Feeding negative volatage to AN1 of PIC micro

[Rohith_elec]

i am using on a 2A load

 

[alexan_e]

I you are using a 2A load (6ohm for 12v) why have you used a 10k load in the schematic?

If the load is so low then even the 1 ohm will introduce a significant error

12v/6 ohm= 2A
12v / (6+1 ohm) =1.714A

Apart from that the 2A means 2v across the 1 ohm resistor , if you amplify it 5 times you should get 10V and with 50 times 100v

A resistor of 0.1 ohm could be used
12v / (6+0.1 ohm) =1.967A
you will still get an error but a smaller one.
Then resistor will give 0.2v for 2A so if you use a gain of 5 you will get 1v

 

[Rohith_elec]

can i make the .1ohm resistor with a wire ?,
Not easy to get a resistor of that rating here .

 

[alexan_e]

No I don't think you can do it with a wire, you need a wire with resistance and you wouldn't be able to measure it either.
Use the 1 ohm resistor for now (or two in parallel to get 0.5 ohm) but just be aware of the error you get in the current reading until you use a proper low value resistor..

---------- Post added at 19:59 ---------- Previous post was at 19:58 ----------

I hope the 1 ohm resistor is a couple of watts becasue you need 1W for each ampere that goes through it so if you use normal 1/4w you will have problems.

 

[Rohith_elec]

so i have to use some 3W 1ohm resistors, i have to buy , isnt it ?

i read 0.8ohm resistance in a wire , so i asked about using that wire as resistance

 

[alexan_e]

I have never used wire to make a resistor, I know that wire-wound resistors have wire internally but I don't think you can make one.

So you don't have any power resistor like 1ohm /1W or 1 ohm/0.5W ?

You can create a resistor using many small resistors in parallel, if you have small 1 ohm 1/4w resistors then you can use 4 of them to get a 0.25 ohm /1w resistor , this resistor will consume I * I * R watts so 2A * 2A * 0.25 = 1W

 

[Rohith_elec]

i am getting only .59V in current measuring circuit

 

[alexan_e]

Using what values in the circuit (gain)?
What value is the shunt resistor and what is the current of the load?
I have no idea what changes you did in the circuit and you just mention the result you got.

 

[Rohith_elec]

for shunt , i used 4 parallel 1ohms
for current gain i used 10K resistors
current of load is 1.75
R5 and R6 are 51K

 

[alexan_e]

OK so the shunt is 0.25 ohm
the gain is 51k/10k = 5.1
and the load current 1.75A

1.75A * 0.25R = 0.4375v * 5.1 = 2.231v this is the voltage you should get from the opamp so something is wrong.

Can you measure the voltage across the shunt resistor to see if it 0.43v so we can verify the current you mention?
Did you use LM358 with a supply of 5v and gnd as in the schematic?

 

[Fragrance]

hi

i used attach input divider for my voltmeter application and i get good result from it hope this will help you

regards
Fragrance

 

[Rohith_elec]

yes,
i am using lm358 with 5V and Gnd as schematic

across shunt, it shows no voltage

if i place multimeter probe there for sometime, the opamp gets heated up

 

[alexan_e]

If you have no voltage across the shunt resistor then there is no current flow, is the load connected?
Did you get the 0.59v opamp output with 0 voltage across the shunt resistor ?

 

[Rohith_elec]

yes,
load is connected

 

[alexan_e]

This doesn't make sense , if the load is getting power through the shunt resistor then you will have a voltage drop on the shunt proportional the the current flow , how can you have the load powered and 0 voltage on the shunt.

 

[Rohith_elec]

or may be the method of reading by me is wrong,

is it from one end of the resistor to the other end
or from one end of resistor to gnd ?


i measured from one end of resistor to other end

 

[alexan_e]

across the shunt resistor means from one end to the other, in this case one side is already connected to ground so you can just use the ground and the other side (not the ground with the ground side).

Do you have the voltmeter set to mV , the measured voltage will be less than 0.5v

 

[Rohith_elec]

yes, i read in mv range

also, the load is a parallel connections of 8 250mA leds from uln2803 from PIC micro, is that because of parallel connection, the load shows no voltage ?

 

[alexan_e]

Are all leds on?

The ULN controls the ground side (low side switch) you shouldn't connect pin9 to the ground directly but to the shunt resistor and then the shunt to the ground so that the output current goes through the shunt resistor.

 

[Rohith_elec]

I corrected the circuit like that,
now voltage across the shunt is 0.246V


i switched on only 6 LEDs , so all togather 1.5A