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In a recent comment I was asked why a diode-connected transistor behaves as it does.

Jonk's answer here addresses this question, and explains that the geometry of the B-E junction in a signal BJT yields closer-to-ideal behaviour (by which I mean it conforms more closely to the diode equation with n=1) than a regular point-junction type diode.

This is fine, but the implication is that you wouldn't then need to join collector and base in order to obtain near-ideal behaviour, and yet we do. Apparently there's a significant benefit to sourcing the bulk of the current from the collector.

My question is, then, why is collector current, which ultimately has to traverse the same B-E junction, apparently not subject to the same rules as current injected directly via the base terminal? It's as if collector current is bypassing the B-E junction altogether (below right), instead of having to traverse that junction (below middle).

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Does it help to note that Vce(sat) < Vbe? The mechanisms explaining this, are relevant to the diagrams. \$\endgroup\$ Feb 11 at 3:49
  • \$\begingroup\$ @TimWilliams That's an underlying question, which I also wonder about. How is it possible for base potential to remain 0.5V higher than the collector, when both are technically connected to the same chunk of P-type? There must be a substantial potential gradient between base terminal contact and the planes of the two PN junctions. I'm still lost. \$\endgroup\$ Feb 11 at 3:55
  • \$\begingroup\$ @SimonFitch The details start with two papers, from Hall and Shockley & Reed, with the authors corresponding with each other during publication. The result is Hall-Shockley-Reed theory. Some is covered better here and here. But it's a global model param and only very approx. \$\endgroup\$ Feb 11 at 6:22
  • \$\begingroup\$ Your question implies that B-E saturation conduction (with C open) have worse ideality. Have you confirmed that this be the case? \$\endgroup\$
    – tobalt
    Feb 11 at 8:41
  • \$\begingroup\$ @tobalt I have not experimented. Are you referring to comparing V/I curves with and without collector connected to base or comparing \$I_S\$? \$\endgroup\$ Feb 11 at 9:20

3 Answers 3

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This question made me think again about the philosophy of this active diode arrangement and for the umpteenth time try to make sense of it. Now I have the feeling that I finally succeeded...

Basic idea

Imagine a 2-terminal device (resistor, diode, transistor input, etc.) through which we pass current and measure the voltage across it. For some reason, we want to pass a small fraction (e.g. 1/100) of the current through the device. So, we shunt the device with a current sink (controlled by the current through the device) which absorbs the rest (99/100) of the current. It acts as a dynamic resistor and the whole circuit as a dynamic current divider.

Implementations

Resistor

This example has no particular practical value in our case; I have added it rather to introduce the concept.

Resistor only: If in a current-supplied (Rb = 1 kΩ) resistor circuit...

schematic

simulate this circuit – Schematic created using CircuitLab

... with linear IV curve...

STEP 1.1

Resistor||resistor: ... we add another (low-resistance) resistor (Rc = 10 Ω) in parallel, a 1/100 of the current will be diverted through Rc; hence the name current divider.

schematic

simulate this circuit

When we sweep Iin, we see that the IV curve of the equivalent resistor Rb||Rc is linear.

STEP 1.2

Resistor||current sink: A more sophisticated solution is to shunt Rb with a current-controlled current source CCCS (Q) instead the humble resistor. There is no real need for it here; I have only used it to introduce the next concept.

schematic

simulate this circuit

STEP 1.3

Diode

Diode only: Then, if in a current-supplied D diode circuit...

schematic

simulate this circuit

... with nonlinear IV curve...

STEP 2.1

Diode||resistor: ... we add a resistor Rc in parallel, a part of the current will be diverted through Rc.

schematic

simulate this circuit

But when we sweep Iin, we see that the IV curve of the equivalent diode-resistor circuit differs significantly from the one above. The simple ohmic resistor does not work here because the diode behaves as a nonlinear "resistor" that changes its "resistance" when Iin varies, and the current divider's ratio varies as well.

STEP 2.2

Diode||current sink: So, to keep the ratio constant, we need a dynamic resistor (CCCS) like in the Schematic 1.3 above to make a dynamic current divider with constant current ratio of 100.

schematic

simulate this circuit

Now the divider's ratio is constant regardless of Iin changes, and the IV curve has its usual shape.

STEP 2.3

Transistor

Unloaded: Let's initially connect a perfect voltmeter as a "load" (open circuit).

Base-emitter junction: Similarly, in a current-supplied base-emitter junction...

schematic

simulate this circuit

... with the same nonlinear IV curve...

STEP 3.1

Base-emitter||collector-emitter junction: ... we can add the collector-emitter part in parallel to the base-emitter one.

schematic

simulate this circuit

Now the divider's ratio is constant (the transistor β), and the IV curve has its usual shape.

STEP 3.2

Loaded: Let's now examine the two circuits above at several loads (10 kΩ, 100 kΩ and 10 MΩ).

Base-emitter junction: When the collector is not connected to the base...

schematic

simulate this circuit

... we see that the output voltage is strongly affected by the load (even a small 10 kΩ load significantly changes the curve shape). This is because a very small base current flows and the base-emitter junction has a very high static "resistance".

STEP 3.2.1

Base-emitter||collector-emitter junction: If we connect the collector to the base...

schematic

simulate this circuit

... we see that the load slightly affects the curve shape (even a significant 1 kΩ load does not change it). The reason is that a high collector current flows, and the collector-emitter part has a very small (β times less) static "resistance". Actually the input current is the same but it is redistributed.

