 # Transistor VCE (collector-emitter voltage)

First i'd like to apologize since it might seem out of topic, it's more towards circuit theory i think.

So my problem here is that, lately I have been studying the BJT transistor and now Im trying to apply the theories in a real circuit. So if what I learnt was correct, in VCC side,

VCC = VRC + VCE + VE

In this configuration since IE = 0, VE = 0

VCC = VRC + VCE

VCC = IC(RC) + VCE

So now lets say I have a supply voltage of 3.7v and I want the IC to be 2A. I must have either RC or VCE. So it got into my mind, if my RC is constant lets say 10 ohm, how can VCE be manipulated or is it constant which depends on what transistor I use?

There is another way around it by finding IB which is IC/β. But, i found out that β or hFE changes depending on IC, and im not sure if the datasheet is accurate so idk. And plus im not sure either whether VBE is actually 0.7V for every transistor.
Here is the datasheet of the transistor im going to use.

Theories without practical approach really confuses me, I don't know if what i said makes any sense either. So summarizing everything i just said, i just wanna know how to find VCE and VBE in transistor datasheet so that i have all the variable to make the calculation.

VE = 0 not because the IE = 0, but because the resistance in the emitter circuit is zero. Emitter and ground potentials are equal.

Your datasheet says that the collector-emitter saturation voltage is no more than 0.5V at a collector current of 2A. The saturation voltage at other currents can be obtained from the Collector-emitter saturation voltage diagram in the datasheet or measured experimentally.
My opinion is that for a transistor in SOT-89 at a current of 2A, you need to think about heat sink.
You can also use a MOSFET with a low gate-source voltage.

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As you raise the current into the base the transistor will start to turn on and Ic will be a value of the DC gain of the transistor multiplied by the base current .
As Ic is increased it will reach a point where it is limited by the battery voltage /Rc and the transistor is approaching saturation. At saturation the voltage Vce is at its minimum , 100mV or whatever your data sheet shows .

Worth a google of how a transistor works

In an amplifier the base would be so biased that the transistor is partially “on” and Vce say around half the supply voltage - then a small change in base current gives a bigger change in collector current .

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IE is not zero, it's IC+IB.
VE is a meaningless name, a voltage requires more than 1 pin.

It may become all clearer if you redraw your circuit so that RC and the transistor form a voltage divider for VCC.

I used that approximation for 30 years of (low power) circuit design. I also assumed an IC/IB of 100, what degrades to 20 or less for power transistors. But this parameter is non-linear and often has a wide range of e.g. 50-250 for a low power transistor type. Narrower ranges are indicated with A, B or C suffix to the transistor name. It means that you only can determine the minimum IB for a given IC.

In your case IB=0.2A should be sufficient for IC=2A. Then also VBE should be assumed higher than 0.7V. For high power circuits the diagrams in the transistor data sheet should be consulted, also for temperature dependencies.

For switching applications you better use MOSFETs with near zero power loss when turned on, so that no or a much smaller heat sink is sufficient.

Oh yah my bad its RE=0

Oh, what about in the dc current gain row, what does it mean, lets say the hFE(1) VCE=2V Ic=1A

Ok, i'll keep that in mind.

Thanks, I think now I understand the concept a lot better.

I'm not sure I'm following, are you're saying it's hard to determine the IC since the gain is not linear, and the only accurate data is the one indicated with the suffix? I'm not sure which indication you are referring to, is it the classification of hFE?

No, the current amplification varies too much. Stable circuits deserve feedback like a RE with RE*IE voltage drop for a constant current source.

