0-10V(IP) dc to 0-3.3V(OP) dc analogue linear conversion circuit?

parth kothiya · ‎2022-02-15

i want to convert 0-10V in to 0-3.3V linearly for analogue input of STM32 MCU so for this application any op-amp circuit which can done this?

Peter BENSCH · ‎2022-02-15

It depends on what you want to do with the output voltage: should it be measured or is it used for further processing with an internal comparator or opamp?

If it is to be measured: does it necessarily have to be converted to 3.3V, or can you also accept a lower output voltage?

To stay with the measurement: both have advantages and disadvantages:

If the maximum output voltage is the same as the reference voltage, you use the full range of the ADC, resulting in maximum resolution.
if the maximum output voltage is smaller than the reference voltage, you can correct the measurement result with a fixed factor, but this reduces the usable resolution

Let's assume for the sake of simplicity that you want to measure the signal converted from 10V to 3.3V. Then you can, for example, reduce the 10V to 3.3V with a voltage divider 67:10 (e.g. 68k+33k or a better optimized ratio), then buffer it with a Unity Gain Buffer Amplifier as described here in Wikipedia, fig 3, and then connect it to the ADC input.

If the opamp is supplied by VDD=3.3V, you have to make sure that it is a RRIO amplifier, i.e. rail-to-rail input and output, like e.g. TSZ121 or TZ181. Of course you can also use other types with a larger offset, but don't forget that the offset is an error in the output signal.

Good luck!

Regards

/Peter

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Peter BENSCH · ‎2022-02-15

It depends on what you want to do with the output voltage: should it be measured or is it used for further processing with an internal comparator or opamp?

If it is to be measured: does it necessarily have to be converted to 3.3V, or can you also accept a lower output voltage?

To stay with the measurement: both have advantages and disadvantages:

If the maximum output voltage is the same as the reference voltage, you use the full range of the ADC, resulting in maximum resolution.
if the maximum output voltage is smaller than the reference voltage, you can correct the measurement result with a fixed factor, but this reduces the usable resolution

Let's assume for the sake of simplicity that you want to measure the signal converted from 10V to 3.3V. Then you can, for example, reduce the 10V to 3.3V with a voltage divider 67:10 (e.g. 68k+33k or a better optimized ratio), then buffer it with a Unity Gain Buffer Amplifier as described here in Wikipedia, fig 3, and then connect it to the ADC input.

If the opamp is supplied by VDD=3.3V, you have to make sure that it is a RRIO amplifier, i.e. rail-to-rail input and output, like e.g. TSZ121 or TZ181. Of course you can also use other types with a larger offset, but don't forget that the offset is an error in the output signal.

Good luck!

Regards

/Peter

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Peter BENSCH · ‎2022-02-15

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/Peter

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parth kothiya · ‎2022-02-16

Can you please suggest me any circuit schemetic for this.

Voltage divider > voltage follower for importance matching.

If I want to add protection here op voltage is must between 0-3.3V because more than 3.3V STM32 ADC pin demage how can I add this protection any circuit schemetic?

Peter BENSCH · ‎2022-02-16

I had already made a circuit suggestion: a voltage divider that divides 10V to 3.3V (or smaller) and a voltage follower whose output is directly connected to the ADC input.

Does it answer your question?

Regards

/Peter

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parth kothiya · ‎2022-02-16

how can i protect ADC ip pin from over voltage >3.3V any suggestion with schematic?

Peter BENSCH · ‎2022-02-16

If you use an operational amplifier supplied with 3.3V for the voltage follower, no higher voltage can occur at the ADC input.

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parth kothiya · ‎2022-02-16

op amp used for voltage follower circuit is rail to rail output right?

any other characteristic OP-AMP ic must have for this application ?

Peter BENSCH · ‎2022-02-16

As I wrote earlier, the opamp must have both a rail-to-rail input and a rail-to-rail output because it operates as a follower with a gain of 1. That means that the output voltage is the same as the input voltage, but the output impedance is much lower.

Other requirements:

single supply, VCCmin ≤3.3V
Ibias depending on your resistors
Voffset <0.2...0.5mV

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parth kothiya · ‎2022-02-16

Thanks you so much now I got this.