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My textbook says that the current through resistor will be same as the collector current when the diode shown matches with the base-emitter diode internal to the transistor. Could somebody kindly explain how is this possible.

enter image description here

So far I have : $$I_R = \dfrac{V_{cc}-V_{be}}{R}$$

I'm struggling a bit in calculating how this current splits into diode and the base of the transistor. Any help is appreciated. Thanks!

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    \$\begingroup\$ Is that all your textbook had to offer? Some assumptions have to be made about the current gain of the transistor for such an equation to hold true. As it is, it is a constant-current sink because the diode locks the Vbe. In real-world conditions the gain of the transistor would create a slightly larger current flow that is fixed as long as the diode maintains its Vdrop. \$\endgroup\$ – user105652 May 20 '16 at 3:26
  • \$\begingroup\$ That is exactly the question that's troubling me. Let me take a screenshot of the textbook page and attach. Thank you :) \$\endgroup\$ – AgentS May 20 '16 at 3:28
  • \$\begingroup\$ Here it is postimg.org/image/oazt0rfb5 \$\endgroup\$ – AgentS May 20 '16 at 3:34
  • \$\begingroup\$ I saw the pictures. They are true current mirrors if the resistor is being driven by another transistor, then that transistors current is duplicated (and inverted) at the collector of the second transistor. The book may call these current mirrors under ideal conditions, but in the real-world R would not connect to +Vcc. Assuming ideal conditions the resistor current and the collector current would be the same or very close. Is that a good enough answer? \$\endgroup\$ – user105652 May 20 '16 at 3:46
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    \$\begingroup\$ Because the b-e junction of the transistor acts a lot like a diode, it creates 2 diodes in parallel, and they share the current under ideal conditions. \$\endgroup\$ – user105652 May 20 '16 at 3:58
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The key to the effect can be found in the photographed text:

If the compensating diode and the emitter diode have identical current/voltage curves

From the schematic, the voltage across the diode and the voltage across the base-emitter junction have identical voltages (they are, after all, shorted together at base and ground). This means that each will have the same current flowing through them. If this were not true, their voltage/current curves would be different.

As it happens, the book is not quite correct. The two currents will not be perfectly identical, since all of the diode current is provided by the anode, while the emitter current is equal to the base current plus the collector current. This means that the collector current \$i_C\$ and diode current \$i_D\$ are related by $$\frac{i_D}{i_C}=1+\frac{1}{\beta} $$ where $$\beta = \text{ transistor gain} $$Since beta is ordinarily large, the difference is small, but it's not zero.

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  • \$\begingroup\$ Thank you! This makes a lot of sense now. Did you get that relation by setting $i_D = i_E$ ? \$\endgroup\$ – AgentS May 20 '16 at 5:23
  • \$\begingroup\$ shouldn't we be setting $i_D = i_B$ instead ? Base emitter voltage develops base emitter current, which is $i_B$ right ? \$\endgroup\$ – AgentS May 20 '16 at 5:31
  • \$\begingroup\$ @rsadhvika - Right. Since diode voltage equals base voltage, diode current equals base current. \$\endgroup\$ – WhatRoughBeast May 20 '16 at 11:15
  • \$\begingroup\$ You're right about the diode current, but the question is about the resistor current, which is slightly larger still. See my diagram. \$\endgroup\$ – Dave Tweed May 20 '16 at 15:33
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If the diode and transistor are perfectly matched, then the thing you can say about them is that the diode current and the emitter current will be the same in this circuit. But it is necessary to also account for the transistor's base current IB:

schematic

simulate this circuit – Schematic created using CircuitLab

Start iwth IB. The collector current is β×IB and the emitter current is (β+1)×IB. If the diode is perfectly matched, then its current is also (β+1)×IB, which means that the total current through the resistor must be (β+2)×IB, which is slightly larger than the collector current by a factor of 1+2/β.

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This is very similar to a simple current mirror.

schematic

simulate this circuit – Schematic created using CircuitLab

Notice how Q2's collector and base are shorted together ? This is effectively making Q2 a diode.

Which makes this circuit just like your circuit, except, rather than a diode, this circuit is using a transistor as a diode.

Collector current is controlled by Vbe by Ebers-Moll equation. Since Q2 sets the Vbe for both Q2 and Q1, then the current in Q1 is controlled directly by Q2 through resistor R_ref.

Similar behavior is observed in the circuit you presented. The voltage at the base of the transistor is determined by the voltage across the diode. You get more accurate results when D and Q2 (base emitter) are have the same i-v curves and characteristics and high beta values for the transistor.

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