I'm trying to solve the following circuit, obtaining all the currents and voltages.

enter image description here

I'm using mesh analysis method, but I'm ending with wrong results.

So I assumed by default a clock-wise loop for \$I_b\$ and the same for \$I_c\$.

The \$\beta=100\$

I setted the equations to obtain base current \$I_b\$:

$$5-50000(I_b-I_c)-0.7 = 0$$

and deriving from the \$beta\$ the collector current \$I_c=100 I_b\$ I obtained with substitutions:

$$4.3 - 50000I_b + 50000(100I_b) = 0 $$

that gives me:

$$I_b=0.87\mu A$$

this result is too small and different from the results obtained from the simulation with the software every circuit.

This problem should be very easy to solve. How can I set the equations correctly to solve this kind of problem?

Thank You.

  • 3
    \$\begingroup\$ The current through R1 is just \$I_b\$, not \$I_b-I_c\$. \$\endgroup\$
    – The Photon
    Commented Aug 2, 2020 at 23:50
  • \$\begingroup\$ @ThePhoton when i used the mesh analysis, the current of the other loop (in this case the one with collector current \$I_c\$) must be subtracted from the current of the first loop (base current). Anyway if I delete the collector current from the equation i will obtain 86uA as a result. But the simulation gives a current of 84.4uA. How can I get that result? \$\endgroup\$
    – Carlo
    Commented Aug 2, 2020 at 23:58
  • \$\begingroup\$ @Carlo 84.4uA and 8.6uA are close enough for this hand calculation. \$\endgroup\$ Commented Aug 3, 2020 at 0:00
  • \$\begingroup\$ @Carlo have you seen my answer? \$\endgroup\$ Commented Aug 3, 2020 at 0:02
  • 1
    \$\begingroup\$ @Carlo if you really want to be precise, you need to get the datasheet of the transistor and check the base current vs. Voltage diagram. The difference between your calculation and simulation should be that the voltage drop is not precisely 0.7 asset this current. \$\endgroup\$ Commented Aug 3, 2020 at 0:02

2 Answers 2


Quick Note

The schematic you show here is from LTspice, I believe. It's default NPN BJT has \$\beta=100\$ and \$I_\text{SAT}=100\:\text{aA}\$. Your "EveryCircuit" link is very unlikely to use the same default model. So LTspice probably will simulate different values. Just FYI.

Nodal Analysis

The nodal method is probably the easiest for solving this problem:

$$\begin{align*} \frac{V_\text{B}}{R_1}+I_\text{B}&=\frac{V_\text{CC}=5\:\text{V}}{R_1}\\\\ \frac{V_\text{B}}{R_1}+\frac{I_\text{SAT}}{\beta}\cdot\left(\exp\left[\frac{V_\text{B}}{V_T}\right]-1\right)&=\frac{V_\text{CC}}{R_1}\\\\ V_\text{B}+\frac{R_1\cdot I_\text{SAT}}{\beta}\cdot\left(\exp\left[\frac{V_\text{B}}{V_T}\right]-1\right)&=V_\text{CC}\\\\\text{conveniently set: }\quad V_{R_1\:\text{SAT}}&=\frac{R_1\cdot I_\text{SAT}}{\beta}\\\\V_\text{B}+V_{R_1\:\text{SAT}}\cdot\left(\exp\left[\frac{V_\text{B}}{V_T}\right]-1\right)&=V_\text{CC} \end{align*}$$

That solves out readily (see Appendix below for details) as:


Spice Comparison

From which, using LTspice parameters and only the simplified portion of the model besides, I get \$V_\text{B}=833.4\:\text{mV}\$ using the above formula. Running LTspice on this I get \$V_\text{B}=829.1\:\text{mV}\$ which I consider quite close since I'm using a highly simplified subset of the model that Spice programs use.

Short Summary

So that's how you solve these kinds of problems with mathematics. (Use Wolfram Alpha to solve the first equation, if you need to. It's not hard to do on paper, though.)

With that base voltage worked out, everything else just falls out readily.

In the case of your "EveryCircuit" simulator, you'd need to find out the model parameter values it uses to get close to its simulation values. That's a different problem. But I'm sure it uses similar techniques to other Spice programs.

