How to measure DC Operating Point of a simple transistor circuit

Question

Could someone explain to me how to find the DC Operating point of this circuit? I need to find it by running a simulation and probing the circuit, not solving for it mathematically, and I'm not sure how to do that.

I know it has to do with the intersection of the DC load line and the Ib line on a Vce vs. Ic graph, but I'm not sure how to simulate that. I can plot the base current, but I don't know how to get the DC load line on the graph to find the intersection.

Thanks in advance.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

This seems to be an academic type of a question. So there are some undefined prerequisites. Are the parameters of the NPN exactly known? If yes, you only have to measure the voltage on Rc or Re. You then know the collector or emitter current and can calculate the base current. As there is no small signal source in your circuit you have already gained the DC operating point.

There's one exception: If Vce derived from your measurements was very small (i.e below 2V), the BJT is probably in saturation. You then have to measure the base current separately, because current amplification doesn't work the normal way, then.

If the parameters aren't known, you have to measure Ib anyway. This can be done by measuring the Voltage on Rb1 or Rb2 and applying KCL.

However, without any signal source the DC operation point is always equal to the current state of this circuit.

Answer 2

I assume you use the base as an input.

The base potential is set by the voltage divider \$R_{b1}\$ and \$R_{b2}\$. \$V_b=V_t\cdot\frac{R_{b2}}{R_{b1}+R_{b2}}=15\cdot\frac{56k}{166k}\approx5.1V\$. The base-emitter junction is like a forward-biased diode, dropping about 0.7 volts, therefore the emitter potential is \$V_e=V_b-0.7=4.4V\$.

Now the emitter bias current (which is approximately the same as the collector current) can be calculated as \$I_{e0}=\frac{V_e}{R_e}=\frac{4.4}{5.1k}\approx0.86mA\$

The voltage drop from the collector to the emitter is approximately the following:

\$V_{ce0}\approx V_t-V_e-I_{e0}\cdot R_c=15-4.4-0.86\cdot5.1=15-8.8=6.2 V\$

Answer 3

I need to find it by running a simulation

So, enter the circuit into your preferred simulation tool (like CircuitLab, for example), and run an operating point analysis.

In CircuitLab you just need to add a 15 V source to your circuit, then go to simulation mode:

1

From here, just press the button that says "Run DC Solver" and wait about 0.05 s. Then you'll be able to move a "probe" cursor around the circuit to see the solution. If you want the branch currents also, you'll need to add in current meter elements on the branches you want to measure.

Answer 4

For a DC operating point solution:

1) Download LTspice, for free, here.

2)Download this file (your LTspice schematic), left-click on it, and then run it by clicking on the running man icon on the menu bar.

The simulation is a DC operating point simulation, and you'll see a table displaying all the DC data for the circuit.

"X" it out and the schematic will come back up.

Mouse the cursor over any wire and the voltage on it will show up at the bottom of the screen or, if you mouse over any component, a hand will come up and the current through that component will be displayed at the bottom of the screen.