1. A transistor amplifies the current so that the IC = IB * hFE. hFE depends on the letter of the transistor. D882R has hFE = 60-120, D882O has hFE = 100-200, D882Y has hFE = 160-320, etc.
You don't know exactly which hFE each specific transistor has. Therefore, the circuit should work if you have a transistor with the lowest hFE.
If you want IC = 2A, you must supply the base current sufficient that any transistor of the selected type will provide the required DC current gain. For example, for D882R transistor hFEmin = 60. If you need a collector current of 2A, you must provide a base current of at least 2000/60 = 33 mA or more.
2. VCE = 2V, IC = 1A indicates the conditions under which the DC current gain measurements were made. The transistor is a highly non-linear element. Under different conditions, we get different values of the hFE.
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If your RC = 10 Ohms, you won't be able to get 2A. If the VCEsat of the transistor is about 0.2-0.4V, then with a battery voltage of 3.7V, you need RC about (3.7-0.3) / 2 = 1.7Ω.
However, this is still a bad solution at low supply voltage. If you mean a Li-ion battery, then its voltage is not stable, it can be about 3-4.2V. Then at Vcc = 3V you will get Ic = (3-0.3) / 1.7 = 1.6A, and at Vcc = 4.2V Ic = (4.2-0.3) / 2 = 2.3 A.
If you need a constant current, you will have to look towards stable current source on the op amp + transistor.

Ok, now I understand what the classification table is for. But then whats the purpose of the hFE diagram, if there are specific range of current gain for each letter, does that mean the diagram wouldn't help? This is the hFE diagram of the transistor anyway.

Ok last example,
Let's say now im using the D882R and i want the IC to be at 0.8A. As you said we can't know exactly the value of hFE so we have to assume IB should be at least 800/60 = 13.3mA or more.
You also said,

So according to the diagram(well idk if im reading it correctly), at 800mA, the VCE supposed to be 70mV?

Sorry for all the trouble, its hard for me to understand since all these variables kinda related to one another in a weird way for me.

This post is a good tutorial by itself.

Plus, there is tolerance (part-to-part variation) and variations with temperature.

FETs & MOSFETs also have tolerances...

You can "experiment" to find the true hfe (under certain conditions) as long as you stay out of saturation (where the current is limited by a resistor or "something else"),

DrDiettrich mentioned feedback. In real-world linear applications negative feedback (corrective feedback) is used so the gain is predictable and stable. This reduces gain so for example, instead of making an amplifier with a gain of 100 you use negative feedback to get gain of around 10, or you might use a 2-transistor circuit with feedback for net gain of 100, etc.

The thing that makes op-amps so easy to use is that they have very-high open-loop gain so the the gain (with feedback) can be entirely defined by two resistors. If the gain needs to be precise we can use precision resistors.

In digital/switching applications we saturate the transistor (or MOSFET) so for design purposes we "assume" low hfe to make darn-sure it's hard-saturated when turned-on. Digital is easier than analog/linear! It would be beneficial to understand the transistor LOAD LINE and Q-point

I think I understand so that's why @DrDiettrich suggested a RE which is to saturate it to a point the IC would be predictable.

The ranking table says that out of all the transistors that were produced, the transistors with hFE = 60-100 received the letter A.
The hFE diagram shows that if we take an average transistor (with hFE ≈ 240), then at average current values it has hFE ≈ 240, at a current of 1A hFE ≈ 200, and at a current of 2A hFE ≈ 140.
You can measure hFE of your spesific transistor if you want.

You are right that based on your diagram, the voltage will be 70-80mV. However, another datasheet D882-R Datasheet shows slightly different values - about 180-200 mV.
Although this transistor is not familiar to me, in my opinion, in reality it will be more, about 200-300 mV.

Forgot to add, so given that we've taken the lowest hFE which means, any IB value greater than 13.3mA should make it likely that the transistor will be in saturation. So I guess that its safe to assume RC = 3.7-0.18/0.8 = 4.38Ω - 4.4Ω based on your given datahsheet.

Yes, this is so, if we provide the necessary base current.
And it is also correct that if you use a 4.4Ω resistor, the current cannot be greater than 0.8 A.

P.S. If you want to know more about how transistors work, read the book "Paul Horowitz, Winfield Hill - The Art Of Electronic" (chapter 2). This is my favorite book on electronics.

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That statement doesn't make much sense.
Hfe is irrelevant when the transistor is in saturation (collector voltage lower than base voltage).
The datasheet shows a 1:10 Ic:Ib ratio for saturation.
The datasheet shows a 1:10 Ib:Ic ratio for saturation.
Leo..

Are you sure that the base current should be 10 times the collector current? This is something new.