Voltage regulator as amplifier

Can I use voltage regulator as a linear amplifier for voltage measurements?

I'm not sure but with use of voltage divider.

Can I use voltage regulator as a linear amplifier for voltage measurements?

No.

What is it that you want to do? If we know that, we can point you to a solution.

An amplifier is a multiplier... If you have an amplifier with a gain of 2 and you feed-in 5V, you get 10V out. If you feed-in 6V you get 12V out.*

A regulator holds the output voltage constant... If you have a 5V regulator and you feed-in 10V, you'll get 5V out. If you feed-in 12V, you still get 5V out.

Most regulators can only regulate-down. If you feed-in 4V, you'll get a little less than 4V out. And, most regulators require a little "extra" voltage. If you feed-in 5V the regulator will "drop out" of regulation and you'll get less than 5V out. The drop-out voltage is specified as the difference, so if you have a 5V regulator with a 1V drop-out, you need to feed-in at least 6V in order for the voltage regulator to operate.

P.S. Amplifiers are usually for signals, not power. An audio power amplifier can "power" a speaker, but you wouldn't use an amplifier to power an Arduino.

  • The amplifier doesn't "make voltage" so if you need 12V out, the amplifier will need a power supply higher than 12V.

OK. Thanks. I wanted to measure voltage changes between 0.6-2.5V and map it to 1-12V proportionally. Was thinking to do it with reference voltage as input.

Was thinking to do it with reference voltage as input.

No. Use an operational amplifier.

I wanted to measure voltage changes between 0.6-2.5V and map it to 1-12V proportionally.

That dosn't make sense. When the voltage is 0.6 you want the output to be 1V, so that is a gain of 1.666, then when the voltage is 2.5V you want the output to be 12V, that is a gain of 4.8. So you can not do this with a linear gain.

This sounds like a X-Y Problem, so what exactly are you trying to do?

Grumpy_Mike: That doesn't make sense.

Yes, it makes perfect sense. This is a "map" function, like the one in the Arduino IDE library. Mathematically, "y = a · x + b". And of course, an op-amp and three or four resistors will do it (given the right op-amp and power connections). "The necessary values are left as an exercise for the reader."

Grumpy_Mike: This sounds like a X-Y Problem, so what exactly are you trying to do?

And indeed it is and Deous needs to explain if he wants to get the job done. Two short sentences does not cut it. :roll_eyes:

Since a linear regulator is an op amp with negative input connected to a voltage reference I think it can be somehow used for signal conditioning. But it depends on real OP’s needs.

Yes I know about the map function, but as the OP wants to use a linear amplifier to a heave this and that is the bit that is a noncense.

The perfectly linear function to map the interval (0.6,2.5) onto the interval (1,12) is y = 5.7895*x - 2.4737

jremington:
The perfectly linear function to map the interval (0.6,2.5) onto the interval (1,12) is y = 5.7895*x - 2.4737

The trick is to translate this to the three resistors for an op-amp. But I don’t feel like calculating the actual resistor values just now (and I can’t find the right diagram to suit - wasted too much time searching). :roll_eyes:

Paul__B: The trick is to translate this to the three resistors for an op-amp. But I don't feel like calculating the actual resistor values just now (and I can't find the right diagram to suit - wasted too much time searching). :roll_eyes:

Probably two opamps, first to apply the gain and then the second to subtract the offset. But wondering what the application is? Tom... :)

But wondering what the application is

Almost certain the OP is not understanding what the application involves.

Can I use voltage regulator as a linear amplifier for voltage measurements?

No.

See reply #1.

TomGeorge: Probably two op-amps, first to apply the gain and then the second to subtract the offset.

Nah, one will do it just fine, takes just three resistors. More op-amps, more noise on the signal and current waste.

But I could not locate a graphic to show, wasted some time looking. If the OP showed engagement with the discussion, I could generate the diagram and even calculate the values. :roll_eyes:

Or, if the OP is shy, they could search the web for "op amp calculculator". This result will calculate 2, 3 and 4 resistor solutions for a single inverting or non-inverting op amp configuration.

Yes, but that is the wrong circuit for his application.

The circuit he wants has the input going directly to the “+”, with a voltage divider to reference voltage and ground and the feedback resistor on the “-”. Total three resistors.

Very frustrating. Cannot find the relevant schematic even using your (corrected) search terms. Equally frustrating, this topic was discussed with all the maths some two plus years ago.



Oh looky! here it is; the search widget on this site does work: :astonished:

The R1 and R2 divider is not required here since we wish to amplify the voltage rather than attenuate it. And of course, the resistor values must be calculated to suit the specific requirement.

if no phase differense AND amplification always >1: 1 opamp, 4-6 resistors (2-3 for signal amplification plus 2-3 for the -value)

if no phase differense ( amplification or division): 2 opamp, 6 resistors (2 X2 for signal amplification/division plus 2 for the -value)