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I bought a dtc144 transistor because it was written in its datasheet that it could do a "Switching circuit" ( as it is written in the Applications section in the datasheet). I though it meant that I could turn on/off a sub-circuit in my PCB design, but when I tested, it didn't worked as I imagined.

Did I misunderstand what switching circuit means? How can I use this transistor to turn on/off a sub circuit?

The datasheet (first page) has the following image:

dtc144_inner_circuit

(if you can not see the image, check the datasheet. I am having problems to upload the image)

I thought that as a transistor, if I put a GND in the "in" pin, the transistor would behave as high impedance between the "out" and the "gnd" pins, and low inpedance if I put VCC instead of GND in the "in" pin. But it seems that I am wrong.

I'd appreciate if you could help me understand. Thanks.

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  • \$\begingroup\$ I see two problems with the second diagram. Not that they have anything to do with the question... \$\endgroup\$ – Ignacio Vazquez-Abrams May 19 '14 at 20:26
  • \$\begingroup\$ OK we hear what you expected, now what did you get? \$\endgroup\$ – Andy aka May 19 '14 at 20:32
  • \$\begingroup\$ I connected 3v3 in "out" pin, in "gnd" pin I connected my circuit and I put 3v3 in "in" pin to close the switch and then I have 1v5 in "gnd" pin instead of 3v3 \$\endgroup\$ – koike Jun 3 '14 at 19:31
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I thought that as a transistor, if I put a GND in the "in" pin, the transistor would behave as high impedance between the "out" and the "gnd" pins, and low inpedance if I put VCC instead of GND in the "in" pin. But it seems that I am wrong.

No, you are correct, except that you left something out. Your description of how the transistor acts as a switch is right on, as long as the "gnd" pin is tied to GND. If you are trying to use the OUT/GND path as a short / open element in a signal path, it won't work. It will only provide a short/open path to ground.

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  • \$\begingroup\$ Hi, thanks for your reply. I was trying to put the transistor switch between the power source (+) and the VCC of my circuit and I was having a big drop of voltage (by half). Now I changed the transistor to rely between the gnd of my circuit and the power source (-) and it works (I have a mini drop of 100 mV). I didn't really understand why, but this is another question. \$\endgroup\$ – koike Jun 3 '14 at 20:35
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You are asking a specific question about a specific circuit, but you're not ready for this yet. You need to understand how a transistor works, at least its first pass approximation. The type of transistor you show is a NPN, which is a type of bipolar junction transistor (BJT).

Surely there are many basic introductions to NPN transistors out there, so I'm not going to go into any detail. Basically, you put a little current into the base and out the emitter, and the transistor can support a lot of current into the collector and out the emitter.

Your top circuit is actually correct in that when IN is grounded, the transistor is off, which essentially open-circuits OUT. When IN is raised to high enough voltage, then current will flow into the base, and the transistor will pull down on OUT. There are of course a lot of details. Without knowing exactly what you did, we can't tell why you didn't observe the behavior as described above.

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Referring to the data sheet the transistor works as a switch if OUT has a higher voltage potential than IN. IN must be 3 volts (value taken from table) or higher than GND for the transistor to conduct from OUT to GND (switch closed). If IN is close to 0 volts compared to GND the transistor is not conducting OUT to GND (switch open).

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