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.
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.
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?
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...
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.....
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?