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lm317 sub?

KrisBlueNZ

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If you can keep both transistors at the same temperature you will see less variation. SMT will make that harder
I thought it would be the opposite. The output transistor is generating the heat, but its VBE is not a big factor; it's the VBE of the feedback transistor that determines the output current, and that transistor doesn't dissipate much power. So keeping them thermally separate should improve regulation, I would have thought...
 

KrisBlueNZ

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Yes, you can use a shunt regulator like a TL431LV instead of a transistor for Q2.

The ON Semiconductor CAT-102 looks like a good option:

http://www.digikey.com/product-detail/en/CAT102TDI-GT3/CAT102TDI-GT3OSCT-ND/3487509

It's USD 1.23 for 1-up and has a reference voltage of 0.6V, reasonably tightly controlled over temperature. Your current sense resistor would need to be 1.2Ω so you could either use a very accurate low-temperature-coefficient 1.2Ω resistor, or several resistors and a trimpot all in parallel, for example.

It is only available in SMT, and needs a DC supply of <18V (which you could provide with a simple zener shunt regulator, as its current consumption is pretty low).

There's a wide range of shunt regulators from various manufacturers in the 1.2~1.3V range. This means more voltage lost across the shunt, so less dropout voltage for the current limiter. Many of these options are available in THT, such as the LT1004: http://www.digikey.com/product-detail/en/LT1004CZ-1.2#PBF/LT1004CZ-1.2#PBF-ND/891840
 

docb

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Do you mean instead of Q1?
I don't really follow what you are suggesting.
 

docb

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Looking at these above, would this be used as is - just a drop in for the old 2222 I was using, with the load just goes between VCC and Isink?

With a 24v supply, is there be over 18v on the cathode/Q2base? I'm not sure I follow why I'd want that zener to create a supply.
 

KrisBlueNZ

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Looking at these above, would this be used as is - just a drop in for the old 2222 I was using, with the load just goes between VCC and Isink?
Right.
With a 24v supply, is there be over 18v on the cathode/Q2base? I'm not sure I follow why I'd want that zener to create a supply.
The CAT-102 that I suggested in post #23 requires a separate power supply connection, and the voltage must not exceed 18V. But it hardly draws any current from that pin, so a simple zener shunt regulator will be fine. Any voltage from 2.2~18V will do; 12V for example.
 

docb

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Ah ok, I now see that's for the CAT-102. The TL431LV looks like a drop in replacement, yes? Any reason why not use that since I'd need no separate supply.
 

KrisBlueNZ

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Ah ok, I now see that's for the CAT-102. The TL431LV looks like a drop in replacement, yes? Any reason why not use that since I'd need no separate supply.
Yes. The CAT-102 has a 0.6V reference voltage. The TL431LV has a 1.24V reference IIRC. So you would be losing 0.64V more across the current sense resistor, which is significant if you're already running close to dropout.
 

docb

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Ah, that is tempting, though I don't see a CAT-102 example circuit anywhere.
 

KrisBlueNZ

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OK, here you go.

270646B.001.GIF

CD should be connected as directly as possible between pins 2 and 5 of U1. If you get oscillation at Q1's base, break the connection at the X and add the compensation capacitor CC and resistor RC as shown.
 

docb

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Using a TL431 , if RS and Vref sets the current, when the VF changes in the LEDs that are the load, Vref changes. If it does, the curent changes. I don't see how this is constant current. What am I missing?
 

KrisBlueNZ

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I'm not sure what you're missing, but here's an explanation of the shunt-regulator-based constant current sink.

shunt regulator current sink.png
Assuming the hFE of the transistor is high, the base current is negligible, so IE and IC (which is the output current IOUT) are almost exactly the same. So the output current produces a voltage drop across RE, i.e. a voltage at VE, which is proportional to the output current.

The TL431 (or whatever shunt regulator you use) "watches" VE and as soon as it reaches the shunt regulator's internal reference voltage VREF (internal to the shunt regulator) it starts to conduct sharply through its anode-cathode path. This draws extra current through RB and makes VB drop. This makes the transistor conduct less strongly, so its collector voltage rises.

An equilibrium is reached where the transistor is conducting just strongly enough to produce enough emitter current to cause VE to match the shunt regulator's internal VREF. Since IC is approximately equal to IE, the output current is controlled to the same value.

Using a TL431 , if RS and Vref sets the current, when the VF changes in the LEDs that are the load, Vref changes.
Which VREF are you talking about? The VREF inside the shunt regulator? That doesn't change. VE changes, but the shunt regulator responds immediately to the change by conducting more or less strongly, so VB changes, and the transistor conducts more or less strongly, to keep VE equal to the shunt regulator's VREF.

The circuit actually regulates VE, and because VE appears across RE and is therefore proportional to IE, IE is regulated. Then because IC is almost identical to IE (the only difference is due to IB, which is, or should be, negligible), IOUT is also regulated.

This is why it's important to use a high gain transistor if you want good accuracy. A MOSFET is even better, because its gate current at DC is zero.
 
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