Calculate the value of RG needed in order to produce the fol

Calculate the value of RG needed in order to produce the following output voltages (peak-to-peak) using the diagram below. a) Vout = 5 volts b) Vout = 7.5 volts c) Vout = 10 volts d) Vout = 10.5 volts

Solution

The op amp amplifies until:

V(+) = V(-) = Vforward(diode) = 1 V

Because one end of Rosc is between half point of 2 equal capacitors, this end of resistance will have a potential V(+)/2 = V(-)/2 =0.5 V

U(Rosc) = V(-) -V(-)/2 =1-0.5 = 0.5 V

I(Rosc) =0.5/330 =1.515 mA

I(Rg) = V(-)/Rg

I(Rf) = I(Rosc) +I(Rg) = I(Rosc) + V(-) / Rg

V(out) -V(-) = I(Rf)*Rf

V(out) = V(-) + [I(R_osc) + V(-) / Rg] * Rf

With the values in the diagram included in your problem (Rg =100 ohm, Rf=1000 ohm, Rosc =330 ohm) one has

V(out) = 1+[1.515 m +1/100]*1000 =1+(1.515 m +10 m)*1000 =1 +11.515 m*1000 =12.515 V

(which is in fact V(out) = 12 V due to voltage sources =+/-12V)

If it is given V(out), one can solve for Rg

V(-)*Rf/Rg = V(out) -V(-) - I(Rosc)*Rf

Rg = [V(-)*Rf] / [V(out) -V(-) - I(Rosc)*Rf]

For Vout = 5 V

Rg = 1*1000 / (5 -1 -1.515 m*1000) =1000/(5-1-1.515) =402.41 Ohm

For Vout =7.5 V

Rg = 1*1000 / (7.5 -1- 1.1.515 m*1000) =1000/(7.5-1-1.515) =200.6 Ohm

For V(out) =10 V

Rg = 1000/(10-1-1.515) =133,6 Ohm

For V(out) =10.5 V

Rg = 1000/(10.5 -1-1.515) =125.23 Ohm