Working of active clamp....?

Question

The text says:

"For the values shown, Vin < +10 volts puts the op-amp output at positive saturation, and VOut= Vin, When Vin exceeds +10 volts the diode closes the feedback loop, clamping the output at 10 volts. "

My inference: When Vin<+10V the opamp increases its output in positive direction in order to equalize voltages at the two inputs..thus as a result the diode become reversed biased and Vin =Vout .. its fine till here... but when Vin>+10V the opamp has to decrease its voltage ..and as the text says that the diode is reversed bias in this case...that means that the opamp outputs some negative voltage (not sure) ..which forwards bias the diode...and the output is clamped to one diode drop below Vopamp.. my problem is ...for Vin>+10V will the opamp output be a negative voltage .... and if yes then suppose Vopamp = -10.7V..for any Vin value greater than 10V.. then -10.7V will experience a voltage drop ....that voltage drop is -0.7V ..

-10.7-(-0.7) = -10V.. ...i.e Vout = -10 V... but it should be 10V...as per text..in order to equalize the inputs....?

please guide..

//Certainly for Vin>+10V....Vopamp cannot be a positive voltage ..that offcourse will reverse biase the diode again.....which doesnt seems possible..//

Answer 1

In order for V_out to be -10V, the op amp would have to output below -20.7V, since the input is at least +10V.

But that's not how op amps work.

In the presence of feedback, the op amp will output the exact voltage required to (approximately) equalize both its inputs. In order to bring the inverting input to at most +10V in order to match the non-inverting input, it needs to output at most +9.3V. The diode will therefore remain forward-biased when the input is greater than +10V, and be reverse-biased or at equilibrium when it is below that.

Answer 2

Actually the diode is reverse-biased when the input signal is < 10V, Vout follows Vin, and the output of the op-amp is saturated at maximum output voltage.

When Vin is > 10V, the output of the op-amp goes to 9.3V and Vout goes to 10V.

The 2k resistor sees the difference between Vin and Vout, and a current flows.

So Vout follows Vin when < 10 V and "clamps" to 10V when Vin is > 10V.