Diode Clipping circuit - How to Diode bias - Page 1
Forum section: electronics / beginners

author: Tomer
author_rank: Regular Contributor
timestamp: April 16, 2018, 02:38:16 pm
content: Hi all, and thank you for reading and helping.I am trying to create a Diode clipping circuit on my breadboard and i am having limited success with the implementation.For reference i used the following article.https://www.electronics-tutorials.ws/diode/diode-clipping-circuits.htmlUsing 2 1N4001-G diodesAC 1Khz and an amplitude of 2VI am able to create a Clipping circuit (Please see pictureshttps://photos.app.goo.gl/2RaHHBYSCUkejGbV2)I am failing with how to connect my breadboard to enable Diode biasing, How can i provide the diode with for example with 1VDC.I think that i understood the theory correctly, but don't know how to implement on the breadboard.Appreciate your time and help.Tom

author: mikerj
author_rank: Super Contributor
timestamp: April 16, 2018, 02:43:04 pm
content: To bias the diodes as shown on that website you need a DC voltage source that is isolated from your AC supply.

author: Tomer
author_rank: Regular Contributor
timestamp: April 16, 2018, 03:03:52 pm
content: Thank youI am using signal generator for the AC sourceand tried to use a DC power bench supply for the DC.I basically don't understand where and how to connect the DC.Tom

author: Paul Price
author_rank: Super Contributor
timestamp: April 16, 2018, 03:57:35 pm
content: You are already clipping the signal waveform without any DC bias. You can see symmetrical clipping at the top and bottoms on each cycle. If you want more clipping, you need only increase the amplitude from your signal generator and insert a series resistor from the generator to the clipper(for instance 100-ohms). If you then gradually increase the output of your signal generator you will then gradually see the waveform turning the sinewave output of your generator into a very clipped square wave.

author: Tomer
author_rank: Regular Contributor
timestamp: April 16, 2018, 05:29:47 pm
content: Hi Paul, Thank you for answering !I am trying to do something a bit different, The diode is clipping the signal when it reaches 0.7V, i want the diode to clip the signal when it reaches 2 volts.I am trying to change the diode from 0.7 to a different higher number and keeping my AC output constant.Tom

author: ferdieCX
author_rank: Regular Contributor
timestamp: April 16, 2018, 05:58:49 pm
content: If you have a dual power supply, you can use the circuit in the picture. The clipping voltage will be 0.7 V + Vz + 0.7 V and,because the zener diode is always polarized, the switching time will be determined by the diodes in the bridge.The AC generator is connected to the left side of R23 and the right side is the output.

author: Tomer
author_rank: Regular Contributor
timestamp: April 16, 2018, 08:18:25 pm
content: Hi FredieCX, thank you for the image and explanation.I am sorry but that got me to be more confusedI am attaching my scheme, is it possible to use that to tell me what to add in order to bias the diodes for 1.7V ?Thank a lotTom

author: Paul Price
author_rank: Super Contributor
timestamp: April 16, 2018, 08:35:30 pm
content: Easy, just use three diodes in series in place of each individual diode. The circuit will then clip the signal when it reaches +/- 2.0-2.1 Volts.Don't forget to have a series resistor(100-ohm to 1k) in series with the signal coming in.Each diode will start to suck up the signal when it begins to conduct, and each diode in series will add to the voltage clip point.Since each diode will start to clip/conduct at about .7V, their series connected voltage adds up, and so for two sets of three diodes in series you get hard clipping at 3 x .7V =+/- 2.1 V.Get this: you don't need to use any D.C. bias supply at all,the diodes do the clipping as you want it all by themselves.

author: Tomer
author_rank: Regular Contributor
timestamp: April 16, 2018, 09:20:55 pm
content: Hi Paul, thank you very much!Can you explain why do i need a resistor in series between the AC signal and Diodes ?Also, can you explain to me how to bias with a DC supply, my purpose is learning and the 2V is just an example, I want to be able to control the bias accurately.Thank youTom

author: Paul Price
author_rank: Super Contributor
timestamp: April 16, 2018, 09:38:39 pm
content: Current flow is controlled by resistance. The reason the diodