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I have an output of 2.5 to 4.5 volts I need to level shift, and amplify it to give me 0 to 10 volts output.

A rail-rail op-amp, using the non-inverting op-amp appears to be the best way to do it. I've noticed a few different ways of doing it. Got very confused with the methods shown.

No AC coupling just a DC voltages in, and out.

I would like to use a Texas Instruments TLV2370IP, I have any regulated voltage that the device will work on (16v maximum).

Any help will be much appreciated.

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    \$\begingroup\$ You sound like you essentially have it sussed. Why don't you post a schematic of what you have come up with so far? \$\endgroup\$ – DiBosco Jan 11 '18 at 15:47
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    \$\begingroup\$ Don't forget to mention what power supply voltage(s) you have available. Not sure what your 2nd last sentence means. Also, you should mention what accuracy you are looking for and if there are any reference voltages available. \$\endgroup\$ – Spehro Pefhany Jan 11 '18 at 15:47
  • \$\begingroup\$ Are you just switching between these two levels, or do you have a continuous range between them? \$\endgroup\$ – brhans Jan 11 '18 at 16:04
  • \$\begingroup\$ To have an answerable question, you need to specify the output current or impedance. \$\endgroup\$ – Chris Stratton Jan 11 '18 at 16:14
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If you have access to a negative rail of 2.5 volts you can use a 1:1 potential divider between the input and -2.5 volts. The centre point of the voltage divider will now be 0 volts when the input is +2.5 volts and it will be +1 volt when 4.5 volts is applied to the input.

So follow this with a gain of ten op-amp and you have a solution. If your negative rail is say -5 volts (or something else) follow this link to a calculator that can help: -

enter image description here

The picture above shows an input at +2.5 volts and the output at 0 volts. Gain is always going to be 5 because that is the output range divided by the input range i.e. 10/(4.5 - 2.5).

You can double check that the output rises to 10 volts when the input is set at +4.5 volts (with no change in R2, R3 or R4).

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  • \$\begingroup\$ I don't especially like this solution because it needs a -V supply and because it has a relatively low input impedance. \$\endgroup\$ – Peter Green Jan 11 '18 at 17:19
  • \$\begingroup\$ @PeterGreen the OP said "Any help will be much appreciated" so feel free! \$\endgroup\$ – Andy aka Jan 11 '18 at 17:23
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Ok, so we need to multiply by 5 and subtract 12.5

An amplifier with a non-inverting gain of 5 has an inverting gain of -4. So we need to bias the feedback divider at 3.125V.

BTW be-aware that "rail to rail" op-amps often have considerablly worse performance as they get close to the rail. This may or may not be a problem for you.

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You can do it like this:

The +5 Vref directly affects the output voltage- a 1mV change in the reference represents a 2.5mV change in the output voltage. You can use a series reference chip or a shunt reference and resistors.

Rail-to-rail i/o is unnecessary. You only need an input common mode range that extends to +4.5V with whatever supply you are using, single supply or rail-to-rail output, and an output range that goes to +10 with the supply you are using. Relaxing that constraint allows you to use cheaper and/or more robust op-amps.

Note that the output has to sink 2.5V/R1 with 2.5V in. Regardless of op-amp type, it won't get all the way down to 0V without a negative supply, but it may get close enough for your purposes.

schematic

simulate this circuit – Schematic created using CircuitLab

If you want to use a different reference voltage or change other parameters, just scratch down the equations and solve them. With the case of a 5V reference, we can simply write down by inspection:

R3 = R1||R2 (zero balance with 2.5V in)

and

R1 = 4*(R3||R2) (gain of +5)

The remaining degree of freedom is scaling so we pick something reasonable.

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    \$\begingroup\$ +1 Logically you have the correct and most direct answer. Subtract the unwanted 2.5 volts. That leaves you with 0 - 2.0 volts which can be multiplied by 5 to get 10 volts full scale. This works as long as the 2.5 volt offset does not change. \$\endgroup\$ – Sparky256 Jan 15 '18 at 6:02

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