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So here is a page from my text book, the Common Emitter Circuit is on the top and on the bottom right you can see the DC equivalent circuit , my question is that by looking at that dc circuit i do not understand why when finding the R-thevenin R1 and R2 are said to be in parallel if , from my perspective, they seem to be in series... and then funny enough when finding the voltage equivalent it does treat them as a series voltage divider....enter image description here

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  • \$\begingroup\$ "From my perspective.." - where exactly is that perspective ? Looking down from Vcc ? After C1 ? Ve ? There are a set "rules" to find thevenin resistance. What are they ? \$\endgroup\$ – efox29 Dec 11 '16 at 1:08
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I think what you are missing is that they are looking for the Thevenin circuit from the base of the transistor to ground. Redrawn, it looks like this:

schematic

simulate this circuit – Schematic created using CircuitLab

The Thevenin equivalent for this circuit would look like this:

schematic

simulate this circuit

When calculating the value of Rth, you have two options:

  1. Using the superposition theorem, treat the voltage source as a short circuit and calculate the equivalent resistance between the two points. In this case, you have two parallel resistors.

  2. Knowing the short circuit current (which is the current in the circuit through R1 if R2 was shorted), calculate what Rth needs to be to achieve the same current using Vth instead of Vcc.

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  • \$\begingroup\$ So is it like applying superposition , where my voltage source gets shorted to ground?and thus R1 and R2 are in parallel "looking" from the base? \$\endgroup\$ – Edwin Fairchild Dec 11 '16 at 2:06
  • \$\begingroup\$ @EdwinFairchild Yes. Calculating the Rth, if you treat the voltage source as a short circuit, you then have a parallel path through the two resistors from the Base to Ground. \$\endgroup\$ – Ben Miller Dec 11 '16 at 2:19
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    \$\begingroup\$ Yup makes sense now, I was just trying to get an equivalent circuit but not using superposition and I couldn't see the path to ground, I appreciate the reply. I built the circuit and did the math and took some measurements and it's all pans out . Thanks. \$\endgroup\$ – Edwin Fairchild Dec 11 '16 at 2:22
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The Thevenin equivalent of the two resistors when connected as shown to the 12V supply is a fixed voltage of 12*R2/(R1+R2) with a series resistor of R1||R2. That's 2.833V with 5.19K in series.

Since this is a linear circuit you can prove this with a couple of points. Open circuit you have the voltage as above. Short it to 0V and the current is 12V/22K = 545uA (no current through the 6.8K, since there is no voltage across it).

If you short the 2.833 to 0, in order to get 545uA the resistor would have to be R = 2.833V/0.545mA = 5.19K.

You can read the Wiki page on Thevenin's theorem which uses superposition and the uniqueness theorem to construct a mathematical proof.

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While one may calculate the equivalent resistance by zeroing the \$12\mathrm{V}\$ source and simply noting that \$R_1\$ and \$R_2\$ are now parallel connected, it is (possibly) instructive to verify this via an explicit calculation.

Looking 'back', from the base, into the node where \$R_1\$ and \$R_2\$ connect, the (DC) open-circuit voltage is given by voltage division since, for the open-circuit condition, \$R_1\$ and \$R_2\$ are series connected:

$$V_{OC} = 12\mathrm{V} \frac{R_2}{R_1 + R_2}$$

By zeroing (grounding) the same node, the short-circuit current involves just \$R_1\$ since \$R_2\$ now has zero volts across and so, has no current through.

$$I_{SC} = \frac{12\mathrm{V}}{R_1}$$

By Thevenin's Theorem, the Thevenin resistance is then

$$R_{TH}= \frac{V_{OC}}{I_{SC}}=\frac{12\mathrm{V} \frac{R_2}{R_1 + R_2}}{\frac{12\mathrm{V}}{R_1}}=\frac{R_1R_2}{R_1 + R_2}=R_1||R_2$$

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