Hi again,
I almost forgot to include the equations for that model which should make the requirement of iOut perfectly clear...
Vout=(G*(R2*V1+R1*V1+ib*R1*R2))/(R2+G*R1+R1)
iOut=(Vout*RL+Vout*R2+Vout*R1-ib*R2*RL)/((R2+R1)*RL)
or more simply:
iOut=(G*RL*V1+G*R2*V1+G*R1*V1-ib*R2*RL+ib*G*R1*R2)/((R2+G*R1+R1)*RL)
where
G is the gain of the amplifier which is drawn as 100000,
ib is the bias current, drawn as 50ua,
R1 is the top resistor as drawn,
R2 is the bottom resistor as drawn,
RL is the load resistor as drawn.
So with G=100000, ib=50e-6, R1=250, R2=750, RL=1000, we get:
iOut=0.010037 amps
which is slightly greater than 10ma, so this would work for any version of the LM317.
I almost forgot to include the equations for that model which should make the requirement of iOut perfectly clear...
Vout=(G*(R2*V1+R1*V1+ib*R1*R2))/(R2+G*R1+R1)
iOut=(Vout*RL+Vout*R2+Vout*R1-ib*R2*RL)/((R2+R1)*RL)
or more simply:
iOut=(G*RL*V1+G*R2*V1+G*R1*V1-ib*R2*RL+ib*G*R1*R2)/((R2+G*R1+R1)*RL)
where
G is the gain of the amplifier which is drawn as 100000,
ib is the bias current, drawn as 50ua,
R1 is the top resistor as drawn,
R2 is the bottom resistor as drawn,
RL is the load resistor as drawn.
So with G=100000, ib=50e-6, R1=250, R2=750, RL=1000, we get:
iOut=0.010037 amps
which is slightly greater than 10ma, so this would work for any version of the LM317.
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