BJT Collector voltage in resistor bias network?

Hello folks!

I'm having a bit of an issue with this:

Values for resistors were selected more or less random, there is no real significance to them at this point, just so that I can ask the question:

  1. why on earth is Vc 3.44V according to this simulation?
  2. same goes for Ve 526mV?

If I replaced the NPN transistor with a wires, the voltage at the point of the transistor would be 2.5V, as I would expect. This leads me to think, there is something I still don't know about how a BJT works.

Is there some kind of feedback going on or maybe some process inside the transistor I've yet to explore?


Assume you see the base resistor voltage divider 500R and 1.5k

The divider sets the base voltage (Vb) to 500 * 5v / 2000 = 1.25v

The voltage drop Vbe is about 0.7v for a silicon BJT.
Therefore the voltage across the emitter resistor (Ve) is 1.25(Vb) - .7(Vbe) ≈.55v okay ≈ 526mV

The emitter current (Ie) is 526mv / 500R ≈ 1.05mA

Therefore the voltage across the Rc ≈ 1.5k * 1.05mA ≈ 1.57v.
Therefore 5v(supply) - 1.57v(Vrc)3.25v(Vc) = voltage at the collector.


The easy thing first: Ube is normally something between 0.6 to 0.7V for silicon. With neglectable Ib the Ub voltage is determined by the voltage divider = 1.25V. Thus Ue is about 1.25-0.65 = 0.6V.

Then 0.6V on Re = 500 Ohm means about 1.2mA. Again with Ib neglected the same current flows through Rc so that the voltage drop on Rc is 1.5k * 1.2m = 1.8V. Remains Uc =5-1.8 about 3.2V.

All calculations by rule of thumb and heavily rounded. Every transistor will introduce some uncertainty, temperature etc. some more. That's why 2 significant digits are usually enough for a practical computation result.

Test: you have 10 minutes to answer the following question:

Your circuit is a constant current source with the load Rc.
What's the minimal and maximal allowed value of Rc in that circuit?

Provided the supply voltage remains constant. To be a true current source, the lower resistor of the voltage divider at the base should replaced with a voltage reference.

The voltage reference is already built into the PSU :wink:

Yes. It's a detail, but a true CCS is independent of supply voltage within its operating limits. Just as a true CVS is independent of the current drawn within its operating limits.

Uhm... Idunno :point_right: :point_left: :thinking: What?

I only got 10 mins to google around and my best guess is that it has something to do with the thermal characteristics of a transistor. But I don't really know how to read the needed into from the BJT I'm using.

Another thought I had was the value of hFe. I imagine, Ic can be controlled by considering hFE @ a given Ib which would give me the Ic(max).

Ehh, I dunno... Want to give me the answer?

Therefore 5v(supply) - 1.57v(Vrc)3.25v(Vc) = voltage at the collector .

What I still don't quite understand is that the potential difference between collector and emitter is Vc-Vb=2.914V. Where does that go?

Yes, I think this is the source of my confusion. I thought that Vce is usually quite tiny if the transistor in on. And therefore I always calculated with Ve = Vc assumption when computing the Rc value. It is clearly not the case, I should look into why that it. What are some of the google search terms I could look into?

No. you don't have to google, only think about resistance and voltage.

If at constant current you increase the resistance then the voltage increases as well. But there is a limit since the supply voltage is only 5V. Now subtract the remaining voltage on Re (0.6V) and the saturation voltage of the BJT (0.4V) and you can compute the maximum Rc for that voltage and the known current.

With similar considerations you also can find the minimal resistance.

And now you also may know the voltage at the transistor. It's the remainder after subtracting the voltage across Re and Rc from Vcc (5V).

If its any help, please refer to the below. Please pardon the handwriting.

The equations are all sort of relative, so the end goal will decide where to start...

1 Like

Take the collector current as fixed and known from preceding calculations.

Specifically, what is your issue with the numbers show? Is it you don't understand the calculations, or think they are wrong?

This circuit would be for a linear amplifier of some sort.

Voltage divider biasing or emitter bias

calculation Details

There is a whole lot to consider when calculating a circuit like this and still missing some pieces. I'll put it together eventually. I got quite a bit of info here but have not yet processed it all.

Basically, I'd like to be able to design such a network with the following criteria:

  • hFE: 170
  • Vcc: 5V
  • Vc: 2.5V
  • Output current at collector: 20mA

And yes, this would be part of a Class A amp. Figured, I should look into them - just for fun.

Here is the class A mode calculation as per your specification...

Let Vc = 2.5V (for max symmetrical voltage swing)
Load = 20mA
Therefore Ic = atleast twice load current for class A mode. Thus Ic = 40mA.

Now, the actual load will be capacitively coupled, and we need to calculate the DC bias conditions.

Ic = 40mA, so Rc = (Vcc - Vc)/40mA = 2.5/40mA = 62.5 ohms.

Ie = Ic (almost) = 40mA.

Assume Vce = 1.5V (not saturated).

So, Ve = 1V, and Re = 1/40mA = 25 ohms.

Vb = Ve + Vbe = Ve + 0.65V = 1.65V (assume Vbe to be 0.65V)

Ib = Ic/hfe = 40mA/170 = 236uA.

Ir1 = 10* Ib = Ir2 (approx) = 2.36mA

So, R2 = Vb/Ir2 = 1.65/2.36mA = 699.1 ohms
And R1 = (Vcc - Vb)/Ir1 = (5-1.65)/2.36mA = 1.42 kOhms

Voltage gain = approx Rc/Re = 6.25/25 = 2.5

This is great, thank you! I can now check my own math against this!

Just a single of follow-up question:

You multiply the Ib by 10. Could you tell me what the reason for that is?

Linked wiki page sais:

In this circuit, the voltage divider holds the base voltage fixed independent of base current, provided the divider current is large compared to the base current.

I assume this could be the reason.

You want to design 10* the base current amount going thru the voltage divider, we could have used 20* 30* etc.

This way the voltage divider will still be effective in creating the required 1.65V base voltage.

Right... Got it!

And where does this assumption coming from? Can this be calculated instead of assuming a value?

There is a data sheet value you can look up.

As a trade off, we want to stay above Vce(sat) so the output does not get distorted.

ex: 2N2222 Vce(sat) ≈.3V

For many small signal BJTs the saturation voltage can be ≈.5V, 1.5V is above this.