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I presume that I can remove the capacitors since the circuit uses DC, but then I don't understand why I can apply a voltage divider in the 30K and 20K resistors since the base is connected between them.

enter image description here

My professor changes the circuit into this: enter image description here

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    \$\begingroup\$ Are they in parallel? Looks more like series to me. \$\endgroup\$ – JRE Jun 26 at 14:08
  • \$\begingroup\$ @JRE I know, it was a mistake. But what my professor does to simply the circuit is, he applies a voltage divider at the 20k resistor which gives -4.8V and then takes the parallel of the 30k with the 20k \$\endgroup\$ – Pedro Jun 26 at 14:11
  • \$\begingroup\$ Your latest comment describes finding the thevenin equivalent of the divider \$\endgroup\$ – sstobbe Jun 26 at 14:17
  • \$\begingroup\$ @sstobbe yes, but I don't understand why I can do that \$\endgroup\$ – Pedro Jun 26 at 14:22
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I don't understand why I can apply a voltage divider in the 30K and 20K resistors since the base is connected between them.

The emitter resistor of 1 kohm projects an impedance onto the base that is beta times higher. So, if beta is 100 and the emitter resistor is 1 kohm, the base will look like a loading impedance of 100 kohm. This loading resistance is seen in parallel with the 20 kohm resistor thus, the effective resistance at the base to ground is 20 kohm || 100 kohm = 16.67 kohm. You can use this to determine a more precise value of base voltage to ground.

I estimate 12 volts * 16.67/(30 + 16.67) = 4.29 volts.

What your prof does is find the equivalent series resistance of the 20 kohm and 30 kohm resistors and to do that you parallel them (12 kohm = 20 k || 30 k). He then calculates the open circuit voltage (base disconnected) as being 12 * 20/(20 + 30) = 4.8 volts.

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  • \$\begingroup\$ It is that process I don't understand, why can I "disconnect" the base and how in turn take the parallel of those two resistors? \$\endgroup\$ – Pedro Jun 26 at 14:29
  • \$\begingroup\$ Thevenin's Theorem states that it is possible to simplify any linear circuit, no matter how complex, to an equivalent circuit with just a single voltage source and series resistance connected to a load. - quote from the web @pedro. So 12 volts, the 30 k and the 20 k can be reduced to a new value of voltage source (4.8 volts) in series with an impedance of 20 k || 30 k. You probably need to go google Thevenin's theorem. \$\endgroup\$ – Andy aka Jun 26 at 14:33
  • \$\begingroup\$ I know how to do the Thevenin equivalent, I've just never done it in the context of a BJT, that is why I'm having trouble visualizing how it works. I only did it for simple RC circuits \$\endgroup\$ – Pedro Jun 26 at 14:40
  • \$\begingroup\$ First thing to do is recognize that 12 volts in series with 30 kohm is exactly the same as a current source of 0.4 mA in parallel with 30 kohm. The 2nd stage then recognizes that the 30 kohm is now in parallel with the 20 kohm giving a resistance of 12 kohm. The third stage is calculating that the modified open circuit terminal voltage is. 0.4 mA into 12 kohm = 4.8 volts. Stage 4 is converting the current source and 12 kohm resistor back into a voltage source of 4.8 volts in series with 12 kohm. \$\endgroup\$ – Andy aka Jun 26 at 14:40

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