How can I manually analyse this simple BJT circuit?

Question

I can calculate Ve and Ie, assuming that Ve = Vb - 0.7 = 14.3 V.

Ie = (14.3 - (-20))/10000 = 3.43 mA.

How can I manually find Vc, Ic and Ib?

Answer 1

You know that \$V_b=+15\:\text{V}\$. You know you have approximately \$35\:\text{V}\$ across the \$10\:\text{k}\Omega\$ emitter resistor, so the emitter current will be slightly less than \$3.5\:\text{mA}\$. A reasonable guess, without model parameters like the saturation current, would have the base to emitter voltage at about \$750\:\text{mV}\$ for such emitter currents. So we can improve the emitter current estimate, slightly, and find \$3.42\:\text{mA}\$, instead. No point in worrying that one any further.

(If you care to confirm the base-emitter voltage, compute out \$26\:\text{mV}\cdot\ln\left(\frac{3.42\:\text{mA}}{1\:\text{fA}}\right)\$. I've no idea if the saturation current is that tiny. It might be 3 or 5 times higher for all I know. But it gets things in the ballpark. The base bulk resistance may be the next factor to consider, as these values are typically in the tens of Ohms. And at these emitter currents this can account for an added \$30\:\text{mV}\$ to \$90\:\text{mV}\$.)

The collector voltage cannot be below the emitter voltage and if the collector current were first taken to be about the same as the emitter current (active mode), then the collector's \$10\:\text{k}\Omega\$ resistor would drop almost \$35\:\text{V}\$. Obviously, that's not happening. So the BJT is saturated, not in active mode. So the collector voltage can be taken to be very close to the emitter voltage -- perhaps just slightly above it by a few tens of millivolts. (Mid-1960's silicon BJTs might be in the hundreds of millivolts.) So the collector resistor will have about \$5\:\text{V}+750\:\text{mV}\$ across it. That means the collector current is about \$575\:\mu\text{A}\$.

The base current is then \$3.42\:\text{mA}-575\:\mu\text{A}=2.845\:\text{mA}\$.

These are rough estimates. But quickly arrived at.

I haven't run this in Spice. But now I will, using LTspice and its model for the 2N5551:

Seems like LTspice found similar answers.

To summarize:

For the emitter current estimate:

1. Start with a guess that \$V_{_\text{BE}}=700\:\text{mV}\$ and estimate the emitter current as \$\frac{+15\:\text{V}-700\:\text{mV}-\left(-20\:\text{V}\right)}{10\:\text{k}\Omega}=3.43\:\text{mA}\$.
2. Sanity check the \$V_{_\text{BE}}=700\:\text{mV}\$ estimate by computing \$26\:\text{mV}\cdot\ln\left(\frac{3.43\:\text{mA}}{1\:\text{fA}}\right)\approx 750\:\text{mV}\$. If you decide this is close enough to (1), then move on. Otherwise, refine (1) to use the new estimate and find the emitter current to be \$\frac{+15\:\text{V}-750\:\text{mV}-\left(-20\:\text{V}\right)}{10\:\text{k}\Omega}=3.425\:\text{mA}\$.
3. Assume \$r_b\approx 10\:\Omega\$ and figure that the voltage drop is then \$10\:\Omega\cdot 3.425\:\text{mA}\approx 35\:\text{mV}\$.
4. Add (3)'s value to \$750\:\text{mV}\$ to find \$785\:\text{mV}\$ as the final estimated \$V_{_\text{BE}}\$ and recompute the emitter current now as \$3.4215\:\text{mA}\$.

There's little point in going any further than this. So proceed to determining if the BJT is saturated or active.

5. Given \$I_{_\text{E}}=3.4215\:\text{mA}\$ and assuming tentatively the idea that \$I_{_\text{E}}\approx I_{_\text{C}}\$ when in active mode, estimate the voltage drop across the collector resistor as \$10\:\text{k}\Omega\cdot 3.4215\:\text{mA}=34.215\:\text{V}\$. If this drop holds then the collector voltage would be \$+20\:\text{V}-34.215\:\text{V}=-14.215\:\text{V}\$. This isn't possible as it is far below the earlier estimated emitter voltage of \$+15\:\text{V}-785\:\text{mV}=+14.215\:\text{V}\$. So the BJT is in saturated mode.

