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If a transistor is in the "active" region and the current to it's base -> emitter is reduced, will the resistance between collector -> emitter increase in a proportional way?

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    \$\begingroup\$ When a BJT is in active mode, the base current directly controls the collector current. The behavior between the Collector-Emitter isn't like a resistor but like a current source whose value is controlled by the base current. \$\endgroup\$ May 6, 2019 at 9:41
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    \$\begingroup\$ @Bimpelrekkie, I'm almost tempted to ask if you're sure (current control), but you probably know my point of view in this relationship. Have you ONE SINGLE proof for current-control? Shall I give you some proofs for voltage control? \$\endgroup\$
    – LvW
    May 6, 2019 at 11:09
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    \$\begingroup\$ @LvW Yes I'm quite sure that, in active mode, if I force a current into the base that that will result in a current of roughly \$\beta\$ times larger being pulled into the collector. But feel free to generate a voltage across the BE which will then result in a base current to which the above also applies. Feel free to provide voltage control examples but in my opinion, that does not explain anything more clearly. But feel free to show that your way is easier/better. \$\endgroup\$ May 6, 2019 at 11:22
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    \$\begingroup\$ I agree with LvW that the BJT is a voltage controlled device, rather than current controlled. The base current is parasitic in nature, and it's unpredictability causes many β-based designs to fail. \$\endgroup\$
    – Bart
    May 6, 2019 at 11:45
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    \$\begingroup\$ Bimpelrekkie and Curd... Up to now - only claims (current-control). No proof. Question: Based on current control - can you explain the following effects: (1) RE-feedback, (2) EARLY effect; (3) current mirror, (4) tempco -2mV/K , (5) translinear loops (from Barrie Gilbert)... Can you? I am curious! \$\endgroup\$
    – LvW
    May 6, 2019 at 14:19

2 Answers 2

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It's much better to think of a bipolar junction transistor (BJT) as being current source, rather than having resistance.

Why you may think? Between the collector and emitter, you can have a voltage across them, and a current flowing between them, and V/I = ohms, right?

The difference between describing a transistor as having resistance, and having a current source (we tend not to make a difference between the terms current source and current sink when using these models), is what stays constant when you change other things. In this case, when you change the collector voltage, the collector current stays more or less constant.

It's not very useful describing the V/I relationship as a resistance, if the resistance varies with the collector voltage. It's always most useful to find something that's constant.

Let's take two good DMMs which, when on the resistance range, one applies 1v to components and the other applies 2v. If we set them to measuring a 1k resistor, they would both read 1k. If we set them to measuring a transistor whose base current was such that the collector current was 10mA, one would read 100 ohms, the other would read 200 ohms.

This is why although a transistor has a defined V and I, and their ratio has units of resistance, we do not say that it has a resistance, because it will be measured differently by different resistance-measuring devices. Two current measuring devices will give substantially the same reading as each other.

However it's just occured to me that DMM tend not to work like that when measuring resistance. They don't apply a voltage and measure the current, they apply a current and measure the voltage. With an excitation current, things get even worse for our putative 'transistor has a resistance' model. Consider two DMMs, one that measures voltage with a 1mA current, and one that uses 2mA. Let's say the transistor IB has been adjusted to pull 1.5mA when the collector has a sufficient bias. The 1mA DMM will be pulled down to VCEsat, somewhere in the 0.1v to 0.3v range depending on the transistor detail, so will show a 'resistance' of perhaps 0.2v/1mA = 200 ohms. The other DMM's 2mA will not be pulled low by the transistor, and most DMMs read OL (overload) when insufficient current is drawn from the test leads to reduce their output voltage.

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  • \$\begingroup\$ But current is also supplied elsewhere from some supply since the transistor cannot "create" current? So if there would be some amps running through the circuit without the transistor, and that current is reduced when the transistor is added, then it must be some "resistance"? \$\endgroup\$
    – user221241
    May 6, 2019 at 10:10
  • \$\begingroup\$ @user221241 Call it a "resistor" if you must. In a BJT in active mode that "resistor" is controlled such that the current remains constant. To me "looking from the outside" that looks and behaves like a current source. If I would assume it is a (constant) resistor and measure it with that assumption, I would not be able to measure a proper resistor value. If I would assume it is a current source, my measurements would confirm that. \$\endgroup\$ May 6, 2019 at 11:25
  • \$\begingroup\$ Would the transistor measure a resistance with a multimeter from C to E? Surely the current is not constant assuming that B -> E changes? \$\endgroup\$
    – user221241
    May 6, 2019 at 12:04
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    \$\begingroup\$ The transistor will have a voltage drop, and a current through it, so a DMM will measure their ratio, and present that as a resistance. However, a different DMM applying twice the voltage would measure twice the resistance, because the collector current barely varies with collector voltage. If two correct reading DMMs read different figures, that's a good argument for saying it's not a 'resistance'. The current varies with VBE, but barely varies with VCE. We assume VBE is constant when we call it a constant current source, better to call it a current source controlled by VBE or IB. \$\endgroup\$
    – Neil_UK
    May 6, 2019 at 13:06
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    \$\begingroup\$ Yes. When we analyse a transistor amplifier as a controlled current source, we assume that it stays biassed so that's true. This is not a full model of a transistor under all conditions, only a restricted linear model. If we set the VCE to zero, then no current flows. More complex models are used in circuit simulators which can represent the behviour over a wider range, but they still have a controlled current source at their core. \$\endgroup\$
    – Neil_UK
    May 6, 2019 at 13:16
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The transistor Iout versus Vout characteristic does have a slope, which we label the Early Voltage and model as a straight line; that line does not pass thru (0,0), but you can describe the incremental change-in-collector-current / change-in-collector-voltage as a resistance.

enter image description here

Because the Early Voltage resistance is part of the small-signal gain computation, you should include this effect in precise modeling of the transistor.

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