uuffff how doess this stupid ting work lol

[cyprio7]

hi guys, im very new to electronics so pls bare with me. Iv been studying diodes and opamps... rectification circuits in particular but im having great trouble understanding how this particular configuration works.

It is a "Superdiode clamping configuration", which supposedly "eliminates" the approx 0.6V volt drop. How it does this i dont understand and iv been staring at the circuit an i cant work it out. Please if anyone knows could you explain why there is not a loss of 0.6V and how the circuit works overall. Iv attached the circuit. Its meant to provide full wave rectification when the capacitor is replaced with a large resistor Rload >> R2+R1. but i think it only provides half wave rectification.

Please explain to me in dummy language so i can understand because im new to electronics.

Thanks guys

 

[williB]

If you are just learning electronics, might i recommend D/L ing, SwitcherCAD III , from linear technology.. its free ..
it takes a little getting used to but is well worth it..
as for your circuit it may well eliminate the 0.6 V diode drop, But as you can imagine , it will not do any usefull rectification ..ie power ..

 

[Styx]

it works because the OPAMP has its closed-loop closed around the DIODE, thus the 0.6V associated with the diode will cause the voltage at the closed part (at the output) to be 0.6V lower then the OPAMP will want and thus it will increase its output to overcome this

have a read-up on OPAMPS and such


A similar kind of cct is used for BJT push-pull ccts.
a NPN-PNP puch-pull is a very good current amplifier but it has a problem at the zero-crossing ie for a signal that is less then +/-0.6v then there is no output

If you use an OPAMP to drive a BJT-pushpull and close the feedback at the output of the BJT then this zero-crossing problem is all but eleimiated


and yer it would be useless for rectification. If you want a lower voltadrop use Schottky diodes

 

[RadioRon]

OK, let's try to walk through how this works. First of all, let's get rid of the capacitor at the output as that complicates things. Next, always remember that in op amp circuit (and most others for that matter) the best way to analyze what the op amp will do is to remember that it will swing its output to whatever it takes to force the negative input to be the same voltage as the positive input. If the output hits its maximum or minimum limit (its rails) while attempting to do this then this rule no longer applies. Ok, so much for basic op amp theory.

Now, imagine that the input is exactly at ground potential, that is the input voltage is exactly 0. This will be our starting point for analysis. Since the input is zero volts, and the op amp will swing its output to force its negative terminal to the same voltage as its positive terminal, we can assume for now that the negative input to the op amp is also at zero volts. By ohms law this means that there is no current flowing through R1. Since the negative terminal of the op amp has an infinite input resistance (at least for the sake of discussion), we can deduce that the current flow through R2 is also zero. So no current is flowing out of the op amp and through the diode, and the output of the circuit is 0 volts. We aren't sure of the output voltage of the op amp at this point other than we know it is not higher than approximately 0.5 or 0.6 volts and we are certain that no current is flowing through the diode, and we know for certain that the output of the circuit is 0 volts.

Next, let's imagine that the input starts to swing negative. At this point the op amp must do something to cause current to flow through R1 in order to insure that the negative input of the op amp remains at zero volts. The only way it can cause current flow from right to left is to force current flow through R2 from right to left. The only way that can happen is if the output of the circuit is starting to swing positive and away from zero. So, the output is going to track the input depending on the ratio of R2 to R1. Let's assume for now that R2=R1 to simplify things. In this case, the output of the circuit would exactly track the input except it would be inverted, that is swinging positive when the input is swinging negative. At this point we can also deduce that in order for the output to swing away from zero, the output of the op amp itself must be swinging above 0.6 volts. Aha! This is the essence of how this circuit gets rid of the diode voltage drop, so make sure you understand this point.

To continue, let the input voltage swing all the way negative and then come back to zero. All the while, the output of the circuit has tracked this volt for volt except the output is positive while the input is negative.

Now, what happens when the input swings positive? Ah, now things work very differently. The moment this swing begins, current flow through R1 starts to go from left to right. So, current flow through R2 must go left to right. The trouble is that this is impossible because the current can't pass through that diode from anode to cathode (i.e right to left). So the op amp output tries to go below 0.6 volts, but cannot get current to flow in the correct direction through R1 and R2 to get the negative terminal to zero volts. But it tries as hard as it can by swinging to its negative rail. It can't go any further so it just sits at that rail. Meanwhile, no current flows through the diode from right to left. You see what has happened now? The op amp is actually disconnected from the output of the circuit because of the diode. Now, if you were to place a resistor from the output of the circuit to ground, you would see that the input voltage would be divided by the values of R1, R2 and the output load resistance. In other words, the input voltage would cause current to flow through R1 from left to right, through R2 from left to right and through the load resistor. Now, if the load resistor is a very high resistance compared to R1 and R2, almost all of the input voltage would be impressed across the load resistor.

Do you see what has happened? A positive swing of voltage at the input shows up directly across the load resistor. And yet, we deduced that a negative voltage swing at the input also shows up as a positive voltage swing at the output. Isnt' that the definition of a full wave rectification? So, the answer is that this circuit does indeed provide full wave rectification if the output load resistance is much higher than R1+R2. And this circuit does indeed eliminate the diode drop.

