Electronic – How to calculate Ic for a very simple NPN transistor circuit

bjtintegrated-circuitnpnsaturationtransistors

I built a very simple circuit with an NPN (2N3904) transistor.
Then, I swept Vce from 0 to 5V and plotted a Vce vs Ic graph (as shown in the picture).
I heard Ic = Is*e^(Vbe/VT), where VT=0.025V, and Is and Vbe are given in the picture
(I also attached other parameters for the transistor as well just in case).

I keep getting Ic=107.4mA using the equation above, but the circuit simulator tells me that Ic has to be around 43mA.

I made sure that Ic/Ib = 100 throughout the whole DC sweep which makes me believe that the transistor is in the active mode.

Do any of you know how to calculate Ic in this case?

enter image description here

Best Answer

In your circuit, you are still not driving the BJT into deep saturation region (switching behavior), since the base voltage (760mV) is slightly larger than the diode's forward voltage (650mV).

If you sweep the base voltage, you would get the following curve. Here you still see that the output current still has some strong dependency on the base voltage range between 700mV and 900mV (your case). I might be wrong here, but the exponential growth of the current comes from the fact, that you do not have any base resistor and the current is mostly a function of the emitter's diode current (formula that you wrote in your question).

sweep

If you do not add any collector resistor, the collector current will be given by:

$$I_C = I_B \cdot \beta = I_S \cdot (e^{\frac{V_D}{n \cdot V_T}}-1) \cdot \beta$$

More to that see this link.