Voltage follower simulation low base current

Question

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?

Answer 1

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

Answer 2

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.