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How can I find the base current of this circuit.
I was trying to find \$\beta \$ from \$(\beta+1)I_B = I_E \$. where \$V_{BE}=0.7V, I_E=0.5mA\$

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ Solve for the node voltages. you have more than enough information to do that. \$\endgroup\$ – Dave Jan 1 '16 at 20:09
  • \$\begingroup\$ This is a weird circuit. Where did you get it? I don't think Vbe will be 0.7V. I suspect you made an error when you copied this. Maybe you meant to use an NPN transistor? \$\endgroup\$ – mkeith Jan 2 '16 at 2:38
  • \$\begingroup\$ @mkeith The solution goes like this, which I can't understand. Applying KVL on emitter-base junction \$4=0.7+I_B R_B+I_C R_C -5\$ ,then assuming \$I_C = I_B\$ the equation is solved . Maybe there is a printing mistake in the question. don't know, will wait some more and delete the question. \$\endgroup\$ – user2332665 Jan 2 '16 at 4:35
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    \$\begingroup\$ The picture is a little blurry. Is the arrow pointing away from the base, toward the emitter? Or is it pointing toward the base? It is possible to use a transistor with collector and emitter swapped. \$\endgroup\$ – mkeith Jan 2 '16 at 7:59
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    \$\begingroup\$ Just to comment the circuits principle: If the circuit serves as a theoretical exercise - it is OK. However, in practice, it is not a good solution (Art of Electronics: Don`t use it.). The reason is that emitter negative feedback does not work efficiently if the base voltage is floating. \$\endgroup\$ – LvW Jan 2 '16 at 9:39
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That PNP is biased off. I suspect you really intended to put an NPN there, and exchange the E & C ?

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  • \$\begingroup\$ Edited likewise. !Imgur \$\endgroup\$ – user2332665 Jan 2 '16 at 7:56
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You know the emitter current. That current through the emitter resistor gives the emitter voltage. Adding the base-emitter voltage to the known emitter voltage gives the base voltage. The difference between the collector voltage (known) and the base voltage (just calculated), divided by the base bias resistor, gives you the base current, and you can now calculate beta.

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Edit, now that you've exchanged the PNP for an NPN. Ve = (Ic * R2)-5V. Vb = Vbe + Ve. Ib = (Vc - Vb)/ R3

IF your drawing is correct, then you're thinking about this too much. Vbe = 0.7V. There's a 100,000 ohm resister that the voltage appears across. I= V/R. That's the base current.

BTW, the current should be flowing out of the base in this case.

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  • \$\begingroup\$ If the drawing is correct, base current = collector current = emitter current = 0, and Vbe = 0. And Ve = 5V. So something doesn't add up. \$\endgroup\$ – mkeith Jan 2 '16 at 6:34

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