amplifiernon-idealnon-invertingoperational-amplifier
This is probably more of an algebra question.
I do not understand where the "1"s come from in equation 1.10 given the much simpler voltage divider equation in the line just above it. And why do they convert a simple resistor division problem, into a division over a division problem?
The second part that loses me is the jump from equation 1.11 to 1.12. I have been trying and trying and I just cant get 1.11 to look like 1.12.
I'm just trying to subtract 10 volts from a voltage source ranging from 12v-60v
Do you have any suggestion on how I could subtract 10 volts from source ranging from 12v-60v
I apologize for quoting comments from the OP rather than taking quotes directly from the question but it appears to me he is clearer in the his requirements in his comments.
Why don't you use a zener diode like this: -
If you then want to process the signal and you think that 50V is too high then the 4k7 can be made a potential divider such as a 1K and a 3k9. This will reduce the 50V to 10.2V and the 2V to 0.408V.
This won't be a perfectly accurate way of subtracting the voltage but you haven't said how accurate this has to be. You can get "zeners" that are very accurate in the form of shunt regulators - they only need a few tens of microamps to work.
The node voltage equations are not so difficult. But it's really a tedious work to type the equations. I got the same result with you, maybe we are both wrong :).
Best Answer
For formula 1.10 they have taken this part of the equation: -
$$\dfrac{R_1}{R_1+R_2}$$
And divided top and bottom by \$R_1\$ to get this: -
$$\dfrac{1}{1+\frac{R_2}{R_1}}$$
Equation 1.11 evolves like this: -
$$V_o = a\left(V_i- \dfrac{V_o}{1+ \frac{R_2}{R_1}}\right)$$
$$V_o\left(1 + \dfrac{a}{1+\frac{R_2}{R_1}}\right) = aV_i$$
$$\dfrac{V_o}{V_i} = \dfrac{a}{1 + \frac{a}{1 + \frac{R_2}{R_1}}}$$
$$\dfrac{V_o}{V_i} = \dfrac{a(1 + \frac{R_2}{R_1})}{1 + \frac{R_2}{R_1} + a}$$
$$\dfrac{V_o}{V_i} = \left(1+\dfrac{R_2}{R_1}\right)\cdot\dfrac{a}{1+\frac{R_2}{R_1}+a}$$
$$\dfrac{V_o}{V_i} = \left(1+\dfrac{R_2}{R_1}\right)\cdot\dfrac{1}{\dfrac{1+\frac{R_2}{R_1}}{a}+1}$$
And that is equation 1.12. So, when a is very large we can approximate 1.12 to this: -
$$\dfrac{V_o}{V_i} = \left(1+\dfrac{R_2}{R_1}\right)$$