Digital Potentiometers (eg. Xicor X9221)

[morizabal]

Hello Everyone,

I am currently working on instrumentation amplifier using digital pots to adjust trim, offset and span. Most digital pots are rated at a ±5V terminal voltage. My question is how do I interpret this?

Does it mean I can put a -5 and 5 on each end of the terminals max, since 5 - (-5) = 10, for example can I put across the terminals 0 to 10, or 15 to 25 as long as I don't exceed 10 V, or does it literally mean I can put more
than -5 or 5 at each terminal.

Thank You,
Mario

 

[MMAshwin]

Hi,

Looking at the data it appears that VH & VL should be no more than +5v , -5v respectivly.

As for you question does it mean the absolute values or just the difference between the two terminals, well I'll answer it at an extreme. What is you want 1000 , 990 across the terminals. Yes the difference is 10v, but lets not forget this is a digitally controlled pot which is done using FET junctions and the like. All of which are powered by the ICs Vcc supply of 5 volts. If you apply the wrong voltage you could damage the IC because of the difference between the supply voltage and the voltage across the pot.

If you want higher voltages then the 9312 goes from 0 - 15 volts


Check out for extra info :-

http://shop.kanda.com/datasheet/x9221.pdf

the above is the data sheet

http://www.xicor.com/pdf_files/design_bulletin_mar_99.pdf

http://www.xicor.com/pdf_files/an88.pdf

the above are application notes Any help ?, yes no let me know.

Regards

 

[morizabal]

Thanks for your help.

The digital pot in question will be operating in a two-terminal mode. The VW (wiper) will be connected to VH. The voltages are as follows. In one extreme the pot will have VH = 10.5 VL = 8.4, at the other extreme VH = 10.5 and VH 10.48 volts.

thanks

 

[MMAshwin]

Hello Again,

Refering to the data sheet (url posted previously) page 8 says that the absolute maximum ratings are ±8v and therefore the maximum voltage across VH & VL is 16v i.e ∆V=(VH-VL).

Now on the fave of it yours conforms to that spec. However it does say on page 8 that the absolute maximum ratings are referenced to Vss which I assume in your case will be 0v.

Now there is a work around however it generates another problem.
1. You could raise Vss and Vcc so that Vss was in the middle of your range. Making Vss 10v and Vcc 15v for example.
However that may require more supplies. But he main problem would be the I²C bus comms. It too is referenced to 0v so some level conversion may be necessary.

2. Is it possible to lower your voltages so that 0v is in the middle of the range. Could you remove the DC through a capacitor or level shift it so that 0v was say equivelent to 10v.

But looking at the data sheet it appears you would have a problem if you connected your voltages as stated. Because looking at the diagram on page 7 the 'wiper' is switched potential division. The switch being an FET. If you place too high a voltage with respect to the gate it may, at worse, destroy it or, at best, the FET will not switch on.

All that aside I'm no expert on these devices although I have used them connected to a PIC micro controller. I never suffered from your dilema because I was controlling 5v and then in effect amplifying the change on the wiper position.

Any help ?, yes no let me know.

Regards

 

[morizabal]

Thanks for the info.

I found a work around that will bring the potential down to acceptable levels. My circuit consist of a simple 3 op-amp instrumentation amplifier with a 10 VDC output. The Pot in question is used as a final trim to bring the voltage around 10 VDC. VL is connected to IN+ on a op amp in a unity gain configuration. Basically I have a voltage divider on the output of the instrumentation amplifier. I decided to cut the initial gain in half and instead of a x1 op-amp gain I opted for a x2 gain. This will keep the voltages in the terminals of the digital pot at < 8 volts referenced to 0 volt ground.

Thanks for you help.