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# ground verses negative voltage?

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#### flightline

##### New Member
Dear Sirs:

I am replacing a 5K rotary linear pot with a digital pot (DS1804) and controller (basic stamp). I will from my computer then be able to control a pot which in turn controls an amplifier's output.

My first problem is I do not understand the difference between negative voltage and ground. The 5K pot is connected to +8v, -8v and the wiper to an input on an amplifier according to schematics. But the digital pot is looking for +6V, a ground (not negative voltage) and input. I must also somehow lower the voltage from 8V to 6V so as not to burn the digital pot and figure out how to ground it. Does anyone have the patients to help me on this one? Thank you very much.

I've been looking over the data sheet for the DS1804, and it sounds like you're a bit confused. First off, the DS1804's lowest resistance range(according to this data sheet any way https://www.electro-tech-online.com/custompdfs/2004/02/DS1804.pdf ) is 10kohms. Second, it seems as though the +8v and -8v are voltages that the 5kohm pot is dividing for the amp input, right? Those wouldn't be the supply voltages for the DS1804, it wants +2.75v to +5.5v along with a plain old ground.
The easiest way to ensure proper grounding for the digital pot is to find another existing ground, one from the amp perhaps, and connect it to that.
Got a part number for the amp or any other additional info?

I don't know what schematic you are looking at, but the Ds1804 is rated to operate from 2.7 to 5 volts and will not tolerate +8 volts anywhere and certainly not any negative volts. Ground is zero volts, the reference plane if you please. There are 99 resistors with selectable taps, so the minimum selectable resistance is 1/100 of 10K or 100 ohms. Do you realize that you have to send pulses to the device to move the tap?

thank you so much

I am always confused...first things first. The DS1804 is looking for 2.5 to 5v, a ground and an wiper output to the amp per DS1804 specs. I want the DS1804 to replace a 5K linear rotatary pot which is currently connected to -8V, +8V and an wiper output at the amp per amp schematics.

The amp is an industrial solenoid controller. We are setting forth on an incredible adventure....building a moving 747 flight simulator. The most expensive part, the hydraulic motion stage now finished!

Thank you for all your help!!!

The hydraulic rams are controlled by solenoids which are in turn controlled by the industrial solenoid amplifier/controller. One way to control the amp (& in turn the solenoids and rams) is with the beforementioned standard 5K pot. 1K Ohms is full down and 3K Ohms is full up for the rams.

A second way is with changing voltage at another place on the hyd. amp. +1 volts to the amp gives full down and -1v for full up on an "input" and "signal ground" line. I might go with changing the voltage instead of resistance. I have built a DA converter that will give me 125 steps from 0-1 v. The problem with this solution is am too stupid to figure out how to invert that voltage to give me 0 to -1v. Voltage inverters only work in the range of input 1.6-8V & output -1.6 to -8v.

Re: thank you so much

flightline said:
The problem with this solution is am too stupid to figure out how to invert that voltage to give me 0 to -1v. Voltage inverters only work in the range of input 1.6-8V & output -1.6 to -8v.

Just use a simple inverting opamp, it will invert perfectly within it's supply rails (but many don't actually quite fully reach the rails). As you are only wanting 0 to -1V, and have +8V and -8V supply rails, there won't be any problem.

I don't have the faintest idea where you got this 1.6-8V idea from!.

I think I am right...what am I missing?

No, No that is an inverter for power. What you need is an operational inverter as below:

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