What is the Rth of the below circuit Given R=R1=R2=10 ohms and g=10 siemens.
Below is my solution.I am getting a value of 0.2 Ohms for Rth.May I know is this correct.
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You need to show us your calculations first. – G36 Jul 16 '23 at 17:17
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Trying test source method but not getting the answer – Hari Jul 16 '23 at 17:28
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This is the proper method, simply show your work. So that we can see where you made the mistake. – G36 Jul 16 '23 at 17:36
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Take a a look at this example https://electronics.stackexchange.com/questions/295771/i-o-resistance-of-common-source-mosfet-with-source-degeneration/295966#295966 – G36 Jul 16 '23 at 17:42
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@newbie How is it that a non-zero current may occur in **R**? – periblepsis Jul 16 '23 at 20:09
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1@G36,Added my solution.I am getting a value of 0.2 Ohms for Rth – Hari Jul 17 '23 at 04:42
1 Answers
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Good job your answer is correct.
Notice that only one equation is needed.
simulate this circuit – Schematic created using CircuitLab
$$V_X = \left( I_X+ (g_m (-V_X) \right )R_2 + I_X R_1 $$
$$V_X = \left( I_X - V_X \:g_m \right )R_2 + I_X R_1 $$ $$V_X = I_X R_2 - V_X \:g_m \:R_2 + I_X R_1 $$ $$V_X + V_X \:g_m R_2 = I_X R_2 + I_X R_1 $$ $$V_X(1 + \:g_m R_2) = I_X(R_2 + R_1)$$ $$V_X = I_X \frac{R_2 + R_1}{1 + \:g_m R_2}$$
And finally $$R_{TH} = \frac{V_X}{I_X} = \frac{R_2 + R_1}{1 + \:g_m R_2} = \frac{10\Omega + 10\Omega}{1 + 10S \: 10\Omega} = \frac{20}{101} [\Omega] \approx 0.2 \Omega$$
G36
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