I was given this circuit for a practice test during one of my classes, but we were never shown how to actually solve it. It's just been bugging me so I wanted to know how others solved for all the different currents and voltages given only the values of the resistors, power supply and the beta dc value(175).(solving for Ie, Vb, Ve, and Vc):enter image description here

I sort of came up with a formula to solve for it by simulating it and then calling R2 a variable along with collector current. From there I just made this up and it works, but I'm not sure how a pro would solve it. I also don't know the name of this configuration as I couldn't find it in my textbook either.

Here is what I figured out I bet someone has already derived this but I couldn't find it, so I thought it was sort of cool. Comes out to R2 = (Vcc-Vbe-Ic*(R3-R1))/(Ic/Hfe) This picture might help with how I got that, although my math could be derp.enter image description here

My calculation came out off by 1k for the base resistor but I assume that's fine because the phone simulator only gave me two significant digits.


Just redrawing your schematic slightly and adding some labels:


simulate this circuit – Schematic created using CircuitLab

Just use KVL to start, following around through the base:

$$\begin{align*} V - I_{Rc}\cdot R_c - I_{Rb}\cdot R_b - V_{BE} - I_E\cdot R_e &= 0 \end{align*}$$

If the BJT is in its active region where your value of \$\beta=175\$ applies (and you'll know one way or another, soon enough), then it also follows that:

$$\begin{align*} I_{Rc} &= I_C+I_B=I_E \\ I_{Rb} &= I_B \\ I_E&=\left(\beta+1\right)\cdot I_B \end{align*}$$

Applying those to the original equation, we get:

$$\begin{align*} V - I_E\cdot R_c - I_B\cdot R_b - V_{BE} - I_E\cdot R_e &= 0 \\ \\ V - \left(\beta+1\right)\cdot I_B\cdot R_c - I_B\cdot R_b - V_{BE} - \left(\beta+1\right)\cdot I_B\cdot R_e &= 0 \\ \\ V &= V_{BE} + I_B\cdot\left[\left(\beta+1\right)\cdot\left(R_c + R_e\right) + R_b \right] \\ \\ I_B &= \frac{V - V_{BE}}{R_b+\left(\beta+1\right)\cdot\left(R_c + R_e\right)} \end{align*}$$

And that pretty much cracks the puzzle.

(It ignores little-re, which might have an impact in some cases but probably doesn't, here. It's impact is probably below 1%. But you could work it back into the equation later, if it matters to you.)

At this point, given your values and using \$V_{BE}=700\:\textrm{mV}\$, I get:

$$\begin{align*} I_B&\approx 10.1\:\mu\textrm{A} \\ I_E&\approx 1.78\:\textrm{mA} \end{align*}$$

So, I'd estimate:

$$\begin{align*} V_E&=I_e\cdot R_e\approx 1.78\:\textrm{V} \\ V_C&=10\:\textrm{V}-I_E\cdot R_C\approx 6.44\:\textrm{V} \\ V_B&=V_E+V_{BE}=V_C-I_B\cdot R_B\approx 2.48\:\textrm{V} \end{align*}$$

Since \$V_{CE} \gt 1\:\textrm{V}\$, the BJT is in its active region and the value of \$\beta=175\$ can be considered to have applied, now that we can check it out. So it is fine to stop at this point and consider the question answered well enough.

Little re is because the base-emitter junction's thermal voltage can be treated as a tiny "battery" at the tip of the BJT emitter. It's always got the thermal voltage just sitting there, which at room temperature will be around \$26\:\textrm{mV}\$. Given a current through it (\$I_E\$), you can turn that into an equivalent resistance. This is called a lot of things, but in just talking I say "little re." In this case, \$re\approx 15\:\Omega\$.

This value adds to \$R_E\$ in the above calculations. Working it out, I find that it impacts the estimated current values by about 0.3%.

Just a note.

  • \$\begingroup\$ Thanks so much, this is exactly what I needed to read. Those are the exact values I found when simulating the circuit. \$\endgroup\$ – Ragecoder Nov 5 '16 at 23:58

To solve this, write equations for each piece you know. If you do that right, you have a number of independent equations that matches the number of variables. Then you solve the set of simultaneous equations.

Things that you should be able to write equations for:

  1. Collector current as a function of base current.

  2. Base current as a function of C-E voltage.

  3. Emitter current as a function of base and/or collector current.

  4. Voltage across R1 as function of collector current.

  5. Voltage across R2 as function of emitter current.

  6. C-E voltage as function of voltages across R1 and R2

Seems like that should do it. Some of these can be simplified by inspection.

  • \$\begingroup\$ Okay, going to run through it again. When you solve for these values do you assume that Ic = Ie? \$\endgroup\$ – Ragecoder Nov 5 '16 at 21:56
  • 1
    \$\begingroup\$ @Ragecoder no, Ie = Ib + Ic. When you solve transistor problems you always assume at first that it is forward biased. So in that case, Ie = Ib + Ic. \$\endgroup\$ – nettek Nov 5 '16 at 21:59
  • \$\begingroup\$ You don't by chance know what the name of this configuration is called do you? \$\endgroup\$ – Ragecoder Nov 5 '16 at 22:10
  • \$\begingroup\$ @Ragecoder The configuration is judged by the fact which of the three transistor electrodes is common to the input and output terminals of a particular circuit (from the point of view of a signal path). Here only the operating point is solved and the signal terminals are not defined, so this question does not give much sense. \$\endgroup\$ – Eric Best Oct 18 '17 at 18:16

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