# NPN Transistor common emitter calculation

I built the following circuit in LTSpice using the default NPN BJT model (beta = 100).

I calculated the Ib to be 115.754uA. This is how I did the calculation:

1. Equivalent voltage at base is:
12V * (800/(800+1000+220)) = 4.75V
2. Equivalent series input resistance = Equivalent Voltage / Equivalent Current:
4.75 / (12v / 1220ohm) = 482.9V
3. This turns the input into equivalent circuit like so:
4. Use KVL to solve Q1 (assuming Vbe is 0.8V):
4.75 - ib * 482.9 - 0.8 - ie * 330 = 0
ie = ib + ic and ic = 100 * ib
ie = ib * 101
4.75 - 0.8 = ib * (482.9 + 101 * 330)
ib = 116.82uA
ic = ib * 100 = 11.68mA
5. but the simulation shows:
ib = 1.3mA and ic = 8.65mA

What is wrong with my calculation?

• Your calculated Ic of 11.69 mA would put the collector voltage way below the base voltage. Your assumption of linear operation isn't true. – glen_geek Mar 1 at 23:03
• Your calculations are correct assuming linear operation, but your transistor is saturated. – StainlessSteelRat Mar 2 at 0:01
• Read this answer electronics.stackexchange.com/questions/367267/… It should help. – G36 Mar 2 at 6:54
• Thank you. When in saturation mode, the Vce,sat of the spice NPN is set to 0.05756V, and the Vbe,sat is 836mV. After using these two values, i am able to solve the circuit. – user97662 Mar 2 at 12:07

## Assuming Active Mode

The Thevenin equivalent of your base divider is:

\begin{align*}V_{\text{TH}}&=V_\text{CC}\cdot \frac{R_5}{R_3+R_4+R_5}\approx 4.7525\:\text{V}\\\\R_{\text{TH}}&=R_5\cdot\frac{R_3+R_4}{R_3+R_4+R_5}\approx 483.17\:\Omega\end{align*}

If you assume that the BJT is in active mode (not saturated) then you come up with the approximate value for the base current this way:

\begin{align*}I_\text{B}&=\frac{V_{\text{TH}}-V_\text{BE}}{R_{\text{TH}}+\left(\beta+1\right)\cdot R_2}\approx 120\:\mu\text{A}\end{align*}

From this, you and work out that the expected voltage drop across the collector resistor would be in excess of $$\\beta\cdot R_3\cdot 120\:\mu\text{A}\approx 12 \:\text{V}\$$. This, alone, tells you that the circuit is saturated.

So the assumption for active mode is wrong.

## Saturated Mode Conclusion

The new circuit, now that you know it is saturated, looks more like this. Using $$\V_\text{SAT}=100\:\text{mV}\$$ and $$\V_\text{BE}=700\:\text{mV}\$$, for example, it's:

simulate this circuit – Schematic created using CircuitLab

(Note that in saturated mode, the collector looks more like a voltage source and not like a current source [as it does when in active mode.])

At this point, use nodal analysis to solve:

\begin{align*} \frac{V_\text{B}}{R_\text{TH}}+I_{V_\text{BE}}&=\frac{V_\text{TH}}{R_\text{TH}}\\\\ \frac{V_\text{C}}{R_1}+I_{V_\text{SAT}}&=\frac{V_\text{CC}}{R_1}\\\\ \frac{V_\text{E}}{R_2}&=I_{V_\text{BE}}+I_{V_\text{SAT}}\\\\ V_\text{B}&=V_\text{E}+V_\text{BE}\\\\ V_\text{C}&=V_\text{E}+V_\text{SAT}\\\\ &\therefore\\\\ V_\text{E}&\approx 3.326\:\text{V}\\\\ V_\text{B}&\approx 4.026\:\text{V}\\\\ V_\text{C}&\approx 3.426\:\text{V}\\\\ \mid\, I_\text{B}\mid=I_{V_\text{BE}}&\approx 1.504\:\text{mA}\\\\ \mid\, I_\text{C}\mid=I_{V_\text{SAT}}&\approx 8.574\:\text{mA}\\\\ \mid\, I_\text{E}\mid=I_{R_2}= \frac{V_\text{E}}{R_2}&\approx 10.078\:\text{mA} \end{align*}

That's the complete analysis for the circuit.

## Summary

For small signal BJTs, $$\V_\text{BE}=700\:\text{mV}\$$ when $$\I_\text{C}\approx 1.5-2.0\:\text{mA}\$$. Clearly, this collector current is more than that range. So I'd expect about $$\26\:\text{mV}\cdot\operatorname{ln}\left(\frac{8.574\:\text{mA}}{1.5\:\text{mA}}\right)\approx 45\:\text{mV}\$$ higher, or $$\V_\text{BE}\approx 745\:\text{mV}\$$. You could go back through the process and come up with a refined quantitative calculation. But the above analysis is already close enough and BJTs vary, anyway. So there's no real need to do more.

(Also note that you are using a default BJT in Spice and I'm using more practical estimates for typical small-signal BJTs.)