Author Topic: Constant current sink using NPN and negative supply.  (Read 1455 times)

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Offline ZeTeXTopic starter

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Constant current sink using NPN and negative supply.
« on: November 20, 2019, 06:18:50 pm »
Hello,



As shown in the following article:

https://www.analog.com/media/en/technical-documentation/tech-articles/lt-journal-article/LTJournal-V24N2-02-df-BenchSupply-Szolusha.pdf



There is a negative supply generator using LTC3632, the reason is to make a constant current sink to draw a few mA's to make sure the regulators regulate down to 0V as there is a minimum load current required.



However, I am not excaly sure how does the current sink work. I have simulated it in LTspice as shown in the following picture:





And as expected, V1 voltage does not matter as R1 always have constant voltage on it, which is ~4.3V (V2 + Vbe) so the current is constant.



How does such a circuit work? I have tried googling around but I haven't find a similar circuit. I would love to hear an explanation.



Thanks.  :)
« Last Edit: November 20, 2019, 06:45:35 pm by ZeTeX »
 

Offline w2aew

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Re: Constant current sink using NPN and negative supply.
« Reply #1 on: November 20, 2019, 06:35:41 pm »
My video on Current Sources and mirrors might help:

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Offline iMo

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Re: Constant current sink using NPN and negative supply.
« Reply #2 on: November 20, 2019, 07:14:15 pm »
Simplified explanation:
It works such the Vbe is always constant (aprox 0.7V for a reasonable Ibe, with -2mV/K TC), the V2=-5V is constant, and the R1 is constant.
The Ibe=(-5+Vbe)/R1 is therefore constant as well .
The transistor's Ice current Ice=beta*Ibe, where beta is the amplification factor (100-500), therefore Ice is constant, with "any" V1.
The Ice will flow through the R1 as well (I_R1=Ibe+Ice), thus increasing the voltage at R1. Increasing the voltage at R1 will cause decreasing the Ibe, however, that is why the Ice decreases as well. So it finds an equilibrium, where
0 = (Ice+Ibe)*R1 + Vbe - 5V
« Last Edit: November 20, 2019, 07:54:50 pm by imo »
Readers discretion is advised..
 
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Offline hermitengineer

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Re: Constant current sink using NPN and negative supply.
« Reply #3 on: November 23, 2019, 08:02:02 pm »
How does such a circuit work? I have tried googling around but I haven't find a similar circuit. I would love to hear an explanation.
In your circuit, the ground reference is the base of the transistor.  If you move the ground reference to the bottom of the resistor instead, you now have a common-collector amplifier.  VB, by that reference, is 5V so VE is, of course, 4.3V.
 

Offline unitedatoms

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Re: Constant current sink using NPN and negative supply.
« Reply #4 on: November 24, 2019, 03:02:59 am »
The circuit can be considered an emitter follower. In case of constant current load it does not make much sense. The resistor alone will do the same work.
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