# Problem with voltage divider simulate this circuit – Schematic created using CircuitLab

I have a problem with a voltage divider circuit. I'm not an electrical engineer, so I am having some trouble finding a solution. I have a circuit with ACS712 30A and another with a precision amplifier (INA125p) and both work properly. The problem is that both circuits give me an analog signal of max 5V. I have a microcontroller (msp432) that accepts an analog signal of max 3.3V. To solve this problem I have done another circuit that implements two voltage dividers. The problem is that when I link the voltage divider circuit, neither of the analog signals have the expected value. In particular when I use the ACS712 circuit, and nothing is attached at the circuit, I measure a voltage of 2.48V, but when I attach the voltage divider circuit, at the same condition, I expect to read a voltage of 1.6V, but in reality I read a voltage of 0.6V. Can someone help me find where I'm wrong?

• Show us your schematic. Feb 13, 2018 at 16:05
• We have no idea what you are doing, how do you expect us to figure out where you did something wrong? Feb 13, 2018 at 16:05
• @brhans at this moment i haven't a schematic for my circuits, however i used a basic circuit that are in the datasheet of acs712 and ina125p Feb 13, 2018 at 16:08
• Use the built in schematic editor to make a schematic of what you have. Feb 13, 2018 at 16:10
• If you'd done exactly what the datasheet(s) show then the likelihood of you having problems is almost zero. So there must be something wrong with what you've done. But you haven't shown us what you've actually done, and there are possibly many different ways of interpreting what you've described. A schematic of what you've done is unambiguous. Feb 13, 2018 at 16:23

This is a common problem with voltage dividers when you feed the value into something which does not have a high enough resistance. simulate this circuit – Schematic created using CircuitLab

As you see in the above circuit, without the load the divider gives you half the applied voltage. With the load in parallel with the bottom resistor, you get 1/3 the voltage.

You can significantly reduce the resistances you use for the divider so they are much smaller than the load resistance to reduce this effect, or you can buffer the voltage with a voltage follower circuit. simulate this circuit

UPDATE:

Now you have added a schematic it is apparent your problem is your resistors are far too small. $232\Omega$ will draw over $20mA$ from 5V, but your IC can only deliver 10mA. As such your divider is pulling down the output.

Use something like 82K and 150K resistors.

• I understand what you tell me! Now i try to do the same thing using another resistor values. Thank you so much Trevor_G! ;) Feb 13, 2018 at 17:22
• PS, at this moment i haven't a 82K and 150K resistor, i only have a 4.7K and 10K resistor. Using this value i obtain a resistor divider with the output value of about 3.4V. Also using your math, and my new resistor value the current is about of 0.3401 mA. I think that this configuration will work, however now i try this, stay tune for my update! ;) Feb 13, 2018 at 17:30
• Using the 4.7K and 10K resistors, the resistor divider work, but only for the ACS712. For the INA125P this setup is wrong. However tomorrow I change the external resistor so that I get the maximum output value near at 3.3V and the problem disapear! In any case, again thank you so much Mr Trevor_G, the your answer solved my problem! :) Feb 13, 2018 at 21:53

The voltage follower is a good start to resolving this issue. But in order to get values that cross 0V (or even just approach it), you'll need a dual supply (+12V, -12V) for the op amp voltage follower. If you don't use a dual supply, you'll get an offset voltage, just enough to throw off your expected output by around .6V to 1V

After this op amp, you'll need a second op amp to scale and balance the output in order to prepare it for the next stage. Generally, there will be four equal resistors in a non-inverting adder configuration, and an offset voltage.

After this, you scale again (a second voltage divider) with the input load impedance included (from the next stage) calculated into your circuit. You can find this on the data sheet, where it shows the input resistance/impedance.

Finally, you can model every bit of this in LTSpice before you build it.