Adjustable Regulator and Digital Potentiometer

[rayhall]

I have a LM2576 adjustable regulator controlled by AD5254BRU1 digital potentiometer.

LM2576 **broken link removed**
AD5254BRU1 https://www.analog.com/static/imported-files/data_sheets/AD5253_5254.pdf‎

The digital pot gets hot and fails. Can you help me with what is wrong. This is the circuit,

Ray.

 

[ronsimpson]

It is hard to find in the data sheet but:
You can not (should not) put a voltage on the digital pot out side the power supply range!
The voltage on A0, W0 and B0 must be near 0 to 5 volts.
Now you have a voltage divider from Vout to feedback to ground. You need to swap places (R27 and POT). The POT needs to set on ground and feedback. R27 needs to connect to Vout.

 

[rayhall]

I have fixed the link to the potentiometer datasheet.

If feedback voltage is greater then the 5v supply to the pot, how will the regulator work if I just send it 0 to 5 volt feedback. I would think the regulator would only then output up to 5 volt ?

Ray.

 

[rayhall]

I have made this change. Will it work now.

 

[ronsimpson]

 

[rayhall]

The last one cannot work. Will this work with no smoke ?

 

[ronsimpson]

I think pin3 (A0) will get above 5V.

R27=1.75k, Pot=1k, Vout=3.0volts
R27=1.75k, Pot=500, Vout=5.62volts
R27=1.75k, Pot=250, Vout=10volts
If my math is right.

 

[rayhall]

Thank you...

Ray.

 

[ronsimpson]

R27 might get hot.
Do you have a pot of a higher value. 10k?

 

[rayhall]

The AD5254 is also available in 5k, 10k and 100k vesrions. I can change to the 10k.

Ray.

 

[JimB]

I have been doing a bit of maths on the digital pot and I think that I have found a fundamental problem with your circuit.

A limitation of the digital pot is that the voltage on the pot cannot exceed Vcc (5volts).
If we put a resistor between the top of the pot and the supply, and make that resistor a value so that we cannot have more than 5volts at the top end of the pot when the regulator is outputting 18 volts, it is then impossible to set the output to 3 volts, there is too much attenuation in the resistor/pot chain.

JimB
(with a cold (man flu?) and feeling sorry for himself, and hpong that he has not made a complete marmelade of the calculations.)

 

[ronsimpson]

Hi Jim,
Look at post #7. The pot only sees 1.25 volts.

 

[JimB]

Ron, you are right (and wrong).

In #7 the link should be between pins 1 and 3 rather than pins 1 and 2.
When that is done, you are correct and there is only ever 1.25v on the pot.

I knew that I had to be confused somewhere, just could not see where.

I was assuming that the pot be used as a potentiometer rather than a varaible resistor.
Then it is possible to put resistors in the top and bottom arms of the pot to restrict the range to an exact 3 to 18 volts.

JimB

 

[MrAl]

Hello,

With the restrictions on the pot as they are i dont see this happening. At least not 3v to 18v. i can see maybe a 12v range but not 3 to 18v.
To set to 3v it takes a maximum resistance, and to set to 18v takes a min resistance. When the resistance is max, the 5v restriction comes into play, and when the resistance is min the 5ma restriction comes into play. That makes it look like these pots just wont work for that large range (3v to 18v is 15v range).
If we go to a higher value pot then the current restriction becomes more severe, so that probably does not help either.

Try it out by working out 3v and 18v at the top of a resistor chain where the pot has to produce 1.25v for either of those values (after adjustment of course).
Maybe you can find a solution. Note that any solution can not put more than 5v on any terminal of the pot nor can it put more than 5ma through the pot for the 1k version, and less for the 10k version, and less for the 100k version, etc. Check the data sheet for the 10k and 100k current limits.

 

[ChrisP58]

I have a circuit that I've used often that should work for you, but I can't seem to get my picture to load.

The fix is to drive the feedback pin with an error amplifier, that compares the output voltage with a control voltage.

Make a resistor divider that makes 5V when the top is at 18V. I used 13.0K and 4.99K. Feed this to the non-inverting input of an opamp.
Use your digital pot to make a 0-5 volt control voltage. Tie that to the inverting input of the opamp.
Tie the output of the opamp through a ~1K resistor to the feedback pin of the regulator.

One advantage to this circuit, is that the control to output voltage ratio is linear.

You may need to make the opamp an integrator. Do this by placing a small cap (10nF) from the output to the inv input. You will also need to add some resistance (10K) between the control voltage source and the inv input.

