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If we increase the collector resistance then how will the collector current change? I understand the circuit behavior if i neglect early effect but when i try to include it that is allow Ic to vary with Vce i get lost because everything changes and it's as if i lose the grip on all variables.

P.S I'm a self studying person solely dependent on internet and books, so I hope that this question isn't blocked because it may look like a homework problem. Thanking you in advance.

Kashmiri
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  • Look here https://electronics.stackexchange.com/questions/355899/how-is-possible-that-with-same-ibase-there-is-more-than-one-vce/355955#355955 – G36 May 23 '20 at 18:04

2 Answers2

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To a first approximation, the transistor is a current source, so changing the collector resistor does not change the current, only the collector voltage

We can model the transistor as either a base current controlled current source (hFE, beta), or a base voltage controlled current source (gm). Which model is most convenient depends on whether we're biassing the transistor, or using it as a small signal amplifier.

To a higher order approximation, the collector current will change slightly with the collector voltage. You've already mentioned the Early effect, so you understand that.

For very high values of Rc, where V+/Rc exceeds the current the transistor is trying to sink, the transistor will be saturated, and neither collector voltage nor current will change.

Neil_UK
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  • Thank you dear Neil, but i was looking for a more detailed description of how Ic changes with Rc. – Kashmiri May 23 '20 at 17:23
  • @YasirSadiq - As Neil stated, basically it doesn't. – Kevin White May 23 '20 at 17:42
  • @Kevin i think it does ,to be exact. – Kashmiri May 23 '20 at 18:17
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    @YasirSadiq - If you are looking into the details then read about the Early Effect. It is a second order effect and for most designs it can be ignored. – Kevin White May 23 '20 at 18:29
  • Thanks but Dear Kevin i have said in the question that I'm trying to understand the process including the early effect. – Kashmiri May 23 '20 at 18:35
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    @YasirSadiq - for design purposes the early effect is included in the apparent output resistance - Ro. Without the early effect it would be infinite. So it acts as a resistor in parallel with the collector to emitter. – Kevin White May 23 '20 at 22:18
  • @Kevin you are referring to its small signal model, I'm talking about the DC bias currents. – Kashmiri May 24 '20 at 04:40
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    @YasirSadiq - agreed but the slope of the current vs collector voltage still acts like a resistance. – Kevin White May 24 '20 at 16:29
  • A reduction in Rc leads to a reduction of the voltage across Rc and then will cause a rise in VCE, now this rise in VCE causes ice to increase due to early effect,then this larger IC current leads to a larger drop via the resistor and hence decreases VCE,which then leads to a decrease in ICE . Now this decreased IC leads to a lesser drop via the resistor hence increasing VCE, Which then lead to an increase in ICE.... And we get stuck in an infinite loop. – Kashmiri May 26 '20 at 04:37
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    @YasirSadiq If you try to solve it strepwise like that, yes, it's an infinite series (not an infinite loop), but like the infinite series 1 + 1/2 + 1/4 + 1/8 ... it sums to a finite result. It's far easier to write down the implicit equation, and then solve it, that's what modelling the effect as an extra resistor on the collector is for. In hybrid parameters, it hoe, the 'output impedance'. – Neil_UK May 26 '20 at 05:13
  • It won't be an infinite loop but an infinite sequence, this was very very helpful for me, Thank you. I would request you if you could explain your next part of modeling as a resistor. Say i change the Rc to 2Rc,and all sources are DC now how would our modeling help here. All i can think is that we've got to use the large signal model and solve those difficult equations as LvW pointed out that it's hard to use math in some cases for exact solutions. – Kashmiri May 26 '20 at 10:46
  • It is solved with Vcc to Rc, and the BJT represented by a current sink from the collector, and by hoe from collector to emitter. The best way to solve this is superposition, solve for voltage and current sources independently, then add the two solutions together. Do the same after replacing Rc with 2Rc. Treating a transistor as a current sink in parallel with hoe is a model, that simplifies its behavior. You can have more complex models, or you can measure a transistor. We tend to use the simplest model that meets our needs. If yours are not met by hybrid parameters, then go more sophisticated – Neil_UK May 26 '20 at 13:08
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Yasir Sadiq...here is my short explaantion:

When modifying the collector resistor (without changing the input signal) the collector voltage - and with it - the voltage Vce changes. For example, a reduction of Rc will increase the voltage VCE across the resistor.

This voltage increase will increase the depletion zone of the reverse-biased B-C pn junction and consequently reduces the width of the remaining base region associated with the forward-biased B-E pn junction. As the result, the E-field within this region will rise (constant B-E voltage) and a larger portion of the emitted charged carriers will arrive at the collector (Ic rises somewhat). This effect is called "base-width modulation" and is the cause of the Early effect.

LvW
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  • Thanks LvW, how can you say that "a reduction of Rc will increase the voltage VCE across the resistor".This is the essence of my question! – Kashmiri May 24 '20 at 00:23
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    .....sorry for my typing error: Of courser, it should read "..across the TRANSISTOR". There is a voltage division between Rc and the C-E path. Hence, a reduction of the voltage across Rc will cause a rise in VCE. – LvW May 24 '20 at 08:32
  • you said " Hence, a reduction of the voltage across Rc will cause a rise in VCE " now this rise in VCE cause ice to increase due to early effect,then this larger IC current leads to a larger drop via the resistor and hence decreases VCE,which then leads to a decrease in ICE . Now this decreased IC leads to a lesser drop via the resistor hence increasing VCE, Which then lead to an increase in ICE.... And we get stuck in an infinite loop. This is what confuses me. I hope I'm clear enough, and thanks for your effort to help dear. – Kashmiri May 24 '20 at 12:00
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    OK...I know what you mean...but: Let us assume that we increase the Rc by 25% and the voltage acros Rc changes also by 25%. When the VCE had before app. the same value, it will increase by 25%. However, the corresponding increase in Ic will NOT as large as 25%...perhaps only 5% ....Have a look on the small slope of the curves Ic=f(VCE)... – LvW May 24 '20 at 12:24
  • I think that does make it clear to a nice degree and i accept it is an answer but I'm wondering how could such a problem be solved exactly using mathematics. And thank you – Kashmiri May 26 '20 at 04:35
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    Yasir S....there are some physical effects which cannot be solved "exactly" using math. – LvW May 26 '20 at 07:40