enter image description here So is it right if I apply Kirchoff's law : Vcc=I1R1+I2R2+I3Re if I1 is the current that flows in R1 resistor, I2 in R2 and I3 in Re

  • 2
    \$\begingroup\$ That is really a closed circuit if you draw the +10V voltage source into the circuit, and you can apply Kirchoff's law if it is ok to ignore the transistor at this part of the analysis. \$\endgroup\$ – Dean Franks Feb 13 at 16:07
  • \$\begingroup\$ @DeanFranks KVL applies around any closed loop, regardless of what else it's connected to, so the transistor doesn't come into it anyway. \$\endgroup\$ – Hearth Feb 13 at 16:11
  • \$\begingroup\$ Equation is correct, but it might not be very helpful for what you're trying to do. \$\endgroup\$ – stretch Feb 13 at 21:11

KVL and KCL are applicable for finding the DC operating point of the entire circuit and not only the specific path you've indicated. To apply KVL to your loop and make the reasoning for it manifest, let me draw up the circuit slightly differently:


simulate this circuit – Schematic created using CircuitLab

Your KVL equation is just as you wrote, but the above circuit shows the loop just a little bit more plainly:

$$\begin{align*}0\:\text{V}+V_\text{CC}-I_{R_1}\cdot R_1-I_{R_2}\cdot R_2-I_{R_E}\cdot R_E&=0\:\text{V}\\\\\therefore V_\text{CC}&=I_{R_1}\cdot R_1+I_{R_2}\cdot R_2+I_{R_E}\cdot R_E\end{align*}$$

If that's all you want to write, you could consider it done. And be right. But most of the time this leaves a lot yet to be desired. We expect that somehow those three currents must be knowable. After all, if you wire up such a circuit something will happen and you should be able to measure voltages between nodes in the circuit or otherwise measure currents with an inserted ammeter somewhere. (And there's more than one KVL loop and certainly some KCL equations that could be also written down, too.)

So just stopping your analysis with obvious but entirely unresolved currents is not very satisfying, even if it's been written down correctly for as far as it goes. (Not far enough, one might imagine.)

Here's a nodal analysis using KCL, for example, that's more satisfying (to me) since it covers the entire circuit and not only one loop in it and therefore promises a bit more:

$$\begin{align*} \frac{V_\text{B}}{R_1}+\frac{V_\text{B}}{R_2}+I_\text{B}&=\frac{V_\text{CC}}{R_1}+\frac{V_\text{E}}{R_2}\\\\ \frac{V_\text{E}}{R_2}+\frac{V_\text{E}}{R_\text{E}}&=I_\text{E}+\frac{V_\text{B}}{R_2}+\frac{0\:\text{V}}{R_\text{E}} \end{align*}$$

If the BJT is in active mode and isn't saturated, then the above covers the complete circuit (despite seeing no equation including a term for \$R_\text{C}\$.) Because, in that case, the collector will be "acting as if" it were a current sink and the collector current is determined by the difference in voltage between the base and the emitter of the BJT. So there's no need for an equation including \$R_\text{C}\$ in order to solve it (assuming an active mode result.)

(The assumption is testable, once you solve it using that assumption. If you attempt the solution and then find that it's impossible for the BJT to be in active mode with all the values worked out, then you've falsified the assumption and therefore the equations and you can then solve the problem using a different approach that assumes the BJT is saturated and not operating in active mode.)

The problem with the two KCL equations above is that you have two equations with four unknowns. But before I grapple with that problem, take a look at the following chart. It's a fuller picture of the base and collector current curves for a typical BJT. It covers three regions, which are labeled as \$I\$, \$II\$, and \$III\$. For most designs you will encounter, the BJT is operated in region \$II\$. Look closely, and I think you can see that the base current and the collector current have a fixed (constant) factor that relates them, in region \$II\$. (The lines are parallel in that region on the log chart.) This factor is called \$\beta\$ and it somewhat useful, despite the fact that different BJTs will have different values for it and that it varies with temperature, because it is fairly constant (at constant temperature) in region \$II\$ for any specific BJT you have in your hand. The other two regions are rarely used, in practice; and part of the reason is because the relationship between the collector and base currents in those other regions isn't nearly as fixed, anymore, and other effects become important (and therefore increase the difficulties involved in managing them along with everything else that needs to be managed.)

enter image description here

Given region \$II\$ active mode operation (the right assumption to make), you know that \$I_\text{C}=\beta\cdot I_\text{B}\$ and that \$I_\text{E}=I_\text{C}+I_\text{B}=\left(\beta+1\right)\cdot I_\text{B}\$.