 

[MrAl]

Hi,

That sounds interesting Chris but i'd be very careful when inserting op amps into the feedback path or any active device because they introduce new poles that could cause bad oscillation. The circuit has to be tested very carefully with various inputs and loads.

At first glance it looks like this should be doable but the extreme restrictions of the digital pot make it a big problem. You're right though it might be done wth an active element of some kind, just with a little caution. It would also be nice to know the output regulation requirement because a transistor would be better than an op amp because it would be much faster than the internal error amplifier so it should not cause oscillation.

 

[alec_t]

a transistor would be better than an op amp

Agreed. But even that may be tricky, because if the digipot is set to give 5V at the wiper when the 2576 is giving 18V, then the wiper will be only 0.8V when the 2576 is giving 3V. 0.8V doesn't allow much headroom for driving the transistor, especially considering Vbe drift with temperature.

 

[ronsimpson]

Ron, you are right (and wrong).

Yes the drawing is wrong. I think you understand the pot is from FEED BACK to ground.

The supply can not output more than about 18 volts because the input is 19V. Even when the pot=0 ohms the output can not get above 18.x volts. Under heavy load it probably can't make it to 18 any way. If you want to stop at 18V then in software limit the pot to some minimum value or add a series resistor.

R27 and (R18+POT) makes a voltage divider for 3V
R27 and R18 makes a voltage divider for 18V.

 

[MrAl]

Agreed. But even that may be tricky, because if the digipot is set to give 5V at the wiper when the 2576 is giving 18V, then the wiper will be only 0.8V when the 2576 is giving 3V. 0.8V doesn't allow much headroom for driving the transistor, especially considering Vbe drift with temperature.

Hi alec,

Yes that's true that's why i was asking about the required spec's. But i think maybe Ron's idea might work if we can use the 10k digital pot. I just checked the spec's on the current and it looks like the 10k pot has a variable max rating from wiper to B but at the top end it is 5ma too, so that might work. I'll get back here with the equations next.

 

[MrAl]

Yes the drawing is wrong. I think you understand the pot is from FEED BACK to ground.

The supply can not output more than about 18 volts because the input is 19V. Even when the pot=0 ohms the output can not get above 18.x volts. Under heavy load it probably can't make it to 18 any way. If you want to stop at 18V then in software limit the pot to some minimum value or add a series resistor.

R27 and (R18+POT) makes a voltage divider for 3V
R27 and R18 makes a voltage divider for 18V.
View attachment 84030

Hi Ron,

Yes that might work, i found out the 10k pot version has a max current of 5ma too at the top end only, but that might be good enough.

Two equations that describe what we need are:
3=(1.25*R27)/(R18+Rp)+1.25 [eq1]
18=(1.25*R27)/R18+1.25 [eq2]

Solving these two simultaneously for R18 and R27 in terms of the pot total resistance Rp we get:
R27=(469*Rp)/300
R18=(7*Rp)/60

Now it is just a matter of selecting an available pot value and testing to see if it meets the specs.

Using a pot value of 1k we get values for R27 and R18 that lead to a maximum current through the pot wiper to B (or A) of around 10ma, which is too high. However, using a pot value of 10k we get:
R27=46900/3
R18=3500/3

or in decimal:
R27=15633.3333
R18=1166.6666

Checking the max current it comes out to around 1.07ma when the output is 18v, so i think this will work without any additional active elements.

LATER:

Usually we can not get the exact component values we want, and often the IC spec's are not exact either. So to insure we can actually get the range 3 to 18v we might want to solve for something like 2.9v to 18.3v and that should give us at least 3.0 to 18.0v in the real life circuit.

If dv1 is the voltage decrease from the 3v spec and dv2 is the increase in voltage of the 18v spec then we have solutions:

R27=((7-4*dv1)*(4*dv2+67)*Rp)/(20*(dv2+dv1+15))
R18=((7-4*dv1)*Rp)/(4*(dv2+dv1+15))

Setting dv1=0.1 and dv2=0.3 we get resistor solutions:
R27=(1023*Rp)/700
R18=(3*Rp)/28

or in decimal with again Rp=10000:
R27=14614.2857
R18=1071.42857

The max current is then about 1.167ma which is still ok.

Finally, given any two voltage set points VL is the lower (like 3v) and VH is the higher (like 18v) the solutions for the two resistor values are:
R27=(Rp*(5-4*VH)*(4*VL-5))/(20*(VL-VH))
R18=(Rp*(5-4*VL))/(4*(VL-VH))

This allows finding resistor values for any two set point voltages. Note we get the same resistor values as previously if we set VL=2.9 and VH=18.3.
There may be other restrictions so you'll have to test after finding the two resistor values for any given circuit.