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Matt Edwards

Sun May 09 2010, 10:30PM

Registered Member #2838 Joined: Fri Apr 30 2010, 07:55PM
Location: tehachapi, CA
Posts: 333

I am thinking about setting up this circuit but have a few questions. It is a bit more complicated then the circuit that I originally intended to use.

They may seem simple questions but I am a still rookie here.
1. What is the -10 on Q1 and Q2. Does that mean that I will have two -10v supplies?
2. Can someone recommend a resistor rating or part number for Rsc? Is a "current sense resistor" the same as a "current shunt resistor"?
3. Do I have to use electrolytic caps for the 0.1uF and 10uF or is there a better option?

hboy007

Sun May 09 2010, 11:35PM

Registered Member #1667 Joined: Sat Aug 30 2008, 09:57PM
Location:
Posts: 374

1) Q1 and Q2 are wired as jFET constant current sources which are quite handy as they eliminate the voltage drop changes across the reference diodes that would otherwise occur during supply voltage variations. I suggest you do not use standard rectifiers but Zener diodes at ~ 5.1V with suitable filtering to cut the noise. At that voltage, avalanche and tunnel effect temperature dependence cancels out quite nicely. You can also use dedicated voltage reference components across the pots (i forget the types, sorry).
2) choose a resistor that creates negligible voltage drop and power dissipation during normal operation. LM317 wants 1.25V across its leads so Rsc creates some negative feedback to limit output current. That's about what you can set at the potentiometer. You want maximum output current when the pot is at the upper limit, so 1.5A should cause 1.25V to drop across Rsc, I think. 1R 2W low inductance resistors will do the job.
3) everything is better than electrolytic caps so if you don't like them, get some MPK foil caps that are impulse rated and add a small ceramic cap as bypass. You can't do anything wrong with these

Matt Edwards

Mon May 10 2010, 12:31AM

Registered Member #2838 Joined: Fri Apr 30 2010, 07:55PM
Location: tehachapi, CA
Posts: 333

Thank you! Do you have any recommendations as to how i should go about the -10v supply? Originally I was going to use a rewound transformer but I am not sure how to get -10v out of it.

So to clarify what you are saying because I am definitely new at this:
1) If I add a 1ohm 2w resistor as Rsc I should be good?
2) I should replace the 1N4001-1N4002 with zener diodes? Or, are you referring to the bridge rectifier from the transformer?
3) I have a few polyester film caps around. This should be decent right?

Matt Edwards

Mon May 10 2010, 01:13AM

Registered Member #2838 Joined: Fri Apr 30 2010, 07:55PM
Location: tehachapi, CA
Posts: 333

I found this diagram on one of the data sheets (figure 26). Can I use a similar formula to determine Rsc? Perhaps I am looking at this wrong but in order to get a close estimate I should be able to use this right?

Vref(1.25v)/Rsc =Iout (=1.5A)

Vref- Is listed as 1.25 average on the data sheet.
Rsc- In the diagram would be the resistor value that I need to determine.
Iout- I would like to be around 1-1.5A.

So I am working with:
1.25v/(x-ohms)=1.5A
(x)=0.83 ohms +/- 0.01

Like I said, I could be completely mistaken. If so, does anyone have a formula that I can use?

radiotech

Mon May 10 2010, 01:18AM

Registered Member #2463 Joined: Wed Nov 11 2009, 03:49AM
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Posts: 1546

I would hang an additional 0.1ufd from adjust1 of LM317-1 to ground too. Call it mistrust of high gain regulators that may oscillate.

hboy007

Mon May 10 2010, 10:24AM

Registered Member #1667 Joined: Sat Aug 30 2008, 09:57PM
Location:
Posts: 374

Oidium45, the formula you mention is the relation I used to get to the 1 Ohm resistor, just derating the current a bit for easy availability of the shunt resistor.

I seem to have failed to clearify what I had in mind with the zener diode. A simple model to describe diode forward voltage drop is the Shockley model

that uses just a majority emission coefficient and the reverse saturation current as fundamental parameters. Inverting the function gives you an idea how temperature drift of the voltage drop across the diode changes your reference voltage as the power supply heats up. To overcome this, the above mentioned effects in Zener diodes are exploited. Have a look at the LM329, it would go well with the < 0.4% drift of the LM317

(see p.6) - at the cost of constructing a temperature stable voltage divider to get back into the 0.. 1.3V range. maybe you can use 1k and 10k linear pots of the same type, one as the actual current dial and the larger one as a shunt. The drift coefficients will match and hence compensate the drift of the voltage divider in total.

Steve Conner

Mon May 10 2010, 10:46AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

To get the -10V supply you'll need another winding on your transformer, with another rectifier and smoothing capacitor.

It's tricky to make a power supply that varies all the way down to zero without the help of an auxiliary negative supply. Here is my own effort.

As you can see, it's quite complex. The circuit you found is about the simplest PSU you can make that will adjust down to zero on both voltage and current.

Temperature coefficient of this circuit won't be very good, you can improve it by swapping each series pair of diodes for a LM385-1.2 voltage reference, then it will be as good as mine.

If you can't get the JFETs, likely any JFET will do as long as the Idss isn't too high, or use a BJT current source.

How to choose Rsc? Size it so that it drops 1.25V across it at your desired current limit, which must be less than 1.5A as that's the regulator's own internal current limit. For instance, if you wanted the current limit variable between 0 and 1.25A, you'd choose 1 ohm. The power rating: P= I^2*R. In this case, 1.56 watts. So you want a 1 ohm, 2 watt resistor. It doesn't need to be something sold as a "current shunt resistor".

hboy007

Mon May 10 2010, 11:13AM

Registered Member #1667 Joined: Sat Aug 30 2008, 09:57PM
Location:
Posts: 374

just a remark on the negative voltage: as the power consumption is low, you might as well get along with another bridge rectifier capacitively coupled to the sec of your transformer. This way it is also quite easy to set up a capacitive voltage divider.

or perhaps not (I'm checkin on that... might be a trap!)

Steve Conner

Mon May 10 2010, 11:22AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

Hey, good idea.

I'll leave you to try to explain it to the original poster, though.

Bjørn

Mon May 10 2010, 12:37PM

Registered Member #27 Joined: Fri Feb 03 2006, 02:20AM
Location: Hyperborea
Posts: 2058

One thing to note is that the regulators usually are not current limited at rated current (which is just a guarantee under specified conditions). The real limit is usually temperature so they can safely deliver a lot more (or less) than their rated current depending on the heatsink. So in some cases it would make sense to go for a 3 A limit instead of 1.5 A.

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