2mV supply for 3 op-amp testing

[carnby24]

Hi guys,

What are the values of resistors do we need to use to gain the output of 2mV supply from 3 resistors and 2 9v batteries?
Thanks ...i really need your advise and how to calculate the resistor values?

 

[audioguru]

Why use 3 resistors when only 2 resistors will give any voltage you want? The voltage will drop as the battery voltage drops because you aren't using a voltage reference IC. Connect the two resistors in series from one battery to the junction (ground) of the two batteries.
100k resistor in series with 22 ohms: 9/100,022= 0.0899mA x 22 ohms= 1.98mV. The polarity is determined by which battery the resistors are across.

 

[AllVol]

Is the output then taken from between the two resistors?

 

[audioguru]

It is a voltage divider, like this:

 

[carnby24]

2mV

I have roughly put the pics here.....How do we derive the calculations? 2-9v batteries and 3 resistors....

 

[audioguru]

I didn't calculate anything.
I selected 100k resistors so the little batteries aren't loaded too much, then simple arithmatic selected the 3rd resistor.

 

[carnby24]

Isnt there a way to prove by calculation to get the resistor values?
I mean the 2-9v batteries are given and output voltage is given....

 

[Hero999]

Read this, and you don't need three resistors you only need two.

 

[audioguru]

2mV/18V= 1/9000. Select two resistors of 100k each for low current and their total is 200k.
200k/ 9000= 22.2 ohms.

Check: 18V/200,022 ohms= 90uA. 90uA x 22 ohms= 1.98mV.

Many resistor combinations will give an output close to 2mV.

 

[Roff]

Perhaps we should as why carnby24 wants these voltages. He may be barking up the wrong tree.

 

[carnby24]

Hey thanks guys.....

 

[carnby24]

Hwy thanks guys.....i wonder why my lecturer wanted 3 resistors when there can be 2 only needed.....thats what confused me.

 

[carnby24]

I am supposed to construct a 2mV supply using 3 resistors and 2 9v batteries to connect to a 3 op amp cct and test its functionality and gain.
I left my studies long time ago and am continuing my studies now.....
Audioguru,

Can i get more specific answer to how you immidiately chose 100k resitors?How do we derieve these values form calculations?

 

[carnby24]

This is the batteries circuit...And where do i connect the ground to?

 

[AllVol]

Could the circuit below satisfy all requirements? I leave the values up to you.

AllVol

 

[carnby24]

its output is meant to be connected to a 3 op amp cct.... iwill upload the cct later.

Thanks to you guys its known that r1=r3.How do i derive the calculations with just the known values like vout=2mV, Vsupply=18V.I need to show how i got these values...

 

[carnby24]

Okay guys ...now i have some more information.....both the outputs need to supply 2 mV each....i guess thats why 3 resistors are used.Now the calculations part.....

 

[audioguru]

This is the batteries circuit...And where do i connect the ground to?

The ground connection is the 0V point of the opamp circuit.

 

[audioguru]

Could the circuit below satisfy all requirements? I leave the values up to you.

AllVol

The resistor R3 in your circuit doesn't do anything and isn't needed.

 

[audioguru]

Carnby,
After 30 replies now you show that you have an instrumentation differential amplifier made with 3 opamps. Why not lookup Differential Amplifier and Instrumentation Amplifier in Google to learn all about them?