Hello all,
I try to replace a mechanical potentiometer 10k with a digital, i use the intersil X9511 (https://jp.intersil.com/content/dam/Intersil/documents/x951/x9511.pdf)
the typical circuit from the Data Sheet it is very simple but i was not work for me. I want to ask if i miss something here, i use the typical circuit with ase store pin used in autostore mode (figure 2). When i use a multimeter to measure resistance in the Vw and Vl or Vh the resistance is about 8-9K, when i try to change the output resistance with a tact switch between PD or PU with ground nothing change and after a while the resistance rise up to 100K and nothing change after that. I think i am doing somthing wrong here..

 

A complete schematic and bill of material for your implementation would be helpful. What kind of switches are you using?

 

If you built the Figure 2 circuit all should be good. I question if you really did since if it only responds by going UP, and will auto store when power goes off. Then you should only see it go up once and once only.

Got a picture of your build? I suspec a mis-wire we may see for you.

(Heck, I mis-wire transistors all the time and they only have 3 leads)

 

Assuming you just put a multimeter across the resistance pins (exactly what I would do)...

I also wonder if the resistor part needs a connection back to the control side to actually drive the FET switches? A battery powered meter provides no connection. Perhaps try grounding Vl to Vss (pins 6 to 4) and see if the reading make more sense.

 

A complete schematic and bill of material for your implementation would be helpful. What kind of switches are you using?

the schematic:

i use electrolytic capacitor 3.3μF and 2 simple tact switches

 

If you built the Figure 2 circuit all should be good. I question if you really did since if it only responds by going UP, and will auto store when power goes off. Then you should only see it go up once and once only.

Got a picture of your build? I suspec a mis-wire we may see for you.

(Heck, I mis-wire transistors all the time and they only have 3 leads)

i was thinking that too, but it also go up when i push the PD switch, but besides that it go up to 100K-150K and the potentiometer is only 10k

 

Assuming you just put a multimeter across the resistance pins (exactly what I would do)...

I also wonder if the resistor part needs a connection back to the control side to actually drive the FET switches? A battery powered meter provides no connection. Perhaps try grounding Vl to Vss (pins 6 to 4) and see if the reading make more sense.

Yes i just put a multimeter across the resistance pins, i will try that, thank you

 

i forgot to say that i try this with 2 new x9511 chip with the same result maybe something wrong with power supply? it is possible that i damage the chip? i check the wires many times...

 

the schematic:

i use electrolytic capacitor 3.3μF and 2 simple tact switches

This is exactly the schematic from the datasheet and as such it is incomplete. It does not show the part number for the schottky diode. It does not show the value of Vcc.

This datasheet is odd in that I cannot find an absolute maximum value for Vcc, or even a typical value. Maybes someone with better eyes can spot it.

 

This is exactly the schematic from the datasheet and as such it is incomplete. It does not show the part number for the schottky diode. It does not show the value of Vcc.

This datasheet is odd in that I cannot find an absolute maximum value for Vcc, or even a typical value. Maybes someone with better eyes can spot it.

in the datasheet it says in 6 page supply voltage 5V +-10%, i find the diode to use from a circuit from internet (ok. that maybe is a problem) but where i can find that information?

 

The diode is not importaint in this test circuit. It is only there should the resistor be connected to a higher potential then Vcc to keep that point from trying to drive everything else on the board.

Since there isn't anything else on the board you can't drive it hence leave the diode out for now.

 

in the datasheet it says in 6 page supply voltage 5V +-10%, i find the diode to use from a circuit from internet (ok. that maybe is a problem) but where i can find that information?

5V - 10% = 4.5 volts
The symbol is a Shottky diode with a forward voltage drop of 0.2 to 0.3 volts. If you use an ordinary diode it could have a forward voltage drop of 0.7 volts. You should measure the Vcc at the Vcc pin on the chip. If it is less than 4.5 volts then you can expect problems. There are several things you can do:

1. Find a Schottky Diode, OR
2. Raise the power supply voltage so that the Vcc at the chip is equal to 5.0V

http://www.diodes.com/datasheets/ds23001.pdf

 

ok. i will try both, thank you very much