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The plan is to have a RTD (Resistance Temperature Detector) measurement circuit, ATTiny85 for temperature reading plus few buttons and an LCD display on the same PCB.

The RTD circuit I've chosen is this one provided by Microchip, which will need 2.5V. ATtiny's operating voltage is 1.8-5.5V and LCD's 3.3V.

I'm thinking of using 3.3V for LCD and ATtiny85, and a voltage regulator like LM1117T-2.5/NOPB to convert 3.3V -> 2.5V.

Is this liniar voltage regulator overkill for a low-power current excitation circuit?

Can i use a simple voltage divider for this task?

Right now I have a 3.3V PCB power supply from Mean Well which I plan to use for everything in my circuit (ATtiny85, LCD, RTD circuit, buttons + 4x50mA leds)

Thank you

EDIT

Ok, from what i understand, the voltage divider ratio can change because is directly proportional with the load so it's a big no-no for a RTD circuit that needs a stable power supply.

Here's the plan B:

220V -> 9V or 12V wall wart

4.75...36V -> 3.3V for ATtiny, LEDs and LCD display (DC/DC 1A converter)

3.3V -> 2.5V for the RTD circuit (DC/DC 500mA converter)

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    \$\begingroup\$ The microchip appnote requires a 2.5V voltage reference, not a 2.5V voltage regulator (high precision, not high current). A voltage reference IC would generally perform better than a voltage divider. \$\endgroup\$
    – MarkU
    Dec 13, 2017 at 9:56
  • \$\begingroup\$ This scheme seems complex. I would use a 5V Walmart power supply, a 3.3V regulator for the LCD and a LM35 temperature sensor run from 5V. Incidentally, my LCD displays run from 5V also, so maybe you could eliminate the 3.3V regulator? \$\endgroup\$
    – user131342
    Dec 13, 2017 at 10:19
  • \$\begingroup\$ @MarkU thanks for the explanation. Please check out my updated post. Two DC/DC high efficiency converters seem like a much better plan. \$\endgroup\$
    – AnfEn
    Dec 13, 2017 at 10:23
  • \$\begingroup\$ The scheme it's not that complex if 2xMCP602 are used instead of MCP609. Everything seem easier to understand. Actually mine is not LCD but an OLED display which requires 3.3V so 2 converters will be needed. Also LEDs are 3V-3.8V. \$\endgroup\$
    – AnfEn
    Dec 13, 2017 at 10:31

2 Answers 2

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You can use a shunt reference such as TL431 from the 3.3V supply. The accuracy and stability of the 2.5V source is not very important because it is also used as the ADC reference, so the result is ratiometric. It also is required to deliver little current, but it should be fairly low noise and have low output impedance because the ADC reference input typically represents a dynamic load. Be sure to read the stability (here we're talking about a different type of stability- so it won't oscillate) section carefully when you pick your bypass capacitance.

There are more accurate references and references that use less power but if you're mains powered anyway an extra mA makes little difference.

If you change that fundamental design feature (ratiometric), your circuit will be much less stable with the same component stability. Resistor stability is important (and little else) so use decent resistors (low tempco, stable). This is one of those situations where a smart design will be more accurate and stable with cheap parts than a dumb design with expensive high performance reference etc. For example, if you used an expensive precision reference for the RTD circuit and the 3.3V supply as the ADC reference you'd be doing as a good job of measuring the lousy regulation and thermal/time drift of the 3.3V regulator as you would be measuring the RTD.

To design the shunt reference, figure out the maximum current draw from the 2.5V reference, add the minimum anode current spec from the shunt reference and calculate the highest resistor that will still deliver both with the 3.3V at the lower tolerance limit and the 2.5V reference at the upper tolerance limit, and use the next lower standard value.

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  • \$\begingroup\$ Thanks a lot for the elaborate answer. Still a bit confusing for a beginer like me but i understood that the result is ratiometric. I've never used shunt regulators, may I ask how to use them? According to the datasheet, catode is connected to Vcc, anode to GND and REF is also connected to catode. Vref can be changed with the value of a resistor placed before catode. I believe that if i choose to use 3.3Vref, the current excitation circuit should also be modified(Rref) in order to obtain that 1mA for the RTD. All the resistors used are metal film THT with 0.1% tolerance. \$\endgroup\$
    – AnfEn
    Dec 13, 2017 at 17:05
  • \$\begingroup\$ The 431 needs 1mA or so. You connect REF to anode and through a resistor to your power supply. It's like a precision zener. The RTD current does not come from the reference, it comes from the op-amp- but some current goes through those resistors. Let's say you pick 2mA then nominally (3.3-2.5)/0.002 = 1.65K so maybe 1.5K. That is probably okay but I didn't really do the calculations- it's your design. \$\endgroup\$ Dec 13, 2017 at 17:27
  • \$\begingroup\$ Now it's clear. There are some atypical values of some of those resistors and whole circuit looks simple on paper but pretty messy on a PCB. For example 25K does not exist so i have to use 24K+1K. I don't have a 431 at the moment, can i test it without it on Arduino? \$\endgroup\$
    – AnfEn
    Dec 13, 2017 at 17:44
  • \$\begingroup\$ @AnfEn I suggest getting some kind of reference diode. You could try a green LED if you feel lucky. TL431s are in many (probably most) switching power supplies and regulated AC adapters- connected to an optocoupler at the low voltage side. \$\endgroup\$ Dec 13, 2017 at 18:31
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The RTD needs a constant current, not a constant voltage. It just happens they show you a circuit that uses a 2.5V reference. Use a band-gap reference diode for that if you need that much precision. Perhaps an LM4040D25FTA.

A 5V wall wart or USB charger should be able to handle all of this, with a cheap 5-3.3V step down. I'd drive the LEDs direct from the 5V though, since those are your biggest current load.

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  • \$\begingroup\$ I've never used band-gaps diodes, thanks for the sugeestion. So you say I can safely put 3.3V on that circuit? I thought they calibrated everything and choosed what components to use based on that 2.5Vref voltage. I'm a bit confused now and still don't know if I should buy both (2.5V and 3.3V) step-down regulators or 3.3V only \$\endgroup\$
    – AnfEn
    Dec 13, 2017 at 13:57
  • \$\begingroup\$ @AnfEn the circuit they indicate generates a constant current calibrated from a 2.5V reference yes. But the 2.5V is just a voltage, and delivers basically zip current. They probably intended the user to just use half the 5V rail with a suitable cap to keep it quiet. You only need one step-down and a reference diode. \$\endgroup\$
    – Trevor_G
    Dec 13, 2017 at 14:07

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