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I need to replace a 5k mechanical potentiometer with a digital one. The hang up is that it will have a maximum of 14V across it. That's too high for most digital pots on the market. The max current should be less than 10mA, so that's not an issue.

These are the main three parts I've narrowed it down to:

  • AD7376 - 128 positions, volatile, 10k, 28V range, TSSOP $2.86
  • AD5290 - 256 positions, volatile, 10k, 30V range, MSOP $1.92
  • AD5292 - 1024 positions, non-volatile, 20k, 33V range, TSSOP $2.62

I think it would be helpful to have it be non-volatile, but it seems like it would be able to initialize to the correct value pretty quickly after initialization, so I don't know if that's too important.

My main question is related to the size. The 7376 has 10k/128 = 78 ohms/tick precision. The 5290 is slightly better at 39 ohms/tick. Finally, the 5292 has 19 ohms/tick.

I'd like to have it be as smooth as possible, so it seems like the 1024 position one would be a good choice. Is there any disadvantage I'm missing with going with a 20k pot when I'm only ever going to use it from 0 to 5k?

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  • \$\begingroup\$ Did you know you can use a MOSFET as a voltage controlled resistor? \$\endgroup\$ – Majenko Dec 3 '11 at 11:28
  • \$\begingroup\$ @Majenko: I think mouche chose the digital pots because of the SPI Interface. \$\endgroup\$ – 0x6d64 Dec 3 '11 at 12:22
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    \$\begingroup\$ Digital pot at 5V + MOSFET + couple of other resistors = digital pot at whatever voltage you want. \$\endgroup\$ – Majenko Dec 3 '11 at 12:48
  • \$\begingroup\$ +1 @ Majenko - Great idea! You should sketch it in an answer. \$\endgroup\$ – SteveR Dec 3 '11 at 13:54
  • \$\begingroup\$ You would be better off showing a circuit you are trying to modify. Just because a potentiometer was used to control something, doesn't it is the only way to do it. In many cases, the pot is there to set the voltage, so there will be many ways you replace it for digital control (PWM or DAC for example). \$\endgroup\$ – Armandas Dec 3 '11 at 17:04
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Replacing a 5 kΩ pot with a 20 kΩ pot and then using just 1/4 of the range isn't the same. The output of a potentiometer is ratiometric, meaning it is relative to the voltage at both ends. It is a linear blending of the two voltages, with the wiper position setting the blending factors from 0 and 1 at one end to 1 and 0 at the other.

The difference between the 5 kΩ and 20 kΩ pots is the output impedance. This is highest in the middle and goes towards 0 at each end. For the 5 kΩ pot, in the middle setting there are effectively two 2.5 kΩ resistor in parallel impedance-wise. That makes its output impedance half that, or 1.25 kΩ. All this scales up by a factor of 4 for the other pot. In the middle, it has two 10 kΩ resistor in parallel, making the output impedance 5 kΩ worst case.

Whether the higher output impedance matters depends on the circuit and is impossible to say with the information you have given us. If the circuit that receives the pot output signal loads it too much, then the result will be non-linear. Whether that matters is again a function of the circuit you have told us nothing about.

However, you are right about the resolution. The 10 bit pot has 1024 unique settings, which will make the adjustment smoother than one with lower resolution. Note that this is only a issue when the pot setting is being changed. A 10 bit pot at a fixed position produces the same output as a 8 bit pot in the same position, but the 8 bit pot can't be set as precisely as the 10 bit pot.

Explain what the application is an show the circuit around the pot, especially where the output signal is going.

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The obvious question is here: Will the resolution suffice, if you only use 25% of the digital pot (let's talk about the AD5292). Let us do a rough estimate for the resulting resolution: 1024/4=256 steps. Compare that to an analog potentiometer: Let us assume that these pots have a working angle of about 300° (is that about right?). That means: you digital pot is accurate like an analog pot set with a precision of 1,2°. Sounds quite ok for me for some applikations.

For a second I thought about wiring a fixed resistor parallel to the pot, but this makes the total resitance quite non-linear and you get large errors esspecially at 0% and 100% pot postion. You would have to compensate these errors in software, this could get a little bit messy.

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As Mr. Lanthrop notes, potentiometers are often used in ratiometric applications. If the outside of a pot is being fed by a low-impedance source (e.g. less than 100 ohms) and the tap is feeding a high-impedance load (e.g. greater than 250K), a 20K pot may work essentially identically to a 5K pot (e.g. a 3/4-scale value on the 20K pot would behave alike a 3/4-scale value on the 5K pot). The higher the source impedance and the lower the load impedance, more the 20K behavior will diverge from that of the 5K. If the source impedance is significant but the output impedance is high, the higher resistance of the 20K pot may be compensated for by simply adding a 6.8K resistor between the ends. If the output impedance is not much higher than the pot resistance, however, it may be necessary to add an amplifier stage with a high-impedance input between the pot and the intended load.

Note that real pots have three components to their resistance: contact point to left terminal, contact point to right terminal, and contact point to wiper terminal. If the pot is being used to feed a high-impedance load, very little current will flow through the wiper terminal, and thus its resistance would not significantly affect circuit behavior even if were large relative to the other two. If, however, it's driving a low-impedance load, and thus has to source or sink current, its resistance will affect its ability to do so. Digital pots have a wiper-terminal resistance which is higher than that of a good mechanical pot, and the resistance characteristics can vary with output current. They are thus not terribly suitable for use in situations where a pot is used to set wiper current rather than a wiper voltage.

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