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I want to use Thevenin's theorem with the two resistors connected to the base.

How can I combine those two resistors with Thevenin? Do you have any other better idea to combine them? (circled with red in the picture) enter image description here

(PS: I don't ask the solution of the question. I'm only asking individually the calculation of those two resistors.)

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  • \$\begingroup\$ Why don't you start by redrawing just the circuit that you want to convert to its Thevenin equivalent. Include \$V_1\$ and the -1 V source. Clearly mark the two nodes where you want to find the equivalence. \$\endgroup\$ Jun 5 at 13:41
  • \$\begingroup\$ I actually drawed like that initially but I could not understand to combine both of them. I already know if both of the resistor nodes are connected to the ground. However in this specific example, one of the node is connected with -1V which confused my mind. \$\endgroup\$
    – emilia
    Jun 5 at 13:42
  • \$\begingroup\$ @ElliotAlderson , what could it be your suggestion please for that -1V node? \$\endgroup\$
    – emilia
    Jun 5 at 13:43
  • \$\begingroup\$ The -1V is measured with respect to ground, so you just need an ideal voltage source between that node and ground. Orient the source and set its value such that you get -1V at the bottom of R2. \$\endgroup\$ Jun 5 at 13:45
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    \$\begingroup\$ Rth is R1||R2 thus if R1 = 30k and R2 is 60k --->Rth = 60k/3 = 20k \$\endgroup\$
    – G36
    Jun 5 at 14:13
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What you need to do at the beginning is to disconnect the input voltage divider from the circuit and find the open voltage (\$V_{TH}\$):

schematic

simulate this circuit – Schematic created using CircuitLab

Now try to find \$V_{TH}\$.

Can you do it?

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  • \$\begingroup\$ @36 (Vi-(-1v))*(60k / 30k + 60k) = ... Is that form correct please? \$\endgroup\$
    – emilia
    Jun 5 at 14:18
  • \$\begingroup\$ No this is not correct. OK, can you find the current value that is flowing in this circuit? \$\endgroup\$
    – G36
    Jun 5 at 14:23
  • \$\begingroup\$ Sure, i can apply voltage divider rule for that: (R1/R1+R2)*(Vi+1V) \$\endgroup\$
    – emilia
    Jun 5 at 14:24
  • \$\begingroup\$ It seems, I really memorized quite everything while i was learning thevenin in the last term. I'm really sorry for that \$\endgroup\$
    – emilia
    Jun 5 at 14:26
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    \$\begingroup\$ No, back to the basics. The current is the circuit is equal to \$I = \frac{V_I - (-1V)}{R_1 +R_2}\$ do you see it? Also, do you notice that Vth is a voltage measured between the middle point and GND? electronics.stackexchange.com/questions/392010/… \$\endgroup\$
    – G36
    Jun 5 at 14:30

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