3
\$\begingroup\$

enter image description here
(Source: Designing Switching Voltage Regulators With the TL494, Page 25)

What does the output of this opamp give? Use approximated values if necessary.


The reference voltage is \$V_R = 5V\$. Denote the opamp output as \$V_e\$.

Considering that the opamp will work in the linear region, we equate the inverting and non-inverting input voltages.

$$ \dfrac{R_9}{R_8+R_9}V_o \tilde= \left[V_e - \dfrac{R_4}{R_3+R_4}V_R \right] \dfrac{R_5}{R_5+R_7} + \dfrac{R_4}{R_3+R_4}V_R \\ \dfrac{R_9}{R_8+R_9}V_o = \dfrac{R_5}{R_5+R_7} V_e + \dfrac{R_4}{R_3+R_4} \cdot \dfrac{R_7}{R_5+R_7}V_R \\ V_e = \dfrac{R_5+R_7}{R_5} \cdot \dfrac{R_9}{R_8+R_9}V_o - \dfrac{R_4}{R_3+R_4} \cdot \dfrac{R_5+R_7}{R_5} \cdot \dfrac{R_7}{R_5+R_7}V_R \\ V_e = \dfrac{101}{2} V_o - 50 V_R \\ V_e = \dfrac{V_o}{2} + 50(V_o - V_R) \\ $$

The result I found doesn't make sense, because the opamp is always saturated to the positive rail voltage.


EDIT: There was a sign error in my original post. MathEE corrected it. Now the result is meaningful. The output is the 50 times the voltage error (excess voltage) at the otput + 2.5 bias level.

\$\endgroup\$

1 Answer 1

3
\$\begingroup\$

I think that you have a sign error in transition from the first to the second line:

$$\begin{eqnarray} \left[\frac{-R_4}{R_3+R_4}\frac{R_5}{R_5+R_7} + \frac{R_4}{R_3+R_4} \right] V_R &= & \left[-\frac{R_5}{R_5+R_7} + 1 \right] \frac{R_4}{R_3+R_4}V_R\\ &=& \left[\frac{R_7}{R_5+R_7}\right]\frac{R_4}{R_3+R_4}V_R\\ \end{eqnarray}$$

\$\endgroup\$
3
  • \$\begingroup\$ You wrote the exact result I came into. I don't see a difference. \$\endgroup\$ May 21, 2014 at 16:03
  • \$\begingroup\$ You have minus the term that I have. In the second equation the minus sign should be a plus. \$\endgroup\$
    – SomeEE
    May 21, 2014 at 16:04
  • \$\begingroup\$ Yeah, I saw it. I'm fixing it in a minute... \$\endgroup\$ May 21, 2014 at 16:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.