This means the collector current has little bearing to the emitter current and that the base current will likely make up an overly large proportion of the emitter current.

6. Estimate the saturated collector current, taking \$V_{_\text{C}}\approx V_{_\text{E}}\$, as \$\frac{+20\:\text{V}-14.215\:\text{V}}{10\:\text{k}\Omega}=578.5\:\mu\text{A}\$.
7. Apply an assumed \$V_{_\text{CE}}\approx 10\:\text{mV}\$ to get \$577.5\:\mu\text{A}\$
8. The base current will be the difference, or \$3.4215\:\text{mA}-577.5\:\mu\text{A}=2.844\:\text{mA}\$.

We are already way past the precision allowed by our assumptions. But close enough and done.

Answer 2

First find out if the transistor is saturated.

By inspection alone, since R1 and R2 are equal, if we assume if the currents through them are approximately equal, \$I_C \approx I_E\$, the voltages across R1 and R2 would be equal too. This is clearly not possible, with the emitter well above 0V. Collector potential has fallen as far as it is possible to fall, and the transistor must be saturated.

More rigourously, if \$I_C \approx I_E\$ (base current is negligible), consider what maximum current can flow in R1 and R2. Maximum current occurs with the transistor just saturated and conducting as well as it ever can do, with near-zero resistance. In that state, current flowing will be, by Ohm's law, the ratio of total potential difference between the supplies, and the total resistance between them:

$$ \begin{aligned} I_{C(MAX)} &= \frac{(+20V) - (-20V)}{R_1 + R_2} \\ \\ &= \frac{40V}{20k\Omega} \\ \\ &= 2mA \end{aligned} $$

In that just saturated condition, emitter potential would be:

$$ \begin{aligned} V_{E(MAX)} &= -20V + I_{E(MAX)}R_2 \\ \\ &= -20V + 2mA \times 10k\Omega \\ \\ &= 0V \end{aligned} $$

Since the emitter has risen way beyond that, to +14.3V, collector potential has fallen to meet it, with \$V_{CE}\$ at a minimum. The transistor is definitely saturated.

What will happen in practice, using the ratio of R1 and R2 shown, as emitter potential rises, the collector falls,and the two eventually meet in the middle, near 0V. If the emitter then continues to rise, as is the case here (dragged upwards by the base-emitter junction, as the base continues to rise), the transistor can't conduct any better, \$V_{CE}\$ stays at a minimum near zero, and the collector must begin to rise also. A simulation shows this behaviour. If I sweep base potential from -20V to +20V, watch what happens to the collector (orange) and emitter (blue) potentials:

The region to the right of the green marker, where collector potential rises again, is where the transistor is saturated.

In saturation, \$V_{CE}\approx 0.1V\$. This means that collector potential \$V_C\$ is:

$$ \begin{aligned} V_C &= V_E + V_{CE} \\ \\ &= 14.3V + 0.1V \\ \\ &= 14.4V \end{aligned} $$

Voltage \$V_{R1}\$ across R1 is:

$$ \begin{aligned} V_{R1} &= (+20V) - V_C \\ \\ &= 20V - 14.4V \\ \\ &= 5.6V \end{aligned} $$

Collector current will be:

$$ \begin{aligned} I_C &= \frac{V_{R1}}{R_1} \\ \\ &= \frac{5.6V}{10k\Omega} \\ \\ &= 560\mu A \end{aligned} $$

Finally, by KCL we know that emmitter current is the sum of base current and collector current:

$$ \begin{aligned} I_E &= I_B + I_C\\ \\ I_B &= I_E - I_C \\ \\ &= 3.43mA - 0.56mA \\ \\ &= 2.87mA \end{aligned} $$