Is this circuit acting as an amplifier and how does it work?

Question

I simulate a diode connected BJT circuit as follows:

And here are the plot for the currents through Rin I(Rin); the wire then collector-base then to base-emitter junction Ic(Q); and through the base-emitter junction Ib(Q):

Almost 99% of the current is flowing through the wire which shorts the collector and the base.

It seems the like majority of the electrons flowing from the ground into the emitter-base junction are pulled by the collector and then all the way to the source. Is this circuit acting like an amplifier? How is the collector pulling the electrons even though its terminals are shorted i.e zero potential difference across CB junction?

Answer 1

What you have is called diode connection configuration. Basically, Vb is always a diode voltage drop higher than Ve. Shorting Vc to Vb only helps with biasing the transistor.

Your circuit is equivalent to this:

^{simulate this circuit – Schematic created using CircuitLab}

Your transistor acts as a diode. This kind of configuration can be used in current mirrors when you want a forward voltage drop from a diode to match the drop from another transistor such as in Wilson current source. But it is useful only from transistors on same die.

You can also use this to replace a diode in a circuit if you will. At low voltage the voltage drop is usually quite small. Yet, you are better off using a proper diode unless this is part of a much more complex circuit where such an arrangement is needed (eg. wilson current source).

Answer 2

VBE is about 0.7v.

That means VCE is also 0.7v. For all reasonable transistors, this means it's in the linear region, as it's above VCEsat, which tends to be in the 0.1 to 0.4v region. It's not the zero VCE that matters, but the 0.7v VBE, and the >VCEsat VCE.

That means the collector current is beta times the base current. So yes, it is acting as an amplifier. A figure of 100 is often used for beta (when no better figure is to hand) so 99% of the current going into the collector is to be expected.

Answer 3

In a diode-connected BJT like this, the BJT is kept right at the edge of active vs saturated mode and so only a small base current is required. Most of the "diode current" flows via the collector to the emitter.

The exact value of \$V_\text{BE}\$ isn't immediately obvious. But it will be whatever is required for the collector current. In this case, from your plot, it appears to be \$V_\text{BE}=800\:\text{mV}-(100\:\Omega+330\:\Omega)\cdot 170\:\mu\text{A}=726.9\:\text{mV}\$.

Suppose that \$\beta=100\$. Then you'd expect to see \$I_\text{C}=\frac{100}{100+1}\cdot 170\:\mu\text{A}=168.32\:\mu\text{A}\$ and therefore the base current would be the remainder, or \$I_\text{B}=170\:\mu\text{A}-168.32\:\mu\text{A}=1.68\:\mu\text{A}\$.

Your plot would appear to be consistent with this idea.