Strange behavior - simulated op-amp rectifier - Page 1
Forum section: electronics / beginners

author: FriedMule
author_rank: Frequent Contributor
timestamp: January 01, 2020, 03:05:44 pm
content: HiI have, as I mostly do, searched and found some article and a schematic that should be great.According to the article, this rectifier should be a wary precise full wave version with nearly no loss.But when I simulate it, it do only rectify the positive AC and on top of that, do it deliver a -2.3V to 9.75V voltage.Info: input 10V 50Hz

author: unitedatoms
author_rank: Frequent Contributor
timestamp: January 01, 2020, 03:43:23 pm
content: https://en.wikipedia.org/wiki/Precision_rectifierI did not look into details, but it looks like the rectifier is working as half wave one. It is intended by design.

author: fcb
author_rank: Super Contributor
timestamp: January 01, 2020, 04:03:05 pm
content: Your circuit consists of two elements.1. Precision halfwave rectifier (R1/R2/D1/D2/IC1)2. Halfwave to fullwave converter (R3/R4/R5/IC2)Split the circuit and test part 1, first without a load - suspect you'll find that R3's virtual earth load on the precision rectifier is the problem, you might need to put a x1 buffer in between D1 and R3.

author: FriedMule
author_rank: Frequent Contributor
timestamp: January 01, 2020, 04:03:48 pm
content: Quote from: unitedatoms on January 01, 2020, 03:43:23 pmThanks for the link. No according to the schematic, the original one on the top most image, it says "full wave rectifier"

author: FriedMule
author_rank: Frequent Contributor
timestamp: January 01, 2020, 04:04:43 pm
content: Quote from: fcb on January 01, 2020, 04:03:05 pmThanks I'll try that now:-)

author: FriedMule
author_rank: Frequent Contributor
timestamp: January 01, 2020, 04:23:40 pm
content: Quote from: fcb on January 01, 2020, 04:03:05 pmThis is what I have done. I have to admit that I have no idea about that buffer you write about, may I please ask if you can explain some more? :-)

author: not1xor1
author_rank: Frequent Contributor
timestamp: January 01, 2020, 05:40:44 pm
content: Quote from: FriedMule on January 01, 2020, 03:05:44 pmThe circuit works works as expected in LTspice. (see attachment)What kind of program are you using?Did you set a proper supply voltage for the opamps (both positive and negative and greater than the AC input peak voltage)?

author: FriedMule
author_rank: Frequent Contributor
timestamp: January 01, 2020, 07:57:58 pm
content: Quote from: not1xor1 on January 01, 2020, 05:40:44 pmI am using Micro-cap to do the simulationI was just about to write that the circuit works fine:http://tinyurl.com/ustlnbxBut many thanks for doing so much out of it, and yes it looks to work fine in LTSpice.I have tried to change the two Op-Amps to the same model as you used, but still with no luck.I have not applied any supply voltage on the Op-amps, since there were not applied any on the original drawing. I tried to remove the supply Voltage on your LT-Spice document and that do remove all the function of the circuit.The function should be a rectifier but if it already demands DC power, then do I need a PSU for that and then a PSU for that circuit, and so on:-)Isn't it posibly to avoid supply Voltage?

author: rstofer
author_rank: Super Contributor
timestamp: January 01, 2020, 11:31:08 pm
content: No, the op amps need a power supply as shown in both of your original schematics.A precision rectifier is used for signal processing, not as some part of a low loss power supply.https://en.wikipedia.org/wiki/Precision_rectifierAnd pay attention to input and output voltage swings. Some op anps can't work within a couple of volts of the power supply rails. Given +-9V supplies, some opamps will only operate to +-7V on the inputs and outputs. Rail-To-Rail opamps work a lot closer but don't assume rail-to-rail is anything more than a marketing term.

