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Old 15th June 2015, 02:33 PM   #1
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Question Some questions about this kind of current source

Hi there.

I saw this kind of current source many times but till now I just know IT IS a current source. I have no idea of how it work, cause it's different from those I am familiar.

So, please tell me how every elements which are circled work.

P.S. This is the first time I post here, I have no idea whether the picture is well added in this message.

[image]http://i4.tietuku.com/7a96770ab7931760.png[/image]
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Old 15th June 2015, 02:51 PM   #2
Mooly is offline Mooly  United Kingdom
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D1 and D2 are in series and generate volt drop of approx 0.65 volts each. So 1.3 volts in total. That forms a reference voltage. Applying that reference to the base of a transistor and adding a resistor to the emitter forms a current sink.

So you have 1.3 volts on the base. The base emitter volt drop is also around 0.65 volts (standard forward biased silicon junction) and so we have 1.3 - 0.65 volts across the emitter resistor.

0.65 volts across 270 ohm is 0.0025 amps (2.5 milliamps). The other transistor has 62 ohms as an emitter resistor and so the current is 0.65/62 which is 10.4 milliamps.

If you replaced the two diodes with a 6 volt zener then the currents would be (6 - 0.65)/220 and (6 - 0.65)/62

Yes
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Old 15th June 2015, 02:53 PM   #3
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It is just two current SINKs which both share same reference.
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Old 15th June 2015, 03:03 PM   #4
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Thank you for you answers. But how R3 and R4 work?
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Old 15th June 2015, 04:29 PM   #5
djoffe is offline djoffe  United States
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R3 supplies about 2.2 mA to bias the diodes and supply Q3 and Q4 base current.

R4 isolates off the base of of Q4 which is important when the amp is driven into clipping at the negative rail. In that case, the Q4 draws enough base current that without the isolation resistor R4, it would zero out (or at least dramatically reduce) the current for the input differential pair.

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Old 15th June 2015, 06:28 PM   #6
cbdb is offline cbdb  Canada
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The current trough a transistor is equivalent to the base emiter voltage , in the same maner as a diode, so tying the two together means the 2 currents will track together. Double the diode curent, doubles the transistor current. If the diode is wired from a transistor wich is matched with the transistor the 2 currents should be identical, and stay identical.
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