# Active clamp circuit confusion....

1. Sep 22, 2015

### brainbaby

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....?

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

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2. Sep 22, 2015

### Merlin3189

Yes the output could try to go negative if it has this capability (split supply.) But it does not need to go negative, only to drop towards 0V. If Vin is 10V, the output only has to drop to about 9.3V to switch the diode on

As soon as the diode switches on, the op amp output is connected to the inverting input (via the diode) and pulls it down, so the op amp output can never fall below about +9.3V

"then suppose Vopamp = -10.7V..for any Vin value greater than 10V."
This is the weak point in your argument! Even if the op amp is capable of outputting -10.7V, which is perfectly reasonable, in this circuit that can not happen. As soon as Vopamp drops to +9.3V, the diode starts to conduct and the feedback ensures that it drops no further.

"//Certainly for Vin>+10V....Vopamp cannot be a positive voltage ..that offcourse will reverse biase the diode again.....which doesnt seems possible"
I'm not sure what you are saying here? For Vin >10V the Vopamp will try to go low and will forward bias the diode. But it can and will be a positive voltage, about +9.3V, which does not reverse bias the diode.
The diode is only reverse biased once Vopamp approaches +10V. The increasing resistance of the diode in the feedback path effectively increases the closed loop gain of the inverting amplifier and drives Vopamp up to positive saturation.

3. Sep 22, 2015

### brainbaby

My inference:
Opamp outputs negative voltage only when split supply ..or negative supply is present..

you told that opamp will go low till 0 V since no spilt supply is present..but what i know that in order to reverse bias a diode its n junction should be connected to positive terminal..of supply..and it this case any voltage till zero volts would be positive ..so technically according to me the diode is reversed biased..but according to you its forward biased...

4. Sep 22, 2015

### Merlin3189

Ok. That paragraph might be a bit confusing, so let's try again.

I did not say the Vopamp will go to 0V, only that it will try to fall towards 0V. But as soon as it falls to about 9.3V, then the diode becomes forward biased and conducts.
This is because the Vin has risen to >10V, which makes the inverting input higher than the non-inverting input, so Vopamp is trying to go low.
Since the anode is connected to the >10V of Vin and the cathode is connected to the 9.3V of Vopamp, then the diode conducts.
Once the diode conducts, current flows through the diode from the inverting input to the Vopamp at 9.3V, which increases the voltage drop across the 2k resistor, so that the inverting input lowers back to 10V and the Vopamp can't then fall any further and settles around 9.3V.

As far as I know; the anode is p-type and cathode is n-type semiconductor; and the diode conducts when the anode is positive relative to the cathode.
In your diagram I think the anode is connected to the inv input and the cathode is connected to Vopamp.
So when inv input (anode) is about +10V or more and the Vopamp (cathode) is +9.3V or less, then the anode is more positive than the cathode, so the diode conducts.

I think you are making a mistake in saying that a diode is reverse biased when its n-type is connected to a positive voltage. The critical point is the difference in voltage between the ends of the diode.
Eg. A diode conducts if the anode (p) is connected to +5V and the cathode (n) to +3V, and it also conducts if the anode is at -20V and the cathode at -21V.

My brain is getting fried by all this detail, so E&OE!

5. Sep 22, 2015

### meBigGuy

I'm not going to try to read and understand what's been posted, just explain it in my own way:

1. The + input is fixed at V+ by the divider (I'm too lazy to compute V).
2. There are two active conditions that exist
A. Vin is less than V+ --- OP amp will saturate positive trying to raise the voltage on the - input to V+, but ..... Diode will block conduction so the output remains at Vin.
B. Vin is greater than V+ --- opamp will go as low as it needs to drop Vout to be equal to V+ through the diode (that will be about Vout - 0.7V which turns on the diode)

The opamp output never needs to go lower than about (V+) - 0.7

Nothing needs to be "negative" as such. If the opamp output is 0.7 below Vout, then the diode will be conducting.

I'm not considering exact diode voltages or leakage currents.

6. Sep 23, 2015

### brainbaby

Nice explanation..thanks for clearing out my obscurity......

7. Sep 24, 2015

### brainbaby

Hey Merlin ...sorry to interupt you again..
I got a little remaining query...

Actually what would be the transient opamp voltage when Vin = 9V just before going in saturation?

should it be +9.7V....??

8. Sep 24, 2015

### Jony130

No, wrong. OpAmp is already in positive saturation region (Vout_op_amp = +15V or so.)
If Vin = V- = 9V and V+ = 10V -----> Vout_op_amp = (V+ - V-)*Aol = (10V - 9V) = 1V * OpAmp open loop gain = positive saturation.

And the opamp will start to come out from saturation for Vin > (" V+" - Vcc/Aol) = 10V - 15V/100 = 9.85V

Last edited: Sep 24, 2015
9. Sep 24, 2015

### brainbaby

Hi jony130 nice to see you again....

As we know that opamp output voltage is responsible for making the diode f.b or R.b ..ok...
when Vin <10 V that mean that its some kind of low voltage associated with inverting terminal..which will force the opamp to output a much more positive voltage ...which R.b the diode and so on.......What i want to ask is that what would be that Vopamp that will make the diode R.b and then the opamp comes to know that the feedback path is open..and hence a perpetual cycle continues which then drive opamp into saturation....because there is some latency time associated with it..which you discussed in my previous thread..." Opamp as an active rectifier and its slew rate limitation" post #9..

hope so that i am relevant.....

10. Sep 25, 2015

### Jony130

This voltage will depend on the your definition of a diode cut-off voltage. If we assume 0.7V then diode is off when Vopamp_output > 9.3V.
But I think that this voltage is irrelevant here. More important is relation between "V+" and "V-". If "V+" is larger than "V-" the op-amp will drive his output into positive direction (towards VCC) and this for sure will R.B the diode. So for any input voltage smaller than 10V op-amp is in positive saturation region.
But as input voltage rising and approach 10V ("V+" - "V-") becoming more smaller and smaller and when it reach point where ("V+" < "V-") the op-amp output will drives his output into negative direction, But as Vopamp_output < 9.3V diode is ON and negative feedback ensures that "V+" - "V-" = 9.3V/Aol and for Aol = ∞ we have "V+" = "V-"

11. Sep 26, 2015

### brainbaby

thanks Jony....