bidirectional dc current sensing with 358 or 324 op amp

dougy83 · Nov 10, 2014

R1/200000 is the gain of the circuit, and includes the effects of the current-sense and other resistors

sahu · Nov 10, 2014

dougy83 said:

Iin is develop by current flow . that's are in milivolt .. so

this EQUATION right for here .
sory for my bad English

dougy83 · Nov 10, 2014

I don't know what you're trying to say. Iin is in Amps.

sahu · Nov 10, 2014

dear sir made this circuit . test it but unexpected result found those are as

alec_t · Nov 10, 2014

What supply voltages did the opamp have? Which opamp did you use?

sahu · Nov 10, 2014

alec_t said:
What supply voltages did the opamp have? Which opamp did you use?

5 volt & opamp LM358

alec_t · Nov 10, 2014

According to its datasheet the LM358 output can only get to within 1.5V of Vcc, so don't expect a full 0-5V output range. This may explain the results you got.

sahu · Nov 10, 2014

alec_t said:
According to its datasheet the LM358 output can only get to within 1.5V of Vcc, so don't expect a full 0-5V output range. This may explain the results you got.

its means , if I use reff volt 1.5 volt then result get expected.
& equation use as
o/p =(1.5 + 10000 * Iin)/200000

sahu · Nov 12, 2014

so, any body not have any sagesition for me .

ronsimpson · Nov 12, 2014

I am lost as to what circuit you are using now.

1) The "V2 2.5" can also be a 1k to +5 and a 1k to ground. If you 2.5V is not exactly right OR your R6 & R2 is not right there will be a offset error. Use 1% resistors. You say that 0 current is 2.352 volts. This is probably because (1/2 of 5V) does not equal 2.5V OR R2 does not equal R6. When things are balanced right there should be 2.5V on both inputs and on the output.

2) You might need a R-R output opamp. The inputs are at 2.5V so you don't a R-R input opamp. These amplifiers will pull to ground and supply better than the old types.

Nigel Goodwin · Nov 13, 2014

ronsimpson said:
2) You might need a R-R output opamp. The inputs are at 2.5V so you don't a R-R input opamp. These amplifiers will pull to ground and supply better than the old types.

I would suggest he's better off using a real split supply (perhaps using a 7660 as my tutorial does - and for the same reason), ensuring it completely reaches 0V.

Also like my tutorial, I would suggest using a 2.5V reference (many newer PIC's have this in-built), thus avoiding limitations approaching the +ve rail as well.

"Rail to rail" opamps are really nothing of the kind, just 'closer' than standard opamps.

Nigel Goodwin · Nov 13, 2014

ronsimpson said:
2) You might need a R-R output opamp. The inputs are at 2.5V so you don't a R-R input opamp. These amplifiers will pull to ground and supply better than the old types.

I would suggest he's better off using a real split supply (perhaps using a 7660 as my tutorial does - and for the same reason), ensuring it completely reaches 0V.

Also like my tutorial, I would suggest using a 2.5V reference (many newer PIC's have this in-built), thus avoiding limitations approaching the +ve rail as well.

"Rail to rail" opamps are really nothing of the kind, just 'closer' than standard opamps.

sahu · Nov 13, 2014

ronsimpson said:
I am lost as to what circuit you are using now.

1) The "V2 2.5" can also be a 1k to +5 and a 1k to ground. If you 2.5V is not exactly right OR your R6 & R2 is not right there will be a offset error. Use 1% resistors. You say that 0 current is 2.352 volts. This is probably because (1/2 of 5V) does not equal 2.5V OR R2 does not equal R6. When things are balanced right there should be 2.5V on both inputs and on the output.

2) You might need a R-R output opamp. The inputs are at 2.5V so you don't a R-R input opamp. These amplifiers will pull to ground and supply better than the old types.

dear sir , here not issue of R6 & R2 . i'm telling here is V(out) = 2.352 v .
i has found both place 2.5 v(- & +) of opamp . where we used 1k resistors (1%)

MrAl · Nov 13, 2014

Hello,

Quick question:
Is it absolutely necessary that the lower shunt lead goes to ground?
If the current to be measured is not common with anything else it does not need to go to ground.

sahu · Nov 13, 2014

MrAl said:
Hello,

Quick question:
Is it absolutely necessary that the lower shunt lead goes to ground?
If the current to be measured is not common with anything else it does not need to go to ground.

yes boos , lower shunt lead to ground.

ronsimpson · Nov 14, 2014

sahu said:
dear sir , here not issue of R6 & R2 . i'm telling here is V(out) = 2.352 v .
i has found both place 2.5 v(- & +) of opamp . where we used 1k resistors (1%)

Just for the fun of it. Make R6 990 ohms and R2-1010 ohms. (-1% and +1%) See what happens.
The "input offset" of the LM358 is "7mV worst case room temperature". This can be seen by changing the 2.5V to 2.507V or to 2.493V. See what happens.

Spice will hand you perfect components. Like 1000.0000 ohms. The LM358 also has imperfections like input offset.

sahu · Nov 16, 2014

i use also like this ,
Where: VO = V1 + V2 − V3
but when v1 = 0, v2=0 & v3=0 .
then vo - 3.5 volt.
i confused with op amp.

ronsimpson · Nov 16, 2014

1) I crossed out the ground symbol that should not be there.
2) If V1=V2=V3=0 volts then Vo= 0 volts. Not 3.5

This could be used on the current sense amplifier talked about above.
Connect V1 to a reference voltage (2.5V).
Connect V2 to ground.
Connect V3 to the current sense resistors. (normal voltage is o volts)
The output will be 2.5V at 0 current. When the current is positive the Vo will drop below 2.5 volts. When the current if negative the voltage will be above 2.5 volts. (there is a inversion there but that is usually OK)

sahu · Nov 16, 2014

ronsimpson said:
1) I crossed out the ground symbol that should not be there.
2) If V1=V2=V3=0 volts then Vo= 0 volts. Not 3.5
View attachment 89293
This could be used on the current sense amplifier talked about above.
Connect V1 to a reference voltage (2.5V).
Connect V2 to ground.
Connect V3 to the current sense resistors. (normal voltage is o volts)
The output will be 2.5V at 0 current. When the current is positive the Vo will drop below 2.5 volts. When the current if negative the voltage will be above 2.5 volts. (there is a inversion there but that is usually OK)

normally adc reading are as

result = 0;
for (i=0;i<20;i++)
{ set_adc_channel(0);
delay_ms(1);
result=result+read_adc();
result = result/20 ; }

if its right

then now

result = result * vout

is righ ?

sahu · Nov 17, 2014

dougy83 said:
I think it's better to do it as shown in the following circuit. Zero amps gives ~2.5V out, -50 A --> 5 V out and +50 A --> 0 V out. Gain can be changed by changing R1.
View attachment 89239

can u ell me , pl what is its Sensitivity ?

bidirectional dc current sensing with 358 or 324 op amp

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