Calculate voltage and current in an op amp

Question

Hi I have a op amp circuit that i need help in calculating some values (see image, Å=A in further description)

The values i have are:

                  RF =  0.4 kΩ

                  RA =  70 kΩ

                  Ri =  0.14 kΩ                          

                  RL =  0.7 kΩ

So for E= 100 mV what are a) Uut and b) IA ?

I have tried doing Uut=EF=E(-RA/(RF+Rk/2))=-6.363 V which was wrong. I thought it was a differential op amp, so i found the F formula in the book, but i don't know if that is correct to use here.

Answer 1

To make things easier for me explain, you have two Ra in the circuit. I will call the lower Ra in the schematic as Ral and the upper as Rau, ok? And I will call the node that connects Ral, Rf and the non-inverting pin of the opamp as V+ and the node which connects Rau, Rf and the inverting pin as V-, ok? So, when the opamp have feedback, it will do everything it cans to make the voltage at V- and V+ to be the same value. It is sometimes called as a "virtual short-circuit". With that in mind, you can imagine a single network made on by E, Ri, Rf and the other Rf, all in series. In such circuit, is pretty easy to calculate the current that will flow through E, Ri and both Rf, right? I will call that current as Ix. Assuming that the opamp is ideal, no current will flow trough the opamp pins. So, you can see that Ix will flow through Ral, and with that information you can calculate the voltage across Ral, right? So, you can calculate the voltage at V+, as one pin of Ral is in the ground (the reference), so V+ has the same value as the voltage across Ral, ok? Remeber that V- and V+ have the same value? So, if you know the value of V+, you know the value of V-. Can you see that Ix will flow through Rau, since there is no other path to it? Can you see that the value of Ia is -Ix? By knowing the value of Ix, can you calculate the voltage across Rau? If you know the voltage of V- related to the reference, can you calculate the voltage on the other pin of Rau related to the reference? Sure you can. And that voltage is... Uut.

I hope now that circuit is easier to understand than my horrible english :)

Answer 2

Classic differential amplifier. Resistors \$R_F\$ and \$R_A\$ connected to non-inverting and inverting inputs of-amp are the same, respectively, so the simplest way to calculate output voltage shall be:

$$ V_{out} = \frac{R_A}{R_F} \cdot (V_+ - V_-) $$ $$ V_{out} = \frac{70k\Omega}{0.4k\Omega} \cdot -0.1V = -17.5V $$

But... there is internal resistance of voltage source \$R_i\$ which is in series with resistor \$R_F\$ connected to non-inverting input, so we need to use a full equation for differential amplifier, and assume that \$R_{Fi}\$ is a sum of \$R_F\$ and \$R_i\$:

$$ V_{out} = \frac{(R_A + R_{Fi})R_A}{(R_A + R_F)R_{Fi}} \cdot V_+ - \frac{R_A}{R_{Fi}} \cdot V_- $$

In above equation \$V_+\$ and \$V_-\$ are potentials referenced to ground. To obtain theme You should remember that both inputs of an op-amp are considered to be on ground level all the time. So \$E\$, its internal resistance and \$R_F\$ resistors create simple circuit with common current flowing through them. This current is:

$$ I_E = E / (R_i + R_F + R_F) = 0.1V / (0.14 + 0.4 + 0.4)k\Omega \approx 106.4{\mu}A $$

Voltage drop at \$R_F\$ is:

$$ V_{R_F} = I_E \cdot R_F = 106.4{\mu}A \cdot 0.4k\Omega = 42.5mV $$

So potential \$V_+\$ (connection point of lower \$R_F\$ and \$E\$) is \$-42.5mV\$, referenced to ground. And potential \$V_-\$ (connection point \$R_i\$ and \$E\$) is \$57.5mV\$, \$100mV\$ higher than \$V_+\$.

Now, we know everything:

$$ V_{out} = \frac{(70k\Omega + 0.54k\Omega)70k\Omega}{(70k\Omega + 0.4k\Omega)0.54k\Omega} \cdot -42.5mV - \frac{70k\Omega}{0.54k\Omega} \cdot 57.5mV \approx -12.97V $$

Current flows through the \$R_A\$ is:

$$ I_A = V_{out} / R_A = -12.97V / 70k\Omega \approx 185.3{\mu}A $$

As user1521378 suggest You can also assume that \$V_-\$ will be at connection point of upper \$R_F\$ and internal resistance of \$E\$, then You should calculate potential at this point in the same manner like for \$V_+\$ and use first simple equation to obtain output voltage.