measuring load current (current sensor)

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canarybird33

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Hi every body
I want design a circuit to measure load current(LOW COST and reliable)
I chose the blow circuit but it doesn't work
(changing the load current doesn't cause voltage variation in a range)
How can I correct this circuit.
load current variation is 3mA to 300 mA and I want to convert it to 0V to 5 ( and the use the ADC of micro-controller)

 


I cant see your circuit which you've mentioned that , came with your post . can you attach it please ?
Anyway ,
Simple put a low value resistor e.g 0.27R or 0.1R in series with ground path and then read the voltage across it , and divide that value by , resistor , then value of the current will be indicated .

Best Luck
Goldsmith
 

300 mA through R2 means 0.6 mA through R1 to give the equal inputs to the op amp. That means 1.2 volts at the voltmeter (across the 2k resistor). Not 5v. So your scaling is off. Also the circuit requires operating with a common mode input voltage very close to 5 volts. Since the 741 is not a rail-to-rail input device, you need to power the op amp with much more than 5 volts.
 
Your circuit cannot operate as shown. Besides the 741 problem, there's no negative feedback to the op amp so it's running open-loop and there is no power to the collector of Q1.

Below is the simulation of a current-measuring circuit using a rail-rail op amp and a single 5V supply. Since a rail-rail op amp cannot go totally to the rail with an output load, I used a gain of 80 to give a nominal maximum output of 4.8V with 300mA current.

Note that the 2kΩ output load R4 just represents whatever load your converter will give and is not needed in the actual circuit

There are many available rail-rail op amps that should work in the circuit besides the one shown.

 
crutschow said:
Your circuit cannot operate as shown. Besides the 741 problem, there's no negative feedback to the op amp so it's running open-loop and there is no power to the collector of Q1..
Click to expand...
I think there is just a small typo in the circuit. If you assume that the point marked R1(1) is actually a positive power supply, then everything makes sense. That's how I interpreted it. Then the negative feedback comes in on the (+) input, but it is still negative feedback because of the inversion provided by Q1. As I see it, the only real problems with the circuit (aside from this typo) is the possible common mode violation and the fact that the scaling is such that 300ma. through the load would only produce 1.2 volts, not 5 volts as the OP was wanting. Changing the 2K resistor to 8333 ohms would fix the scaling problem.
 

Tunelabguy said:
Changing the 2K resistor to 8333 ohms would fix the scaling problem.
Click to expand...

How can I calculate the R2 resistor?
what about R1 resistor?can I change that?
I designed new circuit(main supply is 24v . Being more sensitive in lower current(4 to 50 mA) is significant to me)
When I increased R3, the out put voltage gets more sensitive to the lower current.I got what I want but let me know what is happening in the circuit?what is the role of Q1?(It makes the feedback negative-what else?)

And another question
crutschow said:
There are many available rail-rail op amps that should work in the circuit besides the one shown.
Click to expand...

what is the difference between the black and white circuit and my design? which of them is more reliable?





I simulated the black and white circuit(R sens= 0.05volt/300mA =0.16 ) But the sensitivity to current changing(in 4 to 50 mA) is bad.what I am suppose to do?
 
Last edited by a moderator:

Okay, I understand the circuit now. Some of the resistors are way too high in value. Try changing R1 to 4.75Ω and R3 to 2kΩ leaving R2 at 0.2Ω. Below is my simulation of a similar circuit.

The current to voltage gain of the circuit (volts per amp) is (R3 / R1) * Ild * R2, so you can change any of the values as needed. as long as it keeps the voltages within the linear range of the amp.

Q1 helps translate the current on the power buss to a ground based voltage. The op amp turns on Q1 to generate a current through R1 until the voltage across R1 equals the voltage across R2. This current through R1 also goes through R3 due to the transistor action and generates a voltage across R3 proportional to the current through R2.


 
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I think 0.2 Ohms is too low for R2. Remember the maximum load current is only 300 ma. That little current through a 0.2 Ohm resistor gives a voltage drop of only 60 mV, and that is at full scale. The input offset voltage of a 741 op amp over the full temperature range is about 6 mV max. So we would have a potential error of 10% at full scale and even more percentage error at lower current levels. Of course a better op amp could be used to lower the input offset voltage, but in this case I think an easier solution is to sacrifice a little more drive voltage. If the load is supplied by a 12 volt power supply, dropping only 60 mV to measure the full scale current is an unnecessary approach to perfection. I would allow maybe 10 times as much voltage drop at full load, so a 2.0 Ohm resistor for R2 would give 600 mV drop at full load and ease the input offset voltage specifications on the op amp. If R2 is 2 Ohms, then pick R1 so that a substantially lower current goes through R1. How about leaving R1 = 100 Ohms? If the full load of 300 mA flows through R2, only 6mA will flow through R1 and R3. Then to make the scaling work out, this 6 ma. has to produce 5 volts across R1, so R1 would be 833 Ohms.
 
