Hi,
I'm using an LM2904 as a current sense amplifier as shown in the attached schematic and I don't understand why when there's no current flowing through R5, the current sense resistor, there is still an output of 2 v?
I tried putting a trimpot in place of R6 to eliminate the offset across the inputs, but it didn't eliminate the 2v on the output.
Thanks!
Craig
Craig,
The LM2904 can only sink ten's of micro-amps and maintain a near zero output voltage. At higher sink current a PNP emitter provides current, but the minimum output voltage is about 0.7V or more.
A rail to rail output op-amp is required to operate properly.
Regards,Ron Michallick
Regards,Ronald MichallickLinear Applications
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Hi Ron,
Thanks for the reply.
I had some INA121's laying around , so I threw in one of those and it works well, even with a single supply.
Would a TL082 do the trick? (They're much cheaper.)
I noticed that the V+ rating is max +/- 18 v, but the INA121 seems to work fine with a 24v single supply.
Thoughts?
The TL082 will have the same problem.
It is OK to use the INA121 at 24V singal supply.
Regards,Ron
Maybe you could recommend a rail-rail device that is pin for pin compatible with the 2904?
Given that I need a device that can handle input voltages around 20 v, I need to run V+ above that, which I currently have at 24v.
What I need is a device that handles the above and can maintain a Vol somewhere much closer to ground, given that I'm using it in a single-supply configuration.
As an option, I will try raising the output impedance of the circuit. Do you happen to know how high I can go with the resistance in the feedback voltage divider for a LM22680? That's what this circuit is driving. According to the LM22680 datasheet, they recommend keeping the voltage divider total resistance less than 10k ohms, which is interesting because WebBench specified a resistor pair that totals 17k ohms, above the recommendation.
Will it cause any problems if I eliminate the feedback resistor voltage divider for the LM22680 and just drive it with the op-amp output, which will typically be below the 1.25v reference input?
One way to make the circuit work is to offset the zero current output voltage. The termination of the differential amp from non-inverting input to ground sets the output voltage when in balance (inputs equal). That way the opamp does not need to go to VCC- (GND). However you have to deal with the output offset.For an answer on the LM22680, please start a new thread with that part number.
Ron.