That's it for the explanation except to mention that once you fully understand it, try putting the capacitor back accross the output and then tell us what that does.

 

[cyprio7]

thanks for that. but im still a little stuck. you said...

""output is swinging positive when the input is swinging negative. At this point we can also deduce that in order for the output to swing away from zero, the output of the op amp itself must be swinging above 0.6 volts. Aha! This is the essence of how this circuit gets rid of the diode voltage drop, so make sure you understand this point.""



The output of the OPAMP itself must be above 0.6volts to obviously allow current to flow through it and subsequently through R2. However what i dont get is... the opamp voltage is above 0.6V... then it meets the diode... then in order to get past the diode... it must "give up" 0.6V.

so lets say output of opamp is 5V. The diode will conduct because output of opamp is above 0.6V. However when it goes past the diode, there is a 0.6V drop.

5V - 0.6V = 4.4V. Then going from right to left, there is 4.4V at the right hand side of R2, and then a 4.4V drop across it to make it go to zero volts at virtual earth

this to me is a volt drop. but the fact is there is no supposed volt drop in this configuration. i cant get my head round it.

please could you explain a bit more just on the volt drop bit reguarding the diode.


thank you for your last replies it helped me alot on other aspects.

 

[cyprio7]

so if my previous schematic provides full wave rectification, does this one provide half wave taking into account the new diode iv added?

also with reguards to the full wave rectification circuit....... is it true that when the input swings positive, because the diode effectivly makes it an open circuit, that means on that positive wave cycle, there is no gain, as it doesnt pass through the opamp because it just goes straight through R1 R2 then a huge Rload as a potential divider. or is there loads of gain because its effectivly an open loop configuration, therefore resistors R1 R2 and Rload must be chosen carefully so that the voltage isnt clipping at Rload?

or am i wrong?

thanks

 

[RadioRon]

"The output of the OPAMP itself must be above 0.6volts to obviously allow current to flow through it and subsequently through R2. However what i dont get is... the opamp voltage is above 0.6V... then it meets the diode... then in order to get past the diode... it must "give up" 0.6V.
Well, yes, if current flows through the diode there will be a voltage drop due to that diode so that the op amp voltage will have to be 0.6V higher than the output voltage when current is flowing in the forward direction through the diode.

so lets say output of opamp is 5V. The diode will conduct because output of opamp is above 0.6V. However when it goes past the diode, there is a 0.6V drop.
If the opamp output were at 5V (and still not hitting a rail) the diode will conduct and current will flow from right to left through R2 and R1. There will be a 0.6V drop, so the output of the circuit (to the right of the diode) will be at 4.4 V. In all of these I am assuming that you realize that we are simplifying the behavior of the diode for the sake of understanding the rest of the circuit. In reality, the diode voltage drop will vary with current especially at very low currents, and also with temperature, so it won't be exactly 0.6V in all cases.

5V - 0.6V = 4.4V. Then going from right to left, there is 4.4V at the right hand side of R2, and then a 4.4V drop across it to make it go to zero volts at virtual earth Yes

this to me is a volt drop. but the fact is there is no supposed volt drop in this configuration. i cant get my head round it. Yes, I agree that this seems contradictory. There is always a voltage drop through a real diode, so we never actually eliminate that and if you were told otherwise then there was a misunderstanding. The diode drop never goes away, but the behavior of this entire circuit is as if it was a diode with zero voltage drop. Perhaps when people describe this circuit they use a description that seems to claim to make the diode different than it is. From the point of view of the outside world looking only at this circuit as a two-port black box it does indeed seem like a diode with no voltage drop.

please could you explain a bit more just on the volt drop bit reguarding the diode. Hope that helps a bit


thank you for your last replies it helped me alot on other aspects.You are welcome. It helps that you are very good at explaining your problem in the first place which helps a lot. Your original request deserved attention because you put it clearly and intelligently. Quite refreshing.

 

[RadioRon]

so if my previous schematic provides full wave rectification, does this one provide half wave taking into account the new diode iv added? Yes, I think this one would be deliver a half wave rectified signal to a high impedance load. However, I note that the output impedance of the circuit is very different when the output is swinging positive compared to when it is sitting at zero volts over the next half cycle. So you should consider how that might affect the charge on the capacitor.

also with reguards to the full wave rectification circuit....... is it true that when the input swings positive, because the diode effectivly makes it an open circuit, that means on that positive wave cycle, there is no gain, as it doesnt pass through the opamp because it just goes straight through R1 R2 then a huge Rload as a potential divider. or is there loads of gain because its effectivly an open loop configuration, therefore resistors R1 R2 and Rload must be chosen carefully so that the voltage isnt clipping at Rload? The former rather than the latter is correct. Because the diode effectively disconnects the op amp output from the circuit output there is no gain and it just goes straight through R1, R2 and the load resistor.

or am i wrong?