I measured Ic variation with Ib when Vce = 15v kept constant for BJT (NPN) in Common Emitter config and also BJT Common Emitter config with series 47 Ohm resistor connected.

From the experiment I observed less temperature change when I blew on the 2nd circuit Transistor than on the 1st. Can someone please explain to me why the second circuit with the same transistor with series 47 ohm resistor with the emitter is more stable than the 1st circuit without the resistor? enter image description here enter image description here

I have now added the schematics for the circuit with emitter directly grounded and for the circuit with the 47 ohm emitter resistance The emmitter resistance provides negative resistance; How does help circuit to be less sensitive to temperature changes?

  • 1
    \$\begingroup\$ You should show schematics for both situations .. We can guess but that's not as much fun as it might sound \$\endgroup\$ – Spehro Pefhany Feb 11 '16 at 6:19
  • \$\begingroup\$ could you also add a "Gain vs Ic" graph for the "without resistor" case. Looking at the Ic vs Ib graph, the gain looks fairly constant but its a bit hard to tell. \$\endgroup\$ – user1582568 Feb 11 '16 at 16:52
  • \$\begingroup\$ @spehro-pefhany: I have added the schematics for both. \$\endgroup\$ – Saavin Feb 12 '16 at 22:18

If I understood your question without schematics, I guess the answer is:

With the resistor connected to the emmiter, when IB rises, IE rises due to the amplification. If IE rises, the voltage drop over RE rises too. Since VBE=VB-VE, VBE will lower and IB too. It works like a negative feedback.

Best regards

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.