# How can I manually analyse this simple BJT circuit?

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?

• 'Manually' emitter and collector current are equal, which would make your Vc = 20 V - 3.43 mA * 10 kOhm = -14.3 V, which is impossible as base and emitter are near +15 V. Jun 3 at 9:48
• Hence collector and emitter currents are not equal in this circuit. Yet the current flows through it. Jun 3 at 9:57
• It looks like a microcap schematic so, why didn't you run a dynamic DC response? It would show all the voltages and currents and clearly indicate that the transistor is in saturation. Then, if you need an explanation you might get a decent one. Jun 3 at 10:28
• @Thomas Anderson - Yes of course, you're right. So the majority of the emitter current flows through the base, not the collector. Expecting saturation would bring Vc at Ve + 0.3V = 14.6 V more or less. Then Ic = (20 - 14.6)/10k = 0.54 mA, and Ib = 3.43 - 0.54 = 2.89 V. Jun 3 at 10:30

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.

1. 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.

1. 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}\$$.
2. Apply an assumed $$\V_{_\text{CE}}\approx 10\:\text{mV}\$$ to get $$\577.5\:\mu\text{A}\$$
3. 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.

• @ThomasAnderson Most likely having to do with the model's saturation current being quite different from that of LTspice. The LTspice model I have uses about 2.5 fA. What does your model say? Default operating temperature could be another reason, though. LTspice uses 27 C as the default. Non-ideality (emission coefficient) could be yet another reason. LTspice uses 1 for that value in this case. But to get a smaller value you'd need a number less than 1. That's not likely. So I don't think that's the cause here. Jun 3 at 12:56
• @ThomasAnderson The usual suspect for such differences is the saturation current. First thing that comes to mind for me. With a difference by a factor of about 8 you'd expect to see a difference of about 50 mV. (26 mV times ln(1/8), or times ln(8), depending on point of view.) There's also the base bulk resistance, too, though. You may want to check the model's value there, as well. My model says 10 Ohms, looking at it. Jun 3 at 23:14
• @ThomasAnderson The fact that you see RB=0 tells me that the model is unrealistic. Normally, you will see on the order of tens of Ohms for RB. Should be around a few tenths of an Ohm for RC and RE. But never zero for any of them. So I'd be very suspicious of that model. That doesn't mean I have any confidence in the model from LTspice. But I don't have any specific reason to worry about it, as I do for the one you appear to see there. I just checked the LTspice model against the one I have from the old ORCAD early 1990's library and it mostly matches up. So it's probably okay. Jun 4 at 11:10
• @ThomasAnderson It's not easy to measure all the parameters for BJTs. It used to be that large BJT customers would buy testing equipment (the Tektronix STS group is an example of a supplier of such instrumentation such a customer might buy) and develop their own models. But today, the BJT manufacturers or the Spice simulator companies (ORCAD used to do this) would develop models from those results. It's only in very rare cases now, my opinion, that a product idea may depend upon matching up discrete BJTs. I think some engineers supporting scientific research, perhaps. But few others. Jun 4 at 11:11
• @ThomasAnderson .model Q2N5551 NPN(Is=2.511f Xti=3 Eg=1.11 Vaf=100 Bf=242.6 Ne=1.249 Ise=2.511f Ikf=.3458 Xtb=1.5 Br=3.197 Nc=2 Isc=0 Ikr=0 Rc=1 Cjc=4.883p Mjc=.3047 Vjc=.75 Fc=.5 Cje=18.79p Mje=.3416 Vje=.75 Tr=1.202n Tf=560p Itf=50m Vtf=5 Xtf=8 Rb=10) * National pid=16 case=TO92 * 88-09-07 bam creation *\$ That's the ORCAD one. The LTspice one adds RE=0.1 though. Jun 4 at 11:12

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}

• Thank you for your answer. Your answer is correct as well, but this site does not allow me to accept two answers and periblepsis has answered the question earlier. I simulated this circuit with Micro-Cap 12, and it gave me Vce < 0.001 V, not about 0.1 V as you say. Is that possible at all, or the simulation is not very accurate? Jun 3 at 11:52
• @ThomasAnderson It's possible to have a lower $V_{CE}$, especially at very low collector current, but 1mV is suspiciously small. Jun 3 at 13:32
• Seems suspicious to me too. pixhost.to/show/428/357624328_circuit.png Jun 3 at 14:58