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I'm using BJT (NPN) that is supplied with 7V DC signal at the collecter, 5V DC signal at the base and need a 12V output at emitter. Right now i'm getting 4.35V at emitter. How can i implement it?enter image description here

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    \$\begingroup\$ You're going to have a problem getting 12V at the emitter with only 7V at the collector and 5V at the base. You're observing a diode drop from base to emitter, which is expected for a BJT in the forward active region. What are you trying to accomplish? \$\endgroup\$ – Null Dec 11 '14 at 17:34
  • \$\begingroup\$ Are you expecting the BJT to amplify the 7V into 12V? \$\endgroup\$ – Dan Laks Dec 11 '14 at 17:35
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    \$\begingroup\$ Brush up on your BJT theory. What you're trying to do won't work with what you have. \$\endgroup\$ – Adam Lawrence Dec 11 '14 at 17:35
  • \$\begingroup\$ No, that voltage drop is not at all unusual. If you want to boost the input voltage, you'll need a lot more than just a transistor. \$\endgroup\$ – Dave Tweed Dec 11 '14 at 17:37
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    \$\begingroup\$ junaid, were you thinking that the BJT would combine collector voltage of 7V + base voltage of 5V to give you 12V at the emitter? That is not how BJT's work. If that is not what you were thinking, then please explain what you were thinking would happen with this circuit. \$\endgroup\$ – mkeith Dec 11 '14 at 17:37
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BJTs don't work how you apparently think they work. Remember that the B-E junction looks like a diode to the outside circuit. You are seeing a 650 mV drop (5.00 V - 4.35 V) from B to E, which is squarely in the expected range.

You are using the transistor in emitter follower configuration. You will therefore get significant current gain, but no voltage gain. Actually the voltage gain will be a little less than 1.

The answer is to use a different topology. A emitter follower simply isn't going to do what you want.

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