author: fcb
author_rank: Super Contributor
timestamp: January 01, 2020, 11:51:27 pm
content: Quote from: FriedMule on January 01, 2020, 07:57:58 pmI guess at this point it might be prudent to ask what you are trying to achieve?

author: FriedMule
author_rank: Frequent Contributor
timestamp: January 02, 2020, 08:31:29 am
content: Quote from: rstofer on January 01, 2020, 11:31:08 pmI believe you, my thought on using this method is that I have seen many examples where other have used it as a power rectifier and read articles where people have praised this method. But on nearly all of the schematics have there not been shown any supply pins, so maybe it's why it do work in some simulators?Is there any way to avoid having to use supply? I have thought about using compactors instead and I have thought of an, properly stupid idea, letting some of the finished DC power, back to the Op-Amps supply?

author: FriedMule
author_rank: Frequent Contributor
timestamp: January 02, 2020, 08:36:45 am
content: Quote from: fcb on January 01, 2020, 11:51:27 pmMy end goal is to build a Optoisolator dimmer, it runs at 1.2V-1.4V at 0.1uA-50uAI am thinking of using a simple three resistor current divider, where one resistor is a pot.To save as much power as possible, do I want to eliminate as much voltage drop as possible.

author: capt bullshot
author_rank: Super Contributor
timestamp: January 02, 2020, 09:17:46 am
content: This circuit works as a precision signal rectifier, intended to e.g. measure AC voltages by converting them to DC voltages. It doesn't deliver any power, and requires an external supply to operate at all, indeed it uses more power from the external supply than it uses from the AC input. It's quite common to not show the supply pins in such schematics, they're just implied to be there and appropriately connected to an appropriate supply.So it's not applicable for your purpose, for minimum voltage drop at low power, use schottky diodes as the rectifier.

author: FriedMule
author_rank: Frequent Contributor
timestamp: January 02, 2020, 10:15:01 am
content: Quote from: capt bullshot on January 02, 2020, 09:17:46 amOkay thanks, just a small question, I have seen some use Mosfets with diodes or even schottky's, why do that?

author: capt bullshot
author_rank: Super Contributor
timestamp: January 02, 2020, 10:26:14 am
content: Quote from: FriedMule on January 02, 2020, 10:15:01 amThis circuit is an "active diode" rectifier (the MOSFETs acting as low voltage drop diodes). It's useful when you need to conserve power in somewhat higher power applications. This means a few amperes of output current, and for this particular circuit input with an voltage range of 5...10V. For lower voltages, the MOSFETs won't turn on, and for higher voltages you'd need a special driver circuit.

author: rstofer
author_rank: Super Contributor
timestamp: January 02, 2020, 04:54:11 pm
content: Quote from: FriedMule on January 02, 2020, 08:31:29 amIt is common practice on op amp circuits to omit drawing the power supply wiring but the device almost always shows the pins and labels them something like VC1, VE1 or +9, -9 as shown in your two schematics.How do you expect all those little transistors inside the op amp to work if they don't have power?See the schematic about half way down:https://en.wikipedia.org/wiki/Operational_amplifier

author: Zero999
author_rank: Super Contributor
timestamp: January 02, 2020, 08:35:53 pm
content: Quote from: FriedMule on January 02, 2020, 10:15:01 amThe circuit you're looking for is an ideal diode. The easiest way to do it is to use an IC.https://www.google.com/search?client=firefox-b-d&q=ideal+diode+controller+ICThe schematic you've posted with the MOSFETs, will not work with a capacitor on the output, so is only good for loads which don't require a smoothed power supply. It is also limited to peak AC voltages of no more than the gate-source voltage rating of the MOSFETs, which is typically around 20V.There are ICs available to make a proper active bridge rectifier which will work with a capacitor load or some ideal diode controller ICs can be configured as a bridge rectifier.https://www.nxp.com/docs/en/data-sheet/TEA2208T.pdfhttp://www.ti.com/lit/ds/symlink/lm74670-q1.pdf