Now recall something I said above about region \$II\$ active mode operation: "the collector current is determined by the difference in voltage between the base and the emitter of the BJT." The equation for this is often just called the Shockley equation and shows up for both diodes and active mode BJTs. It's an important equation to memorize. So you should. For region \$II\$ active mode NPN BJTs, it's:


The meaning of \$I_\text{SAT}\$ appears in the chart I gave above. It's the point you get when you apply a ruler on the region \$II\$ curve and draw a hypothetical line backwards to the y-axis. Different BJTs will have different values for it and this is an important model parameter. For small signal BJTs, the value is very small: perhaps a dozen or two femptoamps. For larger signal BJTs, it may be several orders of magnitude larger. Smaller values of \$I_\text{SAT}\$ mean larger base-emitter voltages for any given collector current. So think of this model parameter as controlling the \$V_\text{BE}\$ of a BJT for any desired collector current.

The meaning of \$V_T=\frac{k\cdot T}{q}\$ is both simple and tricky. The tricky part is that it's determined by fundamental principles in statistical thermodynamics and doesn't only apply in electronics fields. It's basic physics stuff, with \$k\$ being the Boltzmann constant and \$q\$ being the charge magnitude of an electron or proton. The simple part is that at room temperature the value is about \$26\:\text{mV}\$ and that's pretty easy to remember. (It moves around by about \$86\:\mu\text{V}\$ per degree Kelvin.)

Of course, for the NPN, \$\mid\:V_\text{BE}\mid\:=V_\text{B}-V_\text{E}\$.

So, now you can move forward with the earlier KCL equations. Look closely below and keep in mind what I wrote above and you should be able to work out the replacements I made here:

$$\begin{align*} \frac{V_\text{B}}{R_1}+\frac{V_\text{B}}{R_2}+\frac{I_\text{C}}{\beta}&=\frac{V_\text{CC}}{R_1}+\frac{V_\text{E}}{R_2}\\\\ \frac{V_\text{E}}{R_2}+\frac{V_\text{E}}{R_\text{E}}&=I_\text{C}\frac{\beta+1}{\beta}+\frac{V_\text{B}}{R_2}+\frac{0\:\text{V}}{R_\text{E}}\\\\ I_\text{C}&=I_\text{SAT}\left(e^{\left[\frac{V_\text{B}-V_\text{E}}{V_T}\right]}-1\right) \end{align*}$$

Now, there are exactly three equations and three unknowns. And that's solvable for the circuit. A closed solution is tricky and would involve the LambertW function. I'll provide a solution below, though, just for completeness. (Although I can refer you to research papers that routinely use the LambertW for problems like this, few engineers have any practice applying it for closed solution forms. And it's not really necessary, since you can solve this iteratively, by hand.)

For an iterative approach, you can assume that the initial values of \$V_\text{BE}=0\:\text{V}\$ and that \$I_\text{C}=0\:\text{A}\$ and then solve for \$V_\text{B}\$ and \$V_\text{E}\$ (which will initially be the same as \$V_\text{B}\$) and then estimate a new value of \$I_\text{C}=\frac{\beta}{\beta+1}\left(\frac{V_\text{E}}{R_\text{E}}-\frac{V_\text{BE}}{R_2}\right)\$. Once you have \$I_\text{C}\$, you can plug that into the Shockley equation to get a new, refined value for \$V_\text{BE}\$. With the new estimated \$I_\text{C}\$ and \$V_\text{BE}\$ you can estimate a new \$V_\text{B}\$ and \$V_\text{E}\$, yielding a new value of \$I_\text{C}\$ and \$V_\text{BE}\$. And this can be repeated over and over until the values settle down.

It's not fast to converge in this case, but the idea works.

The closed solution using the LambertW function is:

$$\begin{align*} i_0&=\frac{V_T}{R_2}&r_0&=R_1 + R_2 + R_\text{E}\\\\ i_1&=\frac{V_\text{CC}}{r_0}&r_1&=R_1+\left(\beta+1\right)R_\text{E}\\\\ k_0&=\beta\cdot\frac{r_0}{r_1}&r_2&=\beta\cdot R_1+\left(\beta+1\right)\cdot R_2\\\\ i_w&=i_0\cdot \operatorname{LambertW}\left(\frac{I_\text{SAT}}{k_0\cdot i_0}\cdot e^{\left[\frac{i_1}{i_0}\right]}\right)\\\\ \text{then,}\\\\ I_\text{C}&=k_0\cdot i_w\\\\ V_\text{B}&=R_1\cdot i_w\cdot \frac{\beta\cdot R_\text{E}-R_2}{r_1} +V_\text{CC}\cdot\frac{R_2+R_\text{E}}{r_0}\\\\ V_\text{E}&=R_\text{E}\cdot\left(i_1+i_w\cdot\frac{r_2}{r_1}\right) \end{align*}$$

In your case, using \$I_\text{SAT}=20\:\text{fA}\$ and \$\beta=150\$, this solves out as \$I_\text{C}\approx 713\:\mu\text{A}\$, \$V_\text{B}\approx 3.208\:\text{V}\$, and \$V_\text{E}\approx 2.576\:\text{V}\$. However, keep in mind that the BJT model is more complex than I've indicated. So if you run this on Spice the computed values will be based upon other model parameters as well and the values will be slightly different. (But not by much.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.