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I want to design a simple circuit that takes 12V from a battery and generates two signals each of range -5V to +5V. These signals are input to a 2-Channel ADC.

I want to be able to change the voltage through the whole range using a pot to test the dynamic performance of the ADC.

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  • \$\begingroup\$ @stevenvh, Analog signals variable by a pot. DC values. \$\endgroup\$
    – Abdella
    Jun 19, 2012 at 14:29
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    \$\begingroup\$ I think by 'signal' you mean a steady voltage? If so, what level of stability, and at what current? And must the two signals change individually or in tandem? - you commented just before I posted :(. But I ams still not sure whether you want a DC signal that can be changed by a pot, or an AC signal (from some source?) that is attenuated by a pot. \$\endgroup\$
    – Wouter van Ooijen
    Jun 19, 2012 at 14:30
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    \$\begingroup\$ Does the battery negative pole have to be the same as the 0V of the delivered output (= can you use a floating ground, from the battery's perspective)? \$\endgroup\$
    – Wouter van Ooijen
    Jun 19, 2012 at 14:55
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    \$\begingroup\$ How do you use a steady voltage to test the "dynamic performance" of an ADC? To test dynamic performance, you need a dynamic signal. \$\endgroup\$
    – The Photon
    Jun 19, 2012 at 16:01
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    \$\begingroup\$ What performance metric of your ADC are you testing? and what is the specification value you are testing to? \$\endgroup\$
    – The Photon
    Jun 19, 2012 at 16:04

2 Answers 2

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I'd put a 10V voltage regulator 12V is just enogh of dropout voltage, so you might look for a low dropput 10V regulator. Then i'd divide 10V by 2 using a high value resistor divider (something like 100K) and then i'd buffer that with opamp - output of opamp is your virtual ground.

enter image description here

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  • \$\begingroup\$ Ok, I'll use a 100K pot to replace R1 and R2, but this way I can't have the whole +-5V range on the same rail (i'll use a bnc connector to connect to ADC), According to the circuit above i should use one connector for 0 to +5V range and another for 0 to -5V. Can we improve the configuration you suggested to be connected to one bnc connector that covers the whole range? \$\endgroup\$
    – Abdella
    Jun 20, 2012 at 15:01
  • \$\begingroup\$ errr, i'm not sure what you mean, but my idea was to connect virtual ground from my schematics to AGND pin of ADC chip, then by connecting two pots to +-5V in my schematics you get two voltages that are +-5V with respect to AGND. \$\endgroup\$
    – miceuz
    Jun 20, 2012 at 16:09
  • \$\begingroup\$ As i understood your original question, you wanted to get two signals from -5 to +5 to connect to ADC two inputs each. So, as long as you connect analog ground of ADC to the middle of 0 to 10V range, ADC will see signals in this range as +5V to -5V. \$\endgroup\$
    – miceuz
    Jun 20, 2012 at 16:14
  • \$\begingroup\$ Thanks a lot , it worked! one last question, for the second channel, should I use the other opamp ? or it's correct to take the output of your schematic to two pots ? \$\endgroup\$
    – Abdella
    Jul 5, 2012 at 20:25
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If all you want is an adjustable DC voltage, just use a linear pot between +5V and -5V with the wiper connected to the ADC input (assuming dual polarity ADC)
The only thing to be careful of is the recommended maximum signal impedance for the ADC, which you can check in the datasheet. Make sure the maximum output impedance from the pot (which is 1/4 of the total impedance, e.g. for a 10K pot the maximum impedance is 2.5K) is lower than this

You could also add a simple opamp buffer if you wish to really mimimise any impedance effects.

EDIT - from reading the comments above I suspect you are planning to use a 12V battery to power this. If this is the case you will need to either:

  • Setup a virtual ground as Wouter mentions.
  • Connect battery between ground and V+ of your circuit, then use an inverter to create your negative rail.
  • Use two batteries in series and take your circuit ground from the middle of them.

If you are not too experienced with this stuff, then I think the two batteries option may be the easiest. 12V is a bit high for the 7805, it will work but waste a fair bit of power, and may need heatsinking if you are drawing a lot of current (see datasheet). I'd use 9V or lower. If efficiency is important then you may wish to use switching regulators.

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  • \$\begingroup\$ The input impedance of the ADC is 6.7K , but the two signals are input to an analog multiplexer before the ADC. If I used a linear pot, how should it be connected to get both +ve and -ve voltages ? Thanks. \$\endgroup\$
    – Abdella
    Jun 19, 2012 at 14:54
  • \$\begingroup\$ Sorry I thought you wanted from 12V to 0V for some reason - for +5V to -5V connect the pot between these two voltages instead. I'm assuming your ADC is dual polarity (otherwise you will need a level shifter) FWIW you can use any pot, I just thought you would prefer the voltage to change linearly for this. \$\endgroup\$
    – Oli Glaser
    Jun 19, 2012 at 15:13
  • \$\begingroup\$ On the pot value, I would use a pot of 5K or below. \$\endgroup\$
    – Oli Glaser
    Jun 19, 2012 at 15:15
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    \$\begingroup\$ Ok, What I understand from you is that I should use +5V regulator and -5V one. Then connect +5V to the input terminal of the 3-Terminal pot and -5V to third terminal, and the output should be taken from the wiper to the ADC, is this correct? How should I connect the gnd ? \$\endgroup\$
    – Abdella
    Jun 19, 2012 at 15:24
  • \$\begingroup\$ Yes that sounds right. Not sure what you mean exactly about ground - you don't need to connect ground to the pot. Are your ADC and multiplexer both dual polarity? (e.g. they are connected to +5V and -5V) It might be worth posting a schematic with part numbers so we can make sure you have things setup right. \$\endgroup\$
    – Oli Glaser
    Jun 19, 2012 at 15:42

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