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Here, the total output resistance is calculated when the Early voltage is completely unspecified and neglected (I suppose it's zero). How has this been done? The formula for terminal output resistance (Ro) for BJT Amplifiers is (Va + Vce)/Ic where Va is the Early voltage, Vce and the collector-emitter voltage and Ic is the collector current. Vce and Ic are calculated via the Q-point. Rout is then calculated by taking Rc and adding in parallel with the calculated Ro. However, my calculation gives me 29.6k Ohms, not 39k Ohms as said in the solution. (Vce/Ic)||Rc gives me 29.6k Ohms. Please let me know where I went wrong. Here are the relevant images:

Solution

SolutionQuestion

enter image description here

What are Av, Ai , Rin, Rout, and the maximum amplitude of the signal source for the amplifier in

Fig. P14.1(g) if RI = 500 , RE = 110 k,

RB = 1 M, R3 = 500 k, RC = 42 k,

VCC = 15 V, −VEE = −15 V? Use βF = 100.

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  • \$\begingroup\$ you should provide what steps you took and intermediate values you got, or this is likely to get closed. \$\endgroup\$ – kolosy Oct 20 '16 at 14:58
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It seems like you assume that you always need to know (and take into account) the Early voltage to calculate the output resistance.

OK, then I will tell you the Early voltage, let's say it is -100 V. Can you now do the calculations and determine Rout ?

If you do the calculation properly you will not get a significantly different answer ! Why is that ?

Look at the output and notice how Rc sits there. How does this Rc relate to the output impedance ? The output impedance can never be higher than the value of Rc.

In the question it is assumed that the output impedance of transistor (which is determined by the Early voltage) is significantly higher than the value of Rc. If the output impedance of the transistor was significant something would have to be specified, like the Early voltage.

Also, since the Early voltage is not even specified, how could you take it into account ? You cannot so you can safely assume that you can ignore the effect that the Early voltage has.

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  • \$\begingroup\$ so how was Rout calculated? \$\endgroup\$ – Carl Oct 20 '16 at 16:14
  • \$\begingroup\$ There is no calculation needed because the transistor does not contribute to Rout. So imagine the transistor gone, now what's left ? \$\endgroup\$ – Bimpelrekkie Oct 20 '16 at 17:37
  • \$\begingroup\$ What's left is a bunch of resistors, which doesn't answer my question. The capacitors are shorted because we are performing AC analysis. \$\endgroup\$ – Carl Oct 20 '16 at 21:22
  • \$\begingroup\$ Well it does answer your question but you don't understand the answer. There will be only one resistor left at the output after ignoring the transistor's output resistance. That one resistor is Rc. No other resistor will have a connection to the output. If you do not understand that then you have to learn about small signal models first. \$\endgroup\$ – Bimpelrekkie Oct 21 '16 at 6:05
  • \$\begingroup\$ Why is Ro neglected? When the voltage gain is less than uf, I know. But why? In which cases can we automatically assume this? \$\endgroup\$ – Carl Oct 21 '16 at 16:09
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"Here, the total output resistance is calculated when the Early voltage is completely unspecified and neglected (I suppose it's zero)............... However, my calculation gives me 29.6k Ohms, not 39k Ohms as said in the solution. (Vce/Ic)||Rc gives me 29.6k Ohms. "

Carl - one error you have made is the following: For neglecting the influence of the Early voltage you have set VA=0. This results in the false value of 29.6kOhms.

However, in contrast to your assumption, for an IDEAL transistor (neglecting the Early voltage) you must set VA to infinite. Only in this case, the slope of the output characteristics Ic=f(Vce) is zero (differential resistance ro=0).

In the calculation they have assumed a rather large value for ro (app. ro=500kOhms) because Rout=Rc||ro=42k||500k=38.7kOhms

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