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

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?

• We know from the Kirchhoff's current law that I_Rin must be equal to I_Re. Therefore (Ib + Ic) = Ie. And electrons prefer to flow through the collector because the base is "too narrow". – G36 Mar 29 '18 at 18:26
• There is electric field between CB and simulation shows potential difference. Okay this pulls up the electrons and creates a path to the source. But CB terminals are also shorted which also means electric field should be zero but it is not. So even terminals are shorted still there is electric field. This was my confusion. – pnatk Mar 29 '18 at 18:30
• But there is electric field between CE also. So electrons can flow to the collector terminal. – G36 Mar 29 '18 at 18:35
• What happens if you short a diode? Is there still electric field across its junctions? There is something called barrier potential across the junction right?. Can that be the field across CB which is pulling up the electrons diffused to the base. Im not sure though. – pnatk Mar 29 '18 at 18:38
• The base is "narrow" and emitter has a lot of "free" electrons (n+ region). And this electrons from the emitter will quickly fill all the holes in the base. So the remaining part must flow to the collector. static.righto.com/images/741/transistor_npn.png. And because you have collector shorted with the base. The collector also "see" the battery "electric field" (voltage). – G36 Mar 29 '18 at 18:45

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).

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.

• Thanks so it acts as an amplifier. Btw in simulation there is a potential diff(causing electric field) across CB junction to pull up the electrons from the base?. But CB terminals are shorted which also means zero potential difference but there is potential difference. That was part of my question. – pnatk Mar 29 '18 at 18:27
• Does your simulation connect collector and base by a wire, or a resistor? If a wire, there will be no voltage betwen them, if a resistor ther will be. Make your mind up. – Neil_UK Mar 30 '18 at 6:04

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.

• In simulation there is a potential(electric field) across CB junction to pull up the electrons from the base?. But CB terminals are shorted which also means zero potential difference but there is potential difference. That was my question. – pnatk Mar 29 '18 at 18:24
• @panicattack In Spice simulation there is NO voltage potential across the CB junction. But do you mean something else? – jonk Mar 29 '18 at 19:45
• Yes sorry I meant in SPICE or empirically there is NO voltage potential across the CB junction.. But we are measuring it from the CB "terminals". Like a disconnected diode. If you tie a disconnected diode's terminals together we might think there is no potential difference across the pn junction. But inside there is still a barrier potential around 0.7V. So I was wondering in my question's case whether this barrier potential across the CB pulling up the electrons to the collector. Because across CB there is zero potential but inside there is barrier field. Is that correct? – pnatk Mar 29 '18 at 19:55
• What pulls the electrons upway to the collector region from the base? Which field/potential? (CB terminal pot. iz zero) Is that the intrinsic barrier pot.? – pnatk Mar 29 '18 at 19:57
• @panicattack That's a different question and it gets into the physical model within the BJT and depletion. There is a LOT of background material (a few chapters' worth) just to get to the point where that question can be meaningfully answered. (My opinion -- others may have different opinions about that.) My answer would have to be very substantially expanded. If you wanted to be able to derive quantitative results, it would need even more expansion and then I'd also have to double-check myself to make sure my own recollections weren't in error, too. More work. – jonk Mar 29 '18 at 21:24