximately zero +/- tens of mV.
For example, the Advanced Linear Devices ALD110800A and similar devices in that series have a nominal zero gate threshold (1uA Ids). The subthreshold behavior is fairly well documented as well.
For your actual situation, use CrossRoads circuit with just about any single supply op-amp or single-supply comparator. LM358 or LM393 will work fine from a 5V supply.
answered Apr 8, 2019 at 12:28
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simulate this circuit – Schematic created using CircuitLab
You could also use an Op Amp, or a Comparator. Set one input to 0.2V, when the other input exceeds that, the output will switch.
Here's one example
https://datasheets.maximintegrated.com/en/ds/MAX9107-MAX9109.pdf
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A well chosen germanium transistor can do this. I have some that switch at .15V. Finding one might be a problem ; I am not offering any of mine, but they still exist. If you need accurateness use one of the other answers, The simplest solution is to prebias a transistor so the base receives enough voltage to turn on when the input is .2V. I have used this trick in non-critical situations.
