Resistor chain substitute

Question

We have to interface with a device that is expecting a chain of 34 Ω resistors with switches connected to the input pins as shown below.

There is no information on the internal schematics of the device, but there is 10 V present on the open circuit and when shorted the current is limited to 5 mA. The documentation mentions "maximum input voltage 34 V", which is weird, since we are not expected to supply voltage to those pins.

My first thought was to simply emulate the suggested circuit with analog switches (like ADG715) and a bunch of resistors. Then I thought of digital potentiometers (like AD5254), that do not need resistors.

The problem, however, is that all these chips have analog voltage limits pretty close to their digital supply lines. So, it seems I need either a way to add a high-voltage-tolerant output stage to the digital pot, or use a completely different approach, like a DAC + VCR (voltage controlled resistor). Unfortunately analog circuits are still my weak spot; I'm not sure I can design a FET-based linear VCR properly.

Any other options I have missed? I should also mention that simplifying the circuit and reducing cost are important to us.

Answer 1

How about this idea?

^{simulate this circuit – Schematic created using CircuitLab}

Choose some suitable MOSFET and, as long as CI is always positive relative to CI return, it'll work. If CI can go negative, you'll need bidirectional switches instead of MOSFETs, for which you could use certain types of solid-state relay or SSR-type optocoupler.

Answer 2

One option to cut down on the number of switches and resistors would be to use a binary chain like this:

^{simulate this circuit – Schematic created using CircuitLab}

Start with all switches closed, then open them in a binary pattern to add in resistance. For example instead of 6 x 34\$\Omega\$ resistors to get 204\$\Omega\$ you would open SW2 and SW3.

The last resistor is 34\$\Omega\$ because the original has 17 steps, so you need one extra 34\$\Omega\$ step than a 4 bit arrangement would give, but it could also be 544\$\Omega\$ to get even more steps.

Answer 3

As for Digital Potentiometers, Microchip have some High-Voltage versions which have an analogue supply voltage of 10V to 36V, in the High Voltage (>10V) Digital Potentiometer Products.

E.g. a MCP41HV51T-502 with 8-bits and a 5 k\$\Omega\$ resistor will give a Step Resistance of 19.6\$\Omega\$. Some issues with using a Digital Potentiometer directly could be:

1. Finding a step size which allows all 34\$\Omega\$ steps to be reliably generated, allowing for tolerances.
2. At reset how the initial resistor string value is set and what effect that has on the device being controlled. This Microchip range are all have Volatile memory. At reset they default to a mid range resistance setting.

Also, since the application is for a wheelchair power system, are there any functional safety requirements which need to be met?

Answer 4

Similar to GodJihyo's answer, but making it possible to use simple logic-level MOSFETs for control:

^{simulate this circuit – Schematic created using CircuitLab}

The idea is to have a +15V supply, which is at least Vgs higher than the maximum voltage put out by the device (10V). I'm assuming logic-level MOSFETs that turn on at 3V and can take a 20V gate voltage (for example PJA3404).

Normally R1-R3 pull up the gates of Q1-Q3, resulting in a direct short. Device sees 0 ohm resistance, equivalent to S1 being closed.

If e.g. IO0 is turned high, Q4 pulls down the Q1 gate and the device sees 34 ohm resistance. Turning on all of IO0, IO1 and IO2 will result in 34 + 68 + 136 = 7 * 34 ohm resistance.

This schematic can be expanded to any number of bits. If infinite resistance is needed as option, one extra MOSFET should be added between the lowest resistor and ground (it can be driven directly by IO pin).