# How does a diode-connected BJT work?

#### dikidera

Joined Sep 6, 2015
14
The information on the wiki is sparse. And the information elsewhere is the same .

All I understood is, the transistor always operates in the active region, and that there is a 0.7v drop.

However I built such a circuit on a breadboard. Here it is, simulated even.

I do not understand what happened . The voltage across the resistor is 11.7v(my real Vcc is 12.3),however if I use the ground as reference, it is 0.68v. Such is the voltage between base and emitter. And base collector are 0v as they are the same potential .

So how did a mere 0.7v drop become a 11.6(11.7)drop? And what happens to the Q point ? Base and collector are the same potential, is the Q point 0.7v as well

#### dl324

Joined Mar 30, 2015
16,110
All I understood is, the transistor always operates in the active region, and that there is a 0.7v drop.
Not when you short the CB junction. In that case, the transistor is just a diode.

To clarify, a transistor has 4 modes of operation and active is just one of them.

#### dikidera

Joined Sep 6, 2015
14
Not when you short the CB junction. In that case, the transistor is just a diode.

To clarify, a transistor has 4 modes of operation and active is just one of them.
Tying collector and base results in a configuration called Diode-connected Transistor

#### crutschow

Joined Mar 14, 2008
33,317
Just a long winded way of saying diode.
A diode connected transistor uses the current gain of the transistor to form a more ideal diode characteristic than just the base-emitter junction by itself.

#### Bordodynov

Joined May 20, 2015
3,116
See

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#### ebeowulf17

Joined Aug 12, 2014
3,307
A diode connected transistor uses the current gain of the transistor to form a more ideal diode characteristic than just the base-emitter junction by itself.
More ideal in which ways? Maybe I'm remembering wrong, but l feel like I've seen a lot of BJTs with low reverse Vbe ratings, like 5V or something. Am I remembering that wrong? Seems like that would be a notable drawback in many situations.

#### Bordodynov

Joined May 20, 2015
3,116
More ideal in which ways? Maybe I'm remembering wrong, but l feel like I've seen a lot of BJTs with low reverse Vbe ratings, like 5V or something. Am I remembering that wrong? Seems like that would be a notable drawback in many situations.
But for logarithmic and atilogarithmic amplifiers, what is needed!
Pick up the transistor 2SD2704K. It has a permissible emitter-base voltage of 25 volts.

#### crutschow

Joined Mar 14, 2008
33,317
More ideal in which ways?
Matching the ideal logarithmic diode current versus voltage curve.

#### WBahn

Joined Mar 31, 2012
29,472
The information on the wiki is sparse. And the information elsewhere is the same .
By tying the collector to the base, you create a two terminal device that behaves like a diode. But it is essentially a self-programming transistor.

A BJT is best thought of, in most circumstances, as a voltage-controlled device in the active region. The collector current is controlled by the base-emitter voltage. But since small changes in Vbe result in large changes in Ic, many circuits can be thought of as current-controlled devices with an essentially constant Vbe.

However, if we connect a BJT as a diode-connected transistor, then we can use external components, such as the resistor in your circuit, to set the collector current and the diode-connection allows negative feedback to result in the proper Vbe being established for that particular current.

If we then connect another well-matched transistor so that they share the same Vbe, the second transistor will attempt to produce the same Ic as the diode-connected one. Thus we can "mirror" the current from the diode-connected transistor into another one. This is this is the basis for countless circuits.

All I understood is, the transistor always operates in the active region, and that there is a 0.7v drop.
It doesn't always operating in the active region. The one region that is pretty much excluded is saturation. But cutoff is certainly on the table.

The drop is not a fixed, invariable 0.7 V. Like a diode, it is a voltage-controlled device. The more voltage across it, the more current through it (and vice versa).

However I built such a circuit on a breadboard. Here it is, simulated even.

I do not understand what happened . The voltage across the resistor is 11.7v(my real Vcc is 12.3),however if I use the ground as reference, it is 0.68v.
Huh? If you use the ground as reference the voltage across the resistor becomes 0.68 V instead of 11. 7 V? How does the choice of reference change the voltage across the resistor?

Such is the voltage between base and emitter. And base collector are 0v as they are the same potential .
Such is the voltage between base and emitter.... what?

So how did a mere 0.7v drop become a 11.6(11.7)drop?
Apples and oranges. The ~0.7 V is across the transistor while the 11.6 V is across the transistor. The sum is 12.3 V, which you stated was your actual Vcc. So KVL is satisfied.

And what happens to the Q point ? Base and collector are the same potential, is the Q point 0.7v as well
The Q point of the transistor is that it has a Vbe = Vce of ~0.7 V and an Ic of about 11.6 mA.[/QUOTE]

#### ebeowulf17

Joined Aug 12, 2014
3,307
Matching the ideal logarithmic diode current versus voltage curve.
Cool! Thanks to you and @Bordodynov for the explanations.

#### OBW0549

Joined Mar 2, 2015
3,566
These graphs of some measurements I made a year or so ago might illustrate the relationship of BJT base-emitter voltage to emitter current, and also show the significance of "ideal diode" behavior.

The first graph plots Vbe vs. Ie on a linear scale for both axes, for two different transistors operated in the "ideal diode" configuration (i.e., base shorted to collector): a 2N3904 general-purpose NPN switching transistor, and a 2N3866 medium-power RF NPN transistor.

It's easy to see how we get the generalization that "Vbe is always about 0.7 volts"; over nearly all of the 0 - 10 mA range of emitter current, the base-emitter voltage is roughly 700 mV. It isn't perfectly constant, since it does increase somewhat with current; and at very low currents the voltage does fall off significantly. But as a crude estimate of Vbe, "0.7 volts" works for some purposes, such as calculating approximate bias voltages in amplifiers.

But look what happens when we take the same data and plot it using a logarithmic horizontal scale for current:

Now it becomes clear that Vbe is actually a logarithmic function of Ie, and that for these two transistors that holds true over more than a 300,000,000-to-1 range of currents, from 30 pA to 10 mA.

This logarithmic relationship is true for the ideal diode configuration but breaks down when Vbe ≠ Vce. First, transistor current gain is generally not constant over that wide a range of collector currents, tending to decrease at the high and low ends of the scale, and this can introduce errors since Ic < Ie. And second, collector-base leakage current tends to dominate at the low end of the scale, introducing even more error. In a diode connected transistor, there is no C-B leakage since Vcb = 0. And third, the Early effect adds even more error.

Anyway, that's the story-- and I'm stickin' to it.