Solution Appendix

The solution steps are that were skipped above are:

$$\begin{align*} V_\text{B}+V_{R_1\:\text{SAT}}\cdot\left(\exp\left[\frac{V_\text{B}}{V_T}\right]-1\right)&=V_\text{CC}\\\\V_{R_1\:\text{SAT}}\cdot\exp\left[\frac{V_\text{B}}{V_T}\right]&=V_\text{CC}+V_{R_1\:\text{SAT}}-V_\text{B}\\\\\frac{V_{R_1\:\text{SAT}}}{V_T}&=\frac{V_\text{CC}+V_{R_1\:\text{SAT}}-V_\text{B}}{V_T}\cdot\exp\left[-\frac{V_\text{B}}{V_T}\right]\\\\\frac{V_{R_1\:\text{SAT}}}{V_T}\cdot\exp\left[\frac{V_\text{CC}+V_{R_1\:\text{SAT}}}{V_T}\right]&=\frac{V_\text{CC}+V_{R_1\:\text{SAT}}-V_\text{B}}{V_T}\cdot\exp\left[\frac{V_\text{CC}+V_{R_1\:\text{SAT}}-V_\text{B}}{V_T}\right]\\\\&\text{swap sides and apply LambertW,}\\\\\frac{V_\text{CC}+V_{R_1\:\text{SAT}}-V_\text{B}}{V_T}&=\operatorname{LambertW}\left(\frac{V_{R_1\:\text{SAT}}}{V_T}\cdot\exp\left[\frac{V_\text{CC}+V_{R_1\:\text{SAT}}}{V_T}\right]\right)\\\\V_\text{CC}+V_{R_1\:\text{SAT}}-V_\text{B}&=V_T\cdot\operatorname{LambertW}\left(\frac{V_{R_1\:\text{SAT}}}{V_T}\cdot\exp\left[\frac{V_\text{CC}+V_{R_1\:\text{SAT}}}{V_T}\right]\right)\\\\&\text{and just solve for }V_\text{B} \end{align*}$$

Final Summary

Most folks just assume a base-emitter voltage (educated guess or good experience) and proceed on that assumption. It's very reasonable to do that, too. (I like to think so because that's how I usually do it.)

But when you ask:

How can I set the equations correctly to solve this kind of problem?

Then you have now opened the door to a different kind of answer, which I've provided here.

There is a way to produce a closed equation as a solution for an equation developed using nodal analysis and a simplified version of the non-linear hybrid-\$\pi\$ equivalent for the Ebers-Moll equation for BJTs. The above shows you how to do that.

(The LambertW function is such that: \$u=\operatorname{LambertW}\left(u\, e^u\right)\$ and is the inverse function for \$f\left(u\right)=u\, e^u\$. In short, it solves or \$u\$ when you know \$u\, e^u\$.)

The basic idea is pretty simple. But when you insert non-linear equations into the mix then some extra skills are required to get a closed solution.

This isn't the way it is solved in Spice programs, though. They use a set of steps, where a linearized version of the non-linear equations are used for each step incrementally, and eventually arrive at a very close (but numerical) solution. They don't try and create closed mathematical answers, as that can be almost impossible as a circuit's complexity increases.

  • \$\begingroup\$ I've work out some examples on a "Fixed Bias" circuit, and the linear equations work well until the values of beta multiplied for base current will exceed the voltage Vcc after that this approach that lacks the non linearity, it is completely wrong (Vce goes negative). @jonk can you provide some material for a correct understanding of these models? Thank You. \$\endgroup\$
    – Carlo
    Commented Aug 4, 2020 at 15:14
  • 1
    \$\begingroup\$ @Carlo I'll need to see your example to be sure of any answer I'd want to provide. Your writing isn't clear enough for me to apprehend the question, yet. (It sounds as though you have a collector resistor, though, which is a different question and involves saturation which your circuit here does not.) I'd like to add two links to examine, though, in the meantime: Level 1 Ebers-Moll Models for BJT and Discussion of Beta in three regions of operation. \$\endgroup\$
    – jonk
    Commented Aug 4, 2020 at 17:40

Current flowing through base will be 5-0.7/50000 = 0.000086A

Current flowing through the collector will IbxG = 0.000086x100=0.0086A

Voltage drop on the transistor (Vcollector - Vemitter) = 5V.

Voltage drop on the resistor will be 4.3V.

Anything else is wrong.If you simulator gives different results then it is wrong.First trust humans then machines.

  • \$\begingroup\$ Yes i saw 5k ohm resistor while it was 50k.I corrected it. \$\endgroup\$ Commented Aug 3, 2020 at 0:24
  • 1
    \$\begingroup\$ "5-0.7" should be in parentheses, though the result is correct. \$\endgroup\$ Commented Aug 3, 2020 at 13:42
  • \$\begingroup\$ @thebusybee the order is addition/substraction then multiplication/division. \$\endgroup\$ Commented Aug 3, 2020 at 14:09

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