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I agree that 0.2Ω is probably too low for R2 unless you want to use a low-offset op amp (incidentally you need a rail-rail op amp for this circuit so a 741 or similar won't work).

But you want 5V across R3, not R1. The op amp forces the voltage across R1 to equal R2 or 0.6V at 300mA for R2 = 2Ω. Thus for 5V output the resistance ratio of R3/R2 is 5V/0.6V = 8.333. So, for an output R3 value of 2kΩ, R2 would be 240Ω.
 

crutschow said:
But you want 5V across R3, not R1. The op amp forces the voltage across R1 to equal R2 or 0.6V at 300mA for R2 = 2Ω. Thus for 5V output the resistance ratio of R3/R2 is 5V/0.6V = 8.333. So, for an output R3 value of 2kΩ, R2 would be 240Ω.
Click to expand...
Correct. That was a typo on my part. Most of my references to R1 near the end of my post should have been R3.
 

hello
I run the below circuit practically
It works well in Proteus but in practical Vout(out put of opamp) is 0.6volt and doesn't change with the changing the current of load.
There is no problem in circuit.I checked it carefully and changed IC but it doesn't work
What is the problem?How can I check or test the circuit?
 

What is the voltage being supplied to R11(1)?

What is the voltage being supplied to pin 4 of the LM324?

Because the LM324 does not have a rail-to-rail common mode input range, you must ensure that the power pin on the LM324 (pin 4) is at least 2 volts higher than any expected input voltage. The input voltage in your circuit is nearly the same as the voltage you supply to R11(1).

Also realize that with only 7.27 ma. flowing in the load, the voltage drop across R9 is only 7.27 mV. This is of the same order of magnitude as the input offset voltage of a LM324 (about 3 mv max). So don't expect to get much accuracy at the low end with this circuit. Even at full load (300 ma?) the input offset voltage is a full 1% of the voltage being measured. You can fix that with a larger R9.

Edit: OK, I just realized that the voltage supplied to R11(1) is 24 volts. So you need to power the LM324 with at least +26 volts, and preferably +28 volts. Can you do that?
 
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That circuit needs a rail-rail op amp.

It also suffers from the problem of high offset voltage from the mismatch in resistor values (change one of the resistors [except R12] by 1% to see this). The circuit with the transistor does not have that problem.
 
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crutschow said:
That circuit needs a rail-rail op amp.
................... The circuit with the transistor does not have that problem.
Click to expand...


thanks guys, the problem in practical was exactly the VCC of lm324.
** An important question: If i change LM324 withe a rail to rail opamp, Is it possible to use a power suuply for both circuit and opamp?(I have to run the LOAD with 24 volts)
When I went to bye a rail to rail opamp I didn't know how to find an opamp which be accessible in shop.and the shopkeeper didn't know abut opamps, too.
So what is the prefix of rail to rail opamps?(I chose LT1637 and other LTs but was not accessible)


If I couldn't find a rail to rail opamp (OR a cost effective one ) What's the best idea to do?
should I use a DC-DC converter? or Using a Capacitor is reliable?
 

There is no prefex or other identification for rail-rail op amps. You just have to search each manfacturer. Fore example, here's a relatively inexpensive rail-rail op amp. What's "cost effective" for you?
 

Op amps are called rail-to-rail in two ways. One is for having an output that can go the full range from one power supply rail to the other. The other way is to have inputs that can operate over the full power supply range. Some op amps have both kinds of rail-to-rail. If it helps you any, you only need rail-to-rail input, not rail-to-rail output. If you can find an op amp that is rail-to-rail for its inputs, that is good enough.
 
crutschow said:
Texas Instrument has several rail-rail types for less then $1 US.
Click to expand...

I did search on https://www.ti.com/ but I couldn't find an input rail to rail opamp which is works at 24volts.
but i decided to use a LM317 to adjust the input voltage 3volts less than supply voltage I will do it and say the result.
have a good time
 

hello everybody
I setup the circuit. Now input pins of opamp is 21volts and supply voltage of opamp is 24 and there is no problem with it but I face with a new problem.
Current limiter circuit which I attached isn't work as I expect
Q2 gets HOT while limiting the current load.it is because of P=Vce*Ice=20volts*300mA=6Watt
What I'm suppose to do?How can I correct this circuit to limit the current(at 300mA)
The Load is in Collector
 

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