Electronics > Beginners
Shunt Resistor with a LM339
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eev_carl:
Hi,

I made a simple series resistive circuit to test out a shunt resistor.  The circuit is shown in the schematic below.  1V DC is sent through a pair of resistors: 0.7ohm and 10kOhm.  The 0.7ohm is a value I measured directly on my MM of 3 "1.5ohm" resistors in parallel.

I'm using an LM339 to amplify the difference between the 1V supply and the junction of the 0.7 and 10k resistors.  I got a current value of 2.23mA which is much higher than the value I measured directly with my MM (187uA).  These are my calculations

A = 1 + 10/1 = 11
Vdiff = 17.6mV / 11 = 1.6mV
I = 1.6*10-3 / 0.7 = 2.23mA

17.6mV is the value I measured on the output of the LM339.  The supply to the LM339 is 9V.  I don't have any other 3 op amp son the LM339 connected.

Feel free to suggest another measuring technique if I'm off base with the comparator.

Thanks,
Carl
glarsson:
LM339 is a comparator, not an op-amp. The feedback from the output should go to the negative input to create a differential amplifier (and replacing the LM339 as well). The top connection of the sense resistor has the higher voltage and the bottom connection has the lower voltage. You wired it the other way.
eev_carl:
Should I run the comparator without feedback into an amplifier?  Is the comparator of any use or should I just use an op amp.

Thanks
glarsson:
First, the 339 has an open collector output. This means that it can't output a voltage, only pull down a voltage provided by a pull up resistor or similar. The 339 flips the output based on which input has the higher voltage. It doesn't act as an amplifier. If you want to amplify the voltage over the sense resistor you need an op amp. Also, the op amp must be able to handle signals close to the power supply. In your application the signals are within 1V of the negative power supply of the op amp, ground in this case. Not all op amps can do this.
eev_carl:
Thanks.  LM339 is the wrong IC for what I'm trying to do
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