7
\$\begingroup\$

I am trying to get a deeper understanding of how transistors actually work so I can design my own circuits around them.

Pictured is an LTspice simulation of a simple voltage follower circuit. I do not understand why the transistor base-emitter current is down in the microamps.

My current, evidently flawed, understanding is that the base-emitter current is equal to the base voltage (minus base drop) divided by the impedance to ground, in this case 1000 Ω. At 10 V base voltage this should yield a base emitter current of 10 mA, and thus a collector current of 10 mA × β should ensue. Yet this is not the case, and the voltage follower does its voltage following, what is the flaw in my understanding?

Current graph, circuit diagram

\$\endgroup\$
10
  • \$\begingroup\$ "My current, evidently flawed, understanding is that: the base current is equal to the base voltage (minus base drop) divided by the impedance to ground, in this case 1000 ohms." Replace "base current" with "emitter current". Then do your beta calculation backwards (i.e. divide instead of multiply). The SPICE parameter here would be BAF, so 200 in this case. \$\endgroup\$
    – Ste Kulov
    Commented Nov 12, 2022 at 6:45
  • \$\begingroup\$ Whoops. I mean BF. Doh! \$\endgroup\$
    – Ste Kulov
    Commented Nov 12, 2022 at 6:51
  • \$\begingroup\$ Sorry I meant Base Emitter current, I'll edit the original question. \$\endgroup\$ Commented Nov 12, 2022 at 6:56
  • \$\begingroup\$ Nope. No such thing. Just emitter current. Here. Do this. Plot I(R1) and right-click on its name in the waveform viewer and change the algebraic expression to I(R1)/200. Then plot Ib(Q1). See how they almost overlap? That's how the two currents are related to each other (mostly). \$\endgroup\$
    – Ste Kulov
    Commented Nov 12, 2022 at 7:03
  • 2
    \$\begingroup\$ As the name suggests this circuit is an emitter follower (voltage follower). Thus, Vout = Vin - Vbe = Vin - 0.6V. And the base current will be equal to \$I_B = \frac{I_E}{\beta +1} \$. So, for example, if \$V_{OUT}\$ is \$5V\$ the emitter current will be \$5mA\$ Thus, the base current will be beta + 1 time smaller than this current \$I_B = \frac{5mA}{\beta +1} = \frac{5mA}{200+1} \approx 25\mu A \$ \$\endgroup\$
    – G36
    Commented Nov 12, 2022 at 7:59

2 Answers 2

7
\$\begingroup\$

Let's first simplify the schematic without any loss of application to your schematic:

schematic

simulate this circuit – Schematic created using CircuitLab

In this case, it's not possible for the base of the NPN BJT to rise above the collector, to the BJT will always be in active mode. This means \$\beta\$ applies. Therefore also \$I_{_\text{E}}=\left(\beta+1\right)\cdot I_{_\text{B}}\$. Simple as that.

(This also strongly suggests that you may want to try sweeping your source towards voltages that are still higher than you tried to see if you detect a transition towards new behaviors. For example, try going up to \$+12\:\text{V}\$ instead of \$+10\:\text{V}\$ and see what happens then.)

Given the base voltage sweep limit to a maximum of \$+10\:\text{V}\$ we can apply the usual KVL of:

$$\begin{align*} V_{_\text{SWEEP}}-I_{_\text{B}}\cdot R_{_\text{B}}-V_{_\text{BE}} - I_{_\text{E}}\cdot R_{_\text{E}}&=0\:\text{V} \\\\ V_{_\text{SWEEP}}-I_{_\text{B}}\cdot R_{_\text{B}}-V_{_\text{BE}} - \left(\beta+1\right)\cdot I_{_\text{B}}\cdot R_{_\text{E}}&=0\:\text{V} \\\\\therefore \\\\ I_{_\text{B}} &= \frac{V_{_\text{SWEEP}}-V_{_\text{BE}}}{R_{_\text{B}}+\left(\beta+1\right)\cdot R_{_\text{E}}} \end{align*}$$

Clearly, \$R_{_\text{B}}\$ is tiny by comparison. So the above can be reduced to \$I_{_\text{B}} \approx \frac{V_{_\text{SWEEP}}-V_{_\text{BE}}}{\left(\beta+1\right)\cdot R_{_\text{E}}}\$. From which you will find that \$I_{_\text{B}}\$ remains small by comparison with the emitter current. It must.