STEP 3.2.2

Applications

Current mirror

The input (current-setting) transistor Q1 of a BJT current mirror is a typical application of this trick. If we connect only its base-emitter junction to the base-emitter junction of the output transistor Q2, the circuit will act as a current amplifier with a gain of β. So we need to drain the excess input current.

schematic

simulate this circuit

For this purpose, we connect the Q1 collector-emitter part in parallel to its base-emitter junction. It diverts the excess input current, and the circuit acts as a current follower (in respect to the magnitude).

schematic

simulate this circuit

So, the role of the active diode Q1 in the current mirror is to attenuate the input (collector) current β times. Then Q2 amplifies it β times (of course, the current direction is reversed which is the goal there). The point of doing this in such an extravagant way is to compensate for various disturbances (temperature, β, etc.)

Conclusions

General

  • In such a current divider configuration consisting of two 2-terminal devices in parallel, the current flowing through the input device is decreased a certain number of (K) times by diverting to the output shunting device.

  • The current divider is dynamic. So when the input current varies and the input device changes its "resistance", the output device also changes its "resistance" so that to keep a constant current ratio.

  • Since the current flowing through the output element is K times higher than the current through the input element, the output "resistance" is K times lower; so low-resistance loads can be connected.

  • Because the input and output elements are connected, there is negative feedback, and the circuit acts as a shunt regulator.

Specific

  • In the active diode configuration, the current flowing through the base-emitter junction is decreased β times by diverting a significant portion through the shunting collector-emitter part. As a result:
  1. Low base current flows through the base-emitter junction, and the IV curve has a good shape.

  2. High collector current flows, and the output resistance is low.

Experimental results

At the beginning of 2017, I participated in such a discussion, and decided to investigate in the laboratory the two circuits - without connection (simple diode) and with connection (active diode) between the base and the collector. For this purpose, I used a rather old data acquisition system MICROLAB (AD periphery connected to an Apple computer) that we had made back in the 80s. The photos are bad, but you can still see the difference between the two IV curves - the second one (with connection) is much steeper, almost vertical.

Base-emitter IV curve

Base-emitter IV curve

Active-diode IV curve added

Active-diode IV curve

Here is a link to the movie taken with phone in hand.

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    \$\begingroup\$ @ Circuitfantasist - "the second one (with connection) is much steeper, almost vertical." Yes, but one should add: Still an e-function. \$\endgroup\$
    – LvW
    Feb 11 at 16:47
  • \$\begingroup\$ @LvW, Yes... The problem is that the graphs are flattened on the left side of the screen because this was a setup for testing various diodes (LEDs, Zener, etc.) some of which had quite a high VF. \$\endgroup\$ Feb 11 at 16:54
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I believe there are two things going on.

  1. The parasitic base resistance is usually much higher that the parasitic collector and emitter resistances, so passing most of the current between collector and emitter reduces the parasitic resistance.

  2. The ideality factor for a diode depends on the recombination mechanism. Transistor collector current doesn't. Minority carriers in the base that find their way to the collector don't recombine: they simply become majority carriers in the collector.

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  • \$\begingroup\$ I agree on #1. I'm not sure about #2. In a pn-junction, the ideality factor \$n\approx 2\$ at low bias is a result of recombination in the depletion region (not in the quasi-neutral regions). Regardless of whether the transistor is diode-connected or the collector is left disconnected, all of the current still passes through the base-emitter junction, so I would expect similar ideality factors at low bias. If someone is willing and able to test a BJT in these two configurations, it would be very interesting to see the comparison. \$\endgroup\$
    – Puk
    Feb 11 at 18:08
  • \$\begingroup\$ @Puk But the part that reaches the collector does not recombine, it merely passes through. \$\endgroup\$
    – John Doty
    Feb 11 at 18:21
  • \$\begingroup\$ Whether a carrier recombines in the base or makes it to the collector shouldn't affect ideality associated with its current contribution. It's like comparing a long diode and a short diode (where the quasi-neutral regions are shorter than the carrier diffusion length). In the absence of recombination in the depletion region, both follow the ideal Shockley equation. \$\endgroup\$
    – Puk
    Feb 11 at 18:44
  • \$\begingroup\$ I would expect that at a given \$V_{BE}\$, the non-ideal current associated with recombination in the base-emitter depletion region to be the same in either configuration. However, since the collector current results in a larger "ideal" current component in the diode-connected case, it would overtake the non-ideal component at a smaller \$V_{BE}\$, so the I-V curve would be ideal (\$n\approx1\$) in more of the bias range. Maybe this is what you mean, and in that case I would agree. \$\endgroup\$
    – Puk
    Feb 11 at 18:44
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Doty is partly right (mechanism #2).

To understand the behavior of collector and base currents the detailed mechanisms of carrier transport need to be understood. It's not possible to explain in a short comment (that's what books on device physics are for). But briefly: the collector current is determined by the electron concentration at the edge of the neutral base, and this is ideally determined by the Boltzmann factor. The base current is a fraction of this current PLUS holes injected to recombine with electrons that find recombination centers in the base depletion region. This second current component is highly non-ideal.

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  • \$\begingroup\$ What's wrong with point #1? \$\endgroup\$
    – John Doty
    Feb 11 at 17:57

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