Feeding negative volatage to AN1 of PIC micro

[alexan_e]

Basically you want to make https://en.wikipedia.org/wiki/Operational_amplifier#Non-inverting_amplifier
using Vcc and GND as supply , is that right?

 

[Rohith_elec]

yes sir

thats right

 

[alexan_e]

OK, that works fine.

 

[Rohith_elec]

I connected, both circuits to 12V input
and
5V and 0V to the opamp power rails,


I get
0.58V for Voltage
and
0.70V for current

please correct me

 

[alexan_e]

post your circuits

 

[Rohith_elec]

sir , the circuit is as 2 and 14 but

the power rails, that is

4th pin to -> 0V.
8th pin to -> 5V.

as below

 

[alexan_e]

The circuit on the left has a gain higher than 1 so it tries to amplify the 12v , I don't see how you intend to use it to feed the ADC

The circuit on the right is an inverting amplifier so when you feed a positive voltage it tries to give a negative in the output, again I don't see how you intend to use it with +12v.

 

[Rohith_elec]

sir, how can i correct it ?

 

[alexan_e]

Select if you want in inverting amplifier or a non inverting one and then select the gain that you want.
The circuit and gain are explained in the wikipedia link I have provided so it is up to you to select what you want to do in your circuit.

The non inverting amplifier is fine if you connect it to a low resistor to amplify the output voltage , in my example I have used a 20m ohm resistor that gave 0,02v/A but if you give 12v it can't work.

The same with the inverting amplifier, it was supposed to be used with a negative input to attenuate and in invert it but you have user a positive 12v input.

Experiment with your simulator to understand the basic operation of the opamp in inverting and non inverting configuration.
Use different inputs and resistor ratio and see the effect in the output.

 

[Rohith_elec]

i made one for current sensing,




can this be made to work ?

 

[Rohith_elec]

for R3 i am using 10K , but in this schematic i wrongly attached as 1K.

---------- Post added at 08:30 ---------- Previous post was at 08:18 ----------

I can get upto 4V by turning the RV1 and can be controlled precisely , but dont know this circuit is ok ?

 

[alexan_e]

The circuit is fine , as it is with the 1K resistors it amplifies the input 50 times so the problem is probably the voltage you are trying to give in the input.
You are using a pot in both the shunt resistor and the load, I have no idea what is the value of the shunt resistor and the load current so I can't calculate the expected output.
If you want to change the resistors then keep R3=R4 and R5=R6 , if you change R3=R4=10K then the gain will be 5

 

[Rohith_elec]

and just this resistor divider network is need for voltage input also isnt it ?

 

[alexan_e]

If you want to reduce the voltage to connect it in the ADC input then your divider is fine, the output is 1/4 of the input

 

[Rohith_elec]

ok , i will build all circuits and test with pic and will post the results,

thanks for such a nice support for me

 

[alexan_e]

I would suggest you post the final schematic with the used values here so we can check if it ok

 

[Rohith_elec]

this is the schematic i have made

 

[alexan_e]

The 10k as a load to 12v will have have a current of 12/10000 = 1.2mA

1.2mA * 1 ohm = 1.2mV

The opamp has a gain of 50 so the output will be 50 * 1.2mV = 60mV
It is kind of low so a 5v 10bit ADC will give a result of 12/1023

---------- Post added at 13:15 ---------- Previous post was at 13:14 ----------

Sorry, you have changed the gain of the opmp to 5 so you will get 5 * 1.2mV =6mV
you can barely get a result for that from the ADC.

 

[Rohith_elec]

so,
can i use the circuit by just increasing the opamp gain to 50 ?

 

[alexan_e]

You can either increase the gain of the opamp or the value of the shunt resistor (from 1 ohm to say 10 ohm).

If you intend to use the circuit with loads of so low current (high resistance load) then a higher shunt resistor value will not be a problem since its value will only be a small percentage of the load resistance and the error it introduces will be very low but as the load resistor gets lower the error will increase.
It all depends on the range you intend to use.