Why am I getting asymmetrical voltages from my voltage divider?

Thread Starter

488 studios

Joined Nov 10, 2023
11
Hey, I'm trying to use an lm35 and an lm3914 as a 0-50 C thermometer. The problem is the lm3914's reference voltage has a minimum of 1.25v, so I need to amplify the lm35's output by 2.5x. I'm using an lm741. I created a voltage divider with a 9v battery to create a virtual ground because I heard the 741's output doesn't go to the supply rails, but my negative rail is at -7.19v, and my positive rail is at +1.571v, even though both resistors measure the same out-of-circuit (about 10 kohms). But when I plug the resistor between ground and negative rail in, it measures 2.428 kohms. Not sure if this is related, but I tried offset nulling, but the output jumps from -5.2v to +.9v instantly, do I need multiturn pots?
 

crutschow

Joined Mar 14, 2008
34,454
Use the commonly available, single-supply LM324 or LM358 which can operate down to 3V.
Before using it, please read its spec sheet, which you apparently didn't do for the 741. :rolleyes:
 
Last edited:

k1ng 1337

Joined Sep 11, 2020
960
Try LM358 with a gain of 2.5. This LTspice simulation is plotting 3 circuit temperatures of 10, 25 and 50C. With a gain of 2.5 at 10mV/C, the outputs voltages are 250mV, 625mV and 1.25V respectively.

Untitled.png
 

crutschow

Joined Mar 14, 2008
34,454
You don't need an amplifier, because you can divide the LM3914 reference and connect a smaller voltage to the resistor chain.
Below is the LTspice sim of a circuit that does that:
The pot U2 adjusts the voltage at the top of the resistor chain to 0.5V, which then lights LED D1 at 50mV and LED D10 at 0.5V input.
The 1kΩ pot along with the values of R1 and R2 give an LED current of about 5mA.
These values can be changed to give a different LED current.
(The LED current is approximately equal to 10 times the current out of REF terminal to ground.)

1703354614913.png
 
Last edited:
Top