Yes, current drawn by the steady light is 0.74A and the voltage accross 0.1 sense resistor is 70mv

Clearly one of those is not true, my bet is that the resistor is not exactly 0.1ohm.
And once again I ask, how exactly do you measure those numbers?

Like I said, connect RS+ to your supply voltage, connect RS- to a resistve divider with a potentiometer such that you can vary the voltage difference between RS+ and RS- from say 0V to 100mV. Set different points and compare the input voltage with output voltage, then you will see if the output function is linear, or there is some offset voltage, or it behaves in some other way.

 

Actually ur right, the resisitor is not exactly 0.1 ohm, its 0.7 ohm. But i ordered for 100m ohm. So ok let me check with potentiometer as u said. thanks

 

@Kubeek: Hey i tried with replacing the sense resistor with the potentiometer ' is that wot u mean?, but i dint get. By using the potentiometer i could see the change in light intensity but wn i tried to calculate the current, i got very different answers compared to the actual load current. I think i did wrong, can u pls edit my circuit and upload?.

 

I took the voltage across the sense resistor and divided that value by sense resistor value to find current drawn by load, is that right?. Does the load resistance doesn't effect ?. I am totally messed up man pls help.

 

Actually ur right, the resisitor is not exactly 0.1 ohm, its 0.7 ohm. But i ordered for 100m ohm. So ok let me check with potentiometer as u said. thanks

You can't measure a 100mOhm resistor very well with a standard meter. The resistance of the leads and connections are more than the resistor itself. If you ordered the resistor from a reliable source, then its probably correct. If you really want to measure it, then remove it from the circuit, source 1A through it, and measure the voltage drop across it. Then calculate the resistance. But I doubt your resistor is the issue.

 

I took the voltage across the sense resistor and divided that value by sense resistor value to find current drawn by load, is that right?. Does the load resistance doesn't effect ?. I am totally messed up man pls help.

Yes, that is correct.

Load current = Voltage across Rsense / Rsense

Remember, the sense resistor is in series with the load, and current is the same throughout a series path.,

 

Have you measured the output of your circuit with no load (zero current) to see if you have an offset?

Have you looked at your signal with a scope to see if there is excessive noise? Look at both the input signal, and the output.

 

@JMac3108: hey, I checked the resistor value as u said, it right 0.1 ohm.

 

I checked the circuit without the lamp (load), input voltage to circuit is 12v so at terminal Rs+ i got 11.99v , at Rs-: 11.99v and at OUT : 0v. So i dont think there is any offset. I checked with normal digital multimeter.

 

@JMac3108 checked with the scope, no noise as well. wot u think ???

 

I checked the circuit without the lamp (load), input voltage to circuit is 12v so at terminal Rs+ i got 11.99v , at Rs-: 11.99v and at OUT : 0v. So i dont think there is any offset. I checked with normal digital multimeter.

Check the output of your MAX4080 with no load. The output should be zero because the load current is zero. If its not zero, then this is an offset.

OOPS, sorry, missed the part of yuor post saying the output was zero. Looks like you don't have an offset.

 

Another possible issue ...

Are you checking the output with a meter or with the ADC in your micro?

The ADC inputs on microcontrollers have an equivalent internal circuit of a series resistor and a cap to ground - an RC low pass filter. This means that there is a required settling time before taking the measurement. Look in the datasheet (or more likely the users guide) for your micro and make sure that you are waiting the correct amount of time.

 

If you have no offset, no noise, and your signal is steady ...

And you measure the voltage across the sense resistor and calculate the load current ...

And you measure the MAX4080 output and apply the appropriate gain factor and this doesn't match the measured current ...

Then perhaps you are not understanding the MAX4080 properly. Are you sure you have the right version (there are several with different gains)? Are you exceeding the full-scale sense voltage for the version you're using?

One more idea - How is the circuit wired? Do you have long wires in series with your sense resistor that would affect the measurement? The sense resistor should be soldered right to the IC pins (no wires, no test leads, etc...).

 

Have you calculated a tolerance analysis?
- Resistor tolerance 1% or 5%?
- Gain tolerance of MAX4080, 1%
- Input offset voltage 1mV x 60V/V = 60mV

 

I havn't calculated the tolerance analysis. I am using MAX4080SASA+ having gain 60v. The sense resistor is connected using single stand wires with quite approx 3cms far from the MAX4080.

 

Tell me one more thing, instead of MAX4080, can i do with just the sense resistor and differential opamp ?. So that the voltage drop accross the sense resistor will be amplified with certain gain and can be present at the out put of an opamp. wot do u think???.

 

wait, Let me just measure it one more time and will give u the readings

 

Here is the final readings
using sense resistor 1 ohm
at Rs+ volts = 11.69v
at Rs- volts = 10.95v
Vout = 11.66v

with sense resistor 0.1 ohm
at Rs+ volts = 12.01v
at Rs- volts = 11.93v
Vout = 5.20v (its decreasing slowely with time in DVM)

Gain for both is 60v

So over all i think i am getting the result with 0.1 ohm but with some variation. Wot u think?

 

3cm is a very long wire to connect to a 100mohm sense resistor. It should be soldered right at the pins (a few mm at the most).

With the 1ohm sense resistor you're measuring 740mA, and with the 100mohm resistor you're measuring 800mA. Have you measured the current with a meter in each case to see how close these are to the real current in your load? I do not know what your load is ... remember that the value of the sense resistor can change the load current depending on how large it is compared to the load.

I don't know why the voltage would be decreasing in time across the 100mohm sense resistor.

Yes, you can use a sense resistor and an op-amp in a differential configuration, but you'll likely get the same results. Its not the MAX4080, its something in your circuit that you're not considering.

 

using sense resistor 1 ohm
at Rs+ volts = 11.69v
at Rs- volts = 10.95v
Vout = 11.66v

Don´t you realize that (11,69-10,95)*60 is 44.4V?
The circuit can´t magically make more voltage than the supply 11.69V, so the output is stuck at 11.66V.
But still for the 0.1ohm resistor the gain quite doesn´t work correctly.