# 5V Current Transducer to 3.3V Analog GPIO Input

I am trying to find 3.3V variant of the 5V Current Transducer LTS_25-NP but so far no success. Is it possible to interface this 5V analog output to 3.3V STM32F407 analog input GPIOs?

Edit: I want to know about the 'FT' (5V tolerant input pins) for the purpose of sensor input when my MCU is powered from 3.3V power supply.

• A simple resistive divider should do the trick. Divide the 2.5V output to 1.65V ...remember to keep the total resistance > 2k Ohm. Commented Feb 18, 2019 at 0:52
• Will this work? R1=5k and R2=10k Commented Feb 18, 2019 at 1:02
• I have to put a 'RC' LPF in between the current transducer output and the analog GPIO input pin. At present the value of R=430 Ohms. Now I have to add the voltage divider also. Please tell how to combine these two together? Commented Feb 18, 2019 at 1:26
• I'd be tempted to use 1% 2k and 4k resistors. The full scale output is (2.5 +/-2)V so maximum swing will be 4.5V - 0.5V on the output. After the divider this should translate to 3 - 0.333V, well within your A/D range. I am assuming they device does not use FET op-amps so will be non-linear or clip much beyond the 4.5 - 0.5V range. Commented Feb 18, 2019 at 1:30
• For your RC LPF simply use 1.333k Ohm as the resistor and calculate your required capacitor. Put that cap in parallel with the lower leg of the divider. Commented Feb 18, 2019 at 1:33

The output swings around a center voltage of 2.5 volts. The minimum load as specified in the datasheet is 2K ohm. What you want is to prevent the 2.5 volt 'center' from going above 3.3 volts at high current levels. Analog inputs are 'fussy' about input resistors staying below 1K if possible, so readings are more accurate.

I am keeping the ratio simple so software can multiple ADC input by 1.33334. Another option is to make both resistors 1K 1%, so the input is divided by 2. To rescale it in software just multiple analog input by 2. You get better resolution at a ratio of 2/3 though it means using floating point to get the 1.3334 multiplier. To multiply by 2 can be done with integers or shift left one place (binary).

If R1 = R2 then R1 should be 1.5K 1% so to stay above the minimum rated load of the sensor without causing sample errors in the analog input.

Input sensitivity is 25mV/A, so you will have to factor that into your scaling routines.

simulate this circuit – Schematic created using CircuitLab

• Very Interesting details to learn the calculations in depth. Please tell me 1k and 2k is better or 5k and 10k is better? Will it effect the linearity of the output by choosing lower or higher individual resistor values which gives the same ratio? Commented Feb 18, 2019 at 1:30
• @scico111 The transducer datasheet says its load resistance should be greater than 2 kΩ. As long as your total resistance is greater than that, it's all dependent on what your analog input can deal with on its inputs. Like Sparky256 says, analog inputs can be fussy about resistances. Commented Feb 18, 2019 at 1:36
• @Hearth Please correct me if I am wrong.. my understanding is that a voltage divider of 1k and 2k resistors will give a load of less than 1k Ohms to the source since their Thevenin equivalent will be parallel equivalent of 1k and 2k. Commented Feb 18, 2019 at 2:35
• @scico111. Incorrect. The source (sensor) will see R1 and R2 as being in series, so it sees a 3K load. The analog input has a few pF of capacitive loading, not much at all.
– user105652
Commented Feb 18, 2019 at 2:38
• @Sparky256 Ok. I got it now. You mean that the source (sensor) will see a series resistance. Please correct me if I am wrong that the analog GPIO pin will see a parallel resistance of R1||R2 at its input? Commented Feb 18, 2019 at 20:07