Current sense circuit

Ghiran,

The schematic you posted did not come through. Can you re-post it?

From a general perspective, our products are great at doing this type of conversion. A small shunt shunt resistor is placed in the line of current you are attempting to measure, and an amplifier measures this voltage and outputs it as a an amplified single ended signal. This video is a great starting point for how this is done and which type of amplifier might work best for your system.  

I need some additional information, such as the common mode voltage and current range you are looking to measure to be able to recommend a device. Can you share what voltages you are present in your system and the amount of current you are looking to measure? Also, Are you wanting the device to pass the current as a voltage back to your microcontroller, or do you want the measurement sent digitally back to the MCU?

Ghiran,

These requirements will be challenging to meet. Regarding the supply voltage, does the supply simply need to exist in this range, or does the part HAVE to have high voltage capability? An main advantage of a current shunt amplifier is that the supply voltage does not need to be as high as the common mode of the signal monitoring it. If you haven't yet, I'd advise you to check out the video I posted above, and make sure you understand which type of amplifier you want to use here and why. 

Here's a few other things to consider:

- a square wave of that frequency should not be too demanding on any amplifier's bandwidth, but CMRR of amplifiers grows worse with A/C frequency. The harmonics of the square wave may show up as slight distortions in the measurement, which will add some error to your system.

- To measure two inputs, you will either need two separate shunts with one amplifier each, or two amplifiers with differing gains monitoring the same shunt. For signals this small, you will need a relatively large gain to make sure the shunt remains small. For example, for a signal of only 12-17mA, if we assume an amplifier with a gain of 50, for your 10V signal, you would need a shunt of 10V/50/15mA = 13.33ohms. This creates a few problems. first, for the low signal, 5mA*50*13.33ohms = 3.33V, which is lower than your desired 5V. Second, a resistor this large is going to add a large amount of error to your measurement due to the bias currents flowing across it. Third, resistors this large will have an impact on voltage seen by the load, as typically R_SHUNT should be much less than R_LOAD. How accurate do you need the measurement to be? do you just need to determine the level difference here?

- Unfortunately, none of our current shunt amplifier products are available in a through hole package. Is this request for ease of prototyping? We do have some leaded more proto friendly packages, such as SC-70 and SOT-23-5/6 if this is an option.

Based on the information so far, I would say with INA199. It is capable of up to a 26V supply voltage, and comes in a leaded SC-70 package.

"How accurate do you need the measurement to be? do you just need to determine the level difference here?"

I need it to be pretty accurate , this is a feedback singal from a motor with wich i control a position . Or i could use theoutput of the amp to power a fast mosfet that has the Vce ranging from 5-10v ?

Ghiran,

The issue here is that the low end of your current range is approaching where the bias currents and offset of a current sense amplifier begin to interfere with the accuracy of the measurement unless you use a rather large shunt. For example, our INA240 is made for motor applications with its PWM rejection feature, but even it will exhibit ~8% error trying to measure 5mA with a 100mΩ shunt:

It also is only capable of a 5V supply, so a maximum ideal Vout is 5V as well, which means you cannot supply your PLC with the needed 10V. Swing limitations would further diminish this.

How are you creating this square wave? It sounds like you are using current out Hall sensors, and then trying to convert that signal back into a voltage? Is that correct? 

"so a maximum ideal Vout is 5V as well"

I thought about it and i would use an arduino witch can communicate with my plc. So the max 5 v Vout sounds appealing .

The signal is coming from a 2 wire hall sensor with current interface. I'm thinking of using a 1ohm resistor since my current is in that range.

Is that a simulation software from witch you got the picture ?