How do these resistors become in parallel for SSEM?

Discussion in 'Homework Help' started by midnightblack, Jun 9, 2012.

Feb 29, 2012
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I am trying to understand small signal equivalent circuits, I am having some trouble understanding this diagram:

The RB1 and RB2 resistors end up in parallel on the SSEM circuit. Is this because we set the DC voltages=0 for SSEM? (And so, the nodes with same voltage are connected?)

What slightly confuses me is that the input voltage will now be across two resistors compared to the macro model where its across RB2 only. Can someone explain this to me?

2. mlog Member

Feb 11, 2012
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The SSEM is an AC circuit. RB1 is connected to the power supply, Vcc, which is effectively "ground" or the common point in the circuit. So Rb1 and Rb2 are in parallel for an AC circuit. Look at Rc on the output. It also is connected to "ground" in the SSEM.

3. Papabravo Expert

Feb 24, 2006
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1,850
Think of it like the DC power supply having a very low impedance to AC (aka small) signals. So from the point of view of the incoming AC signal it sees both resistors and is indifferent to which one the current flows through. Naturally more current will go through the one with the lower value. That is what resistors in parallel do.

Feb 29, 2012
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Ok.. Thats something like what I had in mind. Many thanks.

5. wmodavis Well-Known Member

Oct 23, 2010
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Another way to think of it in your mind is to replace the Vcc power supply with a resistor representing it's ac impedance. Placing the resistors, Rb1 & Rb2, in your diagram in parallel makes the assumption that the ac impedance of Vcc is zero ohms. In actuality it will be greater than zero - maybe 0.1 Ohms, but it depends on the power supply you are using. So let's say it is 0.1 Ohms for discussion. If you replace the Vcc supply with a 0.1 Ohm resistor, the top end of Rb1 would then be connected to circuit common (gnd) through that 0.1 Ohm resistor making an equivalent resistence for Rb1 of Rb1+0.1 Ohms. If Rb1 is 10 Ohms or greater (which it in all likelyhood is, infact probably greater than 1K) the impedance of Vcc is negligible, or ~0 Ohms, so that the equivalent circuit is essentially Rb1 in parallel with Rb2. It is based on the mostly correct assumption that Vcc has an equivalent AC impedance much less than other circuit components, which it usually does.

Feb 29, 2012
31
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That helped me understand even more clearly. Many thanks!

7. WBahn Moderator

Mar 31, 2012
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While these explanations are useful, the real underlying reason why they are in parallel is that the small signal model of a circuit is an application of the principle of superposition. When you studied superposition (I'm assuming you have) and had a circuit with two voltage sources, you analyzed the circuit twice, once with the first source turned on and the other turned off, and the second time with the first source turned off and the second source turned on, right? The total response of the circuit was then just the sum of the two.

When you calculate the DC bias (the "large-signal" behavior) of this circuit, you are leaving the DC supplies turned on and turning off the signal voltage. All the "small-signal" solution is is the second part of the superposition analysis in which you now have a non-zero signal voltage but you turn off the DC supplies. Setting Vcc to 0V then means that the top of Rb1 is at zero volts, just as the bottom of Rb2 is, so these two ends are mathematically shorted together. Note that everything that was said regarding the source impedance works out just find, since turning the Vcc to 0V leaves the impedance in place.

The other big assumption that is made is that the signal is small enough so that the response of the circuit can be considered to be a linear response about the operating point; this is essential since superposition only works for linear systems. The entire "small-signal model" of a transistor is based on a model that describes the changes in the voltages and currents across and through it as a result of the changes in voltages and currents applied to it.