Given the very low value for \$R_{_\text{B}}\$, the base is effectively driven directly by your DC sweep voltage. (The bulk base impedance of the BJT is typically an order of magnitude larger, in fact!) So the current in \$R_{_\text{E}}\$ is rather predictable. Discounting the low base current it is just \$I_{_\text{E}} \approx \frac{V_{_\text{SWEEP}}-V_{_\text{BE}}}{R_{_\text{E}}}\$.

Given that your sweep voltage spans from \$1\:\text{V}\$ to \$10\:\text{V}\$, which is just one order of magnitude at 1:10, I'd tend to initially expect the variation in \$V_{_\text{BE}}\$ to be about \$\pm 30\:\text{mV}\$ around the mid-point.

But that ignores the fact that \$V_{_\text{BE}}\$ is large compared to \$1\:\text{V}\$ at the low end of the sweep and takes up perhaps 2/3rds of what's available. So the low end is really about 3 times smaller, making the range 1:30. So an improved prediction is more like \$\pm 45\:\text{mV}\$ around the mid-point.

Assuming a small-signal BJT here, the midpoint should be close to \$V_{_\text{BE}}\approx 700 \:\text{mV}\$. So this places things such that: \$655 \:\text{mV} \le V_{_\text{BE}}\le 745 \:\text{mV}\$ (granting for now the midpoint value.) So \$345 \:\mu\text{A} \le I_{_\text{E}} \le 9255 \:\mu\text{A}\$. It looks like your \$\beta\approx 200\$, assuming I read the chart as \$I_{_{\text{B}_{\left(10\:\text{V}\right)}}}\approx 47.5\:\mu\text{A}\$, so then \$I_{_{\text{B}_{\left(1\:\text{V}\right)}}}\approx 1.73\:\mu\text{A}\$, which isn't far from the mark.

Your simulator is fine.

\$\endgroup\$
5
  • \$\begingroup\$ Thanks for your answer. I can see how how the application of KVL could solve for this simple case. Encouraged by your suggestion, I allowed the sweep voltage to go above the battery voltage and did encounter strange behaviour, where current on the order of 0.1A began to be drawn through the base. I saw similar effects when adding a resistor to the collector which caused this effect to begin at lower voltages. Could you recommend any complete resources (books or websites) that explain in-depth what is going on here? I'm also interested into why KVL method no longer applies here*. Thanks. \$\endgroup\$ Commented Nov 13, 2022 at 3:49
  • \$\begingroup\$ *I presume this has something to do with the transistor leaving the active region of base voltages? PS: I do not fully understand what you are talking about in the last few paragraphs. Would you mind clarifying for me? \$\endgroup\$ Commented Nov 13, 2022 at 3:51
  • \$\begingroup\$ @SirStrategic Yes, it's because the transistor becomes saturated and no longer is running in active mode. This is because there is another diode in the BJT -- the one between the base and the collector. When the voltage rises a few hundred millivolts above 10 V, then this BC diode becomes active lots of current then floods from the base towards the collector before turning around and heading back towards the emitter. So the base current is very high now without the collector current changing much. \$\endgroup\$
    – jonk
    Commented Nov 13, 2022 at 4:01
  • \$\begingroup\$ @SirStrategic An advanced book that will cover your future needs is Ian Getreu's Modeling the Bipolar Transistor. It's unique and used to be only available to Tektronix employees and customers. But I worked with Ian to get it into Lulu.com (I receive nothing if you buy a book; just Ian gets anything.) Another book I do like, and it is perhaps right at your level, is The Art of Electronics, 2nd or 3rd edition, by Horowitz & Hill. If you do get it, make sure you also get the "Student Manual" (to pair with the 2nd edition) or the "Learning the Art of Electronics" (to pair with the 3rd edition.) \$\endgroup\$
    – jonk
    Commented Nov 13, 2022 at 4:06
  • \$\begingroup\$ Thank you very much for answering my questions and for your suggestions, I'll definitely look into "the art of electronics". \$\endgroup\$ Commented Nov 13, 2022 at 4:37
7
\$\begingroup\$

Your faulty understanding is the first step to come to the correct conclusion:

Assume you had that large base current. It would cause a commensurate (actually much higher) current from collector to emitter.

This large collector current would raise the emitter voltage, even way above the base voltage. So in that scenario the base-emitter junction would become blocking and no base current would flow. But this would also cause the collector current to stop. Rinse and repeat.

This means that the system will either oscillate between the two states or reach an equilibrium, in which all of these conflicting rules are satisfied. The latter is the case where the emitter voltage is raised to such level that the current through the base-emitter-diode equals the collector current divided by the transistor current gain.

If you cut the connection between the collector and the battery, you inhibit any collector current and, as a result, reproduce your initial expectation.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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