2mV supply for 3 op-amp testing

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carnby24

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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?
 
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.
 
It is a voltage divider, like this:
 

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2mV

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

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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.
 

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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....
 
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.
 
Perhaps we should as why carnby24 wants these voltages. He may be barking up the wrong tree.
 
Hwy thanks guys.....i wonder why my lecturer wanted 3 resistors when there can be 2 only needed.....thats what confused me.
 
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?
 
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This is the batteries circuit...And where do i connect the ground to?
 

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

AllVol
 
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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...
 

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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.....
 
AllVol said:
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.
 
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?
 
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