BJT question. What is effective collector load?

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milan.rajik

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I read the book "Principles of Electronics" by V K Mehta

I am having trouble understanding effective collector load. In that book chapter 8 page nos 172 - 173 (see attached images) he says that Ro is in parallel with Rc. I don't understand this. Can you please explain how Ro which is in series with Rc becomes parallel? There is two paths Rc to ground through CE and Rc to ground vis Ro.

In single stage amplifier if there is Coupling capacitor at collector then varying collector current flows through it and DC is blocked. This current will create a voltage drop across Ro or (RL, load resistor)

Please explain how he says Rc is in parallel with Ro in single stage amplifier and Rc is parallel with Ro and Ri in multistage amplifier. Please I need to understand this clearly.


Is it because battery can be replaced by its internal resistance and as internal resistance is small it can be neglected and circuit can be reswrawn as shown in fig iii of third image?
 

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man,in that u have to do the small signal analysis ....so the Vcc would be made to zero potential ....it means that they will be in parallel.refer to conditions of small signal analysis
 

I know that for small signal analysis +Vcc is replaced by GND. If it is a large signal analysis then what happens? Does the Rc and Ro still considered as parallel? Why Collector resistor Rc is grounded in small signal analysis? Is there any expalnation for that or is it just assumed like that without any reasons?
 

Hi milan.rajik

First of all something about your latest post :
I know that for small signal analysis +Vcc is replaced by GND
It means you don't know principles of circuit analysis . who told it is replaced by GND ? who told just in small signal ? in fact the DC source is short circuit to the ground instead of AC sources and it is because of its pretty low intrinsic resistance . ok ?
he says that Ro is in parallel with Rc. I don't understand this. Can you please explain how Ro which is in series with Rc becomes parallel?
Who told ro is in series with RC ? a circuit which called series circuit won't have any other node ok ? so it is not series in DC analysis because there is another path which is collector path .
Furthermore , ro is in parallel instead of AC analysis because there is a common node of collector ( common with RC ) and another common point with Supply side . every DC PSU is shorted instead of AC signals .
Any question ?

Best Wishes
Goldsmith
 
+Vcc is replaced by ground means connecting the Rc end which is connected to +Vcc to ground. It is the same thing.

Who told ro is in series with RC ? a circuit which called series circuit won't have any other node ok ? so it is not series in DC analysis because there is another path which is collector path

Ok. In AC analysis one end of Rc is connected to groung but what is the reason for it?

See in DC analysis there is 2 paths from Rc to ground. One is from Rc and Ro, Ground path and another is Rc and Collector-Emitter-Ground path.

In AC analysis if Rc one end is connected to Ground then Rc, Ro, and collector-emitter are all in parallel.

Now what is the reason for connecting Rc to ground in AC analysis. What is the purpose and reason?
 

+Vcc is replaced by ground means connecting the Rc end which is connected to +Vcc to ground. It is the same thing.
Hi milan.rajik

No replace has not the same meaning because it is not dealing with the concept of DC circuit analysis . in concept we are going to consider VDC is shorted to the ground . it is important that we tell the physical truth behind each effect . ok ? :wink:

Ok. In AC analysis one end of Rc is connected to groung but what is the reason for it?
Just one of it's heads . ok ?
About what is the reason behind that , did you read my former post precisely ? i've explained it in there as well . isn't that clear ? if no i can show it to you by a handwriting .

See in DC analysis there is 2 paths from Rc to ground. One is from Rc and Ro, Ground path and another is Rc and Collector-Emitter-Ground path.

A question : are you familiar with equivalent circuit of a BJT ? do you know how it looks like ? it is important to know because thus i can say how that circuit can be analyzed as well .

Now what is the reason for connecting Rc to ground in AC analysis. What is the purpose and reason?
I highly suggest you to read my former post accurately again !

Best Luck
Goldsmith
 

Please explain the small signal analysis in detail with an example.
I don't think it can be possible in detail but i can write something for you . so wait some days , i'll prepare something for you . the reason i say some day is because i'm horribly busy these days . but i'll make them ready for you .
But you also can refer to some text books too if you are in hurry .

Good Luck
Goldsmith
 

I think you can send Vcc to ground, because in small signal analysis, the circuit becomes linear, so you can apply superposition principle. Considering the input as DC + AC, when you work with the AC input you have to "turn off"(superposition principle) the DC sources.
Another way of see this could be, since in small signal analysis the voltages and currents are actually increments, then as the VDD is always constant, its increment is zero so, its an "ac ground" .
 

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