This is by far the most understandable explanation on this subject I have found so far. One thing to mention (you do so in a follow-up video, but I don't think it is here): this works on a dual supply op-amp, at least the negative rail needs to be able to get down to the Vf of the diode (I got all weird results with my LM358 connected to V+ and GND, but as a result I learned a lot). Thanks for this great video!
This is very useful. That "super-diode" circuit with a single diode is often shown as being correct, but the big mistake here is that the reverse-biased diode will open the feedback loop, and the op-amp might even misbehave if one of the inputs is out of the common-mode or differential mode range. Also, a too high impedance in the negative feedback loop is always a problem, and should be avoided at all costs, because the op-amp might oscillate (even if it apparently doesn't have any kind of positive feedback).
Wow what a great explanation. When you explained the use of the summation amp to make it a full wave rectifier, it immediately clicked in my head “oh duh” 😂
Note: one doesn't absolutely need precision resistors - 1 trimpot used in the feedback loop of the rectifier opamp *or* one trimpot to adjust the input weights into the inverting adder is enough to balance out any error in the other resistor pair.
you should show a op amp rectifier circuits witch uses only one Op and only like 5 resistors, to make it also very useful it only works on single rail supply which is very handy. This is interesting for Applications with a single rail supply, like in Sensors etc. one disadvantage is that the output is only 1/2 of an Ideal full way rectifier, if u are interested i can tell you i don't think it is a common known circuit, nice vid btw!
Great explanation with practical demonstration. Which book you are referring in the video. Can you provide the book details. Also some suggested reading materials for experiment engineers.
The drawing is different comparing to the schematic in the book. There is only one input connection at the beginning of the circuit. Where to connect the other wire, to ground? Same on the output.
Thanks for the video! At some point you said the fully rectified wave has double the original frequency. That's very interesting! I was looking for some way of obtaining half the frequency, maybe rectifying two halves up followed by two halves down. I think I would need some sort of memory trigger. Would you have some hint on how to achieve that?
The easiest way to divide the frequency of a signal by 2 is with a flip-flop: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-OL52f-k3iIE.htmlm28s . The output will be a square wave, so if you need a sine wave output, this would have to be low-pass filtered to remove the higher frequency content of the square wave.
+davidbear1961 "Applications of Operational Amplifiers, 3rd Generation Techniques", by Jerald G. Graeme (Burr-Brown). It's an oldie, but a goodie; out of print unfortunately, but available as a PDF on archive.org: archive.org/details/ApplicationsOfOperationalAmplifiers-3rdGenerationTechniques
Correct, mis-matches in the resistor ratios will cause error in the rectified output, as will imperfections in the op amp itself, such as input voltage offset errors.
Really great video, thanks for sharing:-) I hope that you'll please answer my noob question. To me it sounds like you have 2V peak to peak as AC input and 1V DC peak. Does that not mean that you loose 1 volt while rectifying?
If i have very low signal below the forward biased voltage of diode , can this configuartion rectify it with unity gain ? Or should i increase the gain then rectify it ?
so r1 and r2 becomes parallel with load resistor and forms a volt divider.r1 and r2 are in series since no current goes to i/p terminal of opamp . right?? NICE VIDEO
+Shawn Micheals Well it's a split supply design, which is usually undesirable for simplicity / low voltage operation. The circuit's primary performance disadvantage is that the slew rate of the op amp limits its frequency response. An alternative circuit that uses an op amp and a comparator helps to overcome both of these drawbacks: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-1TLZBWx6BCc.html
Nice. very good. what is the name of the book. and if you know a full wave rectifier with out diode and by op amps and can explained in tueb. thenk you a very mach.
what if you use twoo opamps in parallel but one will use the non inverting input the sum of both in the output would be a fully rectified sinus to (just 2 cheap diodes extra to use)
You could do that, but you'd lose a diode of voltage on the output; if you're dealing with smaller signals, the diodes won't even turn on. Even if you're dealing with larger signals, the actual voltage drop across a diode is not precisely controlled, so you'd lose precision.
Hi! Great video! Just discovered your channel it's great. I tried this circuit with a friend. The inverting half wave rectifier (in blue) seems to saturate 1.5V before reaching V-, even with a rail to rail OPamp. Is there a way to obtain such a swing? Maybe with your other circuit using a comparator and a relay? Thanks for reading.
+David Bernfeld Are you trying to run this circuit from a single supply? It's meant to be used with a dual supply; the rectifier circuit from my other video is a better choice for single supply operation. With that said, I'm not sure how a rail-to-rail op amp could be saturated if its output is 1.5V above V-? Could you provide some more details on your circuit and what you're seeing on the scope?
Thanks for the reply! It was in deed a single supply configuration. However, we did create a Vmid=Vcc/2 with 2 resistors and a buffer. The circuit should work exactly the same or are we missing something? We can't get closser to V- than 2 diode voltage drops. We tried your other circuit, works LIKE A CHARM! Thanks!
do i miss something? do you mean the half wave rectifier only suit for low resistance load? as the signal will have direct voltage division to the ouput
If you want half wave rectification with only one op amp and one diode as initially shown in the video, then yes, your load has to be much smaller than the series combination of R1 and R2. However, as shown at 7:56, the addition of a second diode fixes that particular issue, and you will get proper half wave rectification with both high and low resistance loads.
With the second diode in place the op amp can hold its inverting input at ground, since that diode is forward biased once the op amp's output drives ~0.6v negative. This makes the junction of R1 and R2 effectively grounded.
One common application is in the feedback loop of an automatic level control or automatic gain control circuit, say for an audio amplifier. By rectifying the audio signal, it is easy to detect the peak voltage of the signal and adjust the gain of the amplifier either up or down in order to maintain a constant amplitude output from the amplifier.
The resistor values at the frequencies typically used for this type of application aren't critical and are usually selected based on your desired input impedance for the circuit. Opamp selection is usually based on your requirements for the circuit; higher frequency signals will obviously need a faster op amp. For the examples here, IIRC, I used MCP601/602 op amps and 10K ohm resistors.
+Yuri Novais Araujo The circuits in this video are designed to be used with dual supplies (e.g., +-10V). I have another video on single supply rectifier circuits here: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-1TLZBWx6BCc.html
