How to 'increase resistance range' of 10 turn bourns potentiometer

[mdijkens]

Hi,

I am trying to build a simple low noise linear power supply based on a LT3083.
I want to be able to set the voltage range at least from 3.3V to 9V but even better from 1V to 12V
According to the datasheet I would need a variable resistor of 20k per volt controlled, so a 220k potentiometer in series with a 20k resistor would be ideal.

However, searching on eBay, I can't find any Bourns 10 turn potentiometers above 200k and only one very expensive ($35) 200k.
Now I do have a good quality high grade 100k Bourns 10 turn potentiometer, but I can't figure out a way to have it behave like a 220k 
I was wondering if there's a simple and easy way to get the desired result with a 100k pot? It just has to provide resistance to ground in such a way that the LT3083 can 'drain' it's 50uA of constant current with the desired voltage drop over the variable resistance (20k per volt)

I'm really looking for a simple solution with a 1 or 2 commodity parts

Any ideas?

[Vtile]

As far as I know there is no such trickery atleast with passive elements.

Since the parallel connected resistances are smaller than the smallest resistance involved you can not use that route and if you connect a fixed series resistor, you can not reach zero.

With purely passive elements you could also apply a coarse and fine pot method.

Since I'm not particularly a electronics wizard I do not try to invent an active method to do the trick, but I'm sure that there is a simplistic solution with a pot and transistor(s)

http://cds.linear.com/docs/en/datasheet/3083fa.pdf
Page 18 last schema... Go from there.

[jeroen79]

You could augment the 50 uA current source with another one so that more current flows through the pot.
Adding another 50 uA would make a 100k pot act as a 200k pot.
(in this particular application)
Of course, it would have to be a sufficiently stable and noiseless current source.

Another option is to switch a 100k resistor in series with the pot to switch to a high range.
Perhaps even have a rotary switch with 12 20k resistors and then a 20k pot.
Or a coarse and a fine pot.

Otherwise use an opamp to buffer the set voltage as in the schematic Vtile reffered to: DAC-controlled regulator on page 18.
Instead of a DAC you would use a voltage reference and pot.

[Vtile]

Hah! Use mechanical solution.. Attach two 110k 10 turn pots to one knob with rubber belt and some pulleys.   

I do not have anything serious to add. Jeroen79s solutions above sounds solid.

[jeroen79]

Another idea: use an opamp to 'amplify' and 'raise' the pot.
http://tinyurl.com/ydgdmsle
The voltage divider on the left should have a much much higher resistance than the pot.

[StillTrying]

As you only need to go down to min. 1V you can raise the 50uA up to anything you want this way.



www.eevblog.com/forum/chat/the-state-of-regulators-for-linear-power-supplies-a-rant-(!)/msg1222220/#msg1222220

[jeroen79]

That would indeed work quite well.
both fixed resistors would end up with the same voltage drop so the extra current could simply be sourced from the output.

[StillTrying]

"the extra current could simply be sourced from the output."

It also saves having to deal with the regulators min. load current when there's no load connected.

Lose about 100mV to 1V across the resistors so that small mV changes between the regs ADJ and OUT don't have much effect on the voltage regulation.

[mdijkens]

StillTrying, Thanks for your advise!
I kind of understand it a little bit 
But when changing the pot, the output voltage changes and so is the current added via the 22 ohm resistor, which makes the voltage even higher? What do I not understand ? 
For my situation, could I just use the 1k, 22 and 100k pot to get the range of 1V to 12V out? I find it difficult to calculate that now since I don't fully understand it and for me the LT3083 is quite an expensive part which I don't want to break experimenting with it 

[StillTrying]

The low value 22R was just to get the min output voltage as low as possible - 50mV, as you don't need the min that low make the 22R and 1K larger.

If you make the current through the output resistor 5.0mA you'll then have 5.05mA flowing through your 10-turn, a 2K2 10 turn will get you very close to 500mV to 12V output. 

[onlooker]

The datasheet says ,

"The LT®3083 is a 3A low dropout linear regulator that can
be paralleled to increase output current or spread heat on
surface mounted boards."

You may parallel two (with the two share the same Rset).  There is an example of parallel of 4 in the datasheet.

[jeroen79]

No, the current through the 22 ohm resistor is determined by the voltage applied to it.
Which in turn is determined by the 1k resistor and the 50uA current from the SET pin.

The pot has no effect on this and will just carry the combined current.

[mdijkens]

But will the current through the output resistor not change when I change the output voltage?
If I set the output to 12V, the current will be 12x higher then when I set the output to 1 volt?

With my 100k pot, having 0.1mA through the output resistor will already give me the 1V to 12V range, right?
But how to calculate the output resistor since the output is not constant

[mdijkens]

Ow... Does this mean the current flows from SET to OUT ?
I figured it the other way around 

So in my case the output resistor should be 500 ohm then? (50mV/0.1mA) ?

[onlooker]

As you only need to go down to min. 1V you can raise the 50uA up to anything you want this way.



www.eevblog.com/forum/chat/the-state-of-regulators-for-linear-power-supplies-a-rant-(!)/msg1222220/#msg1222220

The formulas are

       V_out = I_set*(R3+R_1k + R3*R_1k/R₂₂)

       I_R3 = I_set(1+R_1k/R₂₂)

If you make R_1k = R₂₂ = 1kR, then you got what you wanted (double "I_set" through R3)

       V_out = I_set*(2*R3+1k)

       I_R3 = I_set*2

[StillTrying]

"Ow... Does this mean the current flows from SET to OUT ?"

No, all the current from SET and OUT flows through your pot, it's a very constant current of 50uA + 5mA.

I've worked out the values for O/P from 0.19V to 12.8V, adjust the R values to suit especially if you're can't get a 2k5 10-turn.

[mdijkens]

Aha, I think I finally got it 

Just to test if I'm ok now:
- the 3k9 on SET gives a Vdrop of 0.195 compared to Vout
- so a 2k7 on OUT would give me 70uA coming from OUT
- so 120uA over 100k pot gives me 12.22 Vdrop
- 12.22 + 0.195 = 12.415V max OUT and 0.195V min OUT
OK?

And the transistor with 220R is just to bring SET down to turn output on/off ?
I don't want to use that part since I have a voltmeter connected to OUT to set the output with the pot...

[bson]

However, searching on eBay, I can't find any Bourns 10 turn potentiometers above 200k and only one very expensive ($35) 200k.

Anything "Bourns" on eBay is likely to be fake anyway.  It's not the place to buy electronic components unless you enjoy getting ripped off.  A quick search found 56 different kinds of Bourns 200k trimmers in stock at Mouser...  Single turn, about $1 ea in qty 1:

http://www.mouser.com/ProductDetail/Bourns/3362P-1-204LF

[StillTrying]

- so a 2k7 on OUT would give me 70uA coming from OUT
- so 120uA over 100k pot gives me 12.22 Vdrop

Don't forget the regs min load current, if your load is very low the output V of the reg. will rise a little bit above the SET voltage. The min load current is 1mA, so the wasted 5mA could be lowered quite a bit, it depends on what value 10-turn you're going to use.
You're right on the rest!

http://cds.linear.com/docs/en/datasheet/3083fa.pdf