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Now I'm trying to design a circuit which have 2 voltage level (12V and 3.3V). What I want to create is simple switch using 3.3V to turn off 12V. I know there are thousand question about transistor on this site, but after spending some time scrolling, I haven't found what I wanted.

So far, this is what I have designed and simulate it on Proteus.

schematic

simulate this circuit – Schematic created using CircuitLab

The question is, do I have a wrong circuit? or perhaps the Proteus simulator is not correct? Thanks in advance. Please ask if this question is not clear. Thanks :D

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  • \$\begingroup\$ Ask yourself, what is C1 doing? What is Q1 doing? What happens to Q2 when Q1 is switched on and off? Do the connections on Q2 make sense? \$\endgroup\$ – tangrs Jan 3 '17 at 9:17
  • \$\begingroup\$ Your problem is C1. That will bock DC. Short that one out and flip R2 so it connects from base of Q2 to 12 V as a pull-up. You need a base resisotr on Q1 as well. \$\endgroup\$ – winny Jan 3 '17 at 9:17
  • \$\begingroup\$ As I understand, you just want to control the 12V lamp with a 3.3V signal, right? Then, you just need a NPN and a resistor. Or you may use a mosfet (something like that). But how did you come up with the above circuit? Where does it come from? What was your logic when designing it? \$\endgroup\$ – dim lost faith in SE Jan 3 '17 at 9:26
  • \$\begingroup\$ I've deleted the C1, shorted Q2-Emitter to ground, switch R2 into pull up resistor from Q2-base to 12V, and add 330 ohm res from SW1 to Q1-base. But still its not working \$\endgroup\$ – dpw Jan 3 '17 at 9:26
  • \$\begingroup\$ Did the bulb just dissappear from your circuit? As @dim said it, if you are fine with floating ground when the bulb is turned off, you can do it with just one NPN and a resistor. \$\endgroup\$ – winny Jan 3 '17 at 9:28
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On your circuit, there are several strange placement of components. R1 and C1 is clearly wrong, also as mentioned in comment by @winny , you should have pull-up resistor to Q2 base. There are so many combination of transistor that can be used as a switch, you can try this

schematic

simulate this circuit – Schematic created using CircuitLab

Correct me if I'm wrong, but maybe here's some explanation:
R1 and R2 is used to limit base current, R3 as pull-up resistor, and C1 to filter the output (optional)

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schematic

simulate this circuit – Schematic created using CircuitLab

This should do the trick. I used the lightbulb as your load/output and the switch(SW1) is now normal-closed Q1 can also be a N-Channel Mosfet.

This should work like this: SW1 is closed so Q1 is active and bring the gate of M1 to ground. If SW1 is open Q1 is closed and the gate of M1 is on 12V

This is a low-cost-solution Suggestions to improve this circuit are welcomed

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  • \$\begingroup\$ What is C1 for? \$\endgroup\$ – Daniel Tork Jan 3 '17 at 10:18
  • \$\begingroup\$ C1 is for debouncing the switch. \$\endgroup\$ – Ben Jan 3 '17 at 10:24
  • \$\begingroup\$ Ok,I understand,well done. \$\endgroup\$ – Daniel Tork Jan 3 '17 at 10:26
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There are a great many things wrong with this circuit. The most significant being that the load is connected across V2. This means the circuit is not switching power to the load. If Q2 ever did conduct it would be a short circuit (or very nearly so) across V2. Again, the "switch" transistor is in parallel with V2. I recommend you start over and use a Darlington pair for the switch, the basic concept is explained here:

https://en.wikipedia.org/wiki/Darlington_transistor

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  • \$\begingroup\$ I'd like to post a regular comment for you (and Ben), but I don't have the rep, so I'll do it here: What is the purpose of SW1? In the text of the message, it says you want to control the output with the 3.3 volt signal. Is that correct? \$\endgroup\$ – Michael Gorsich Jan 3 '17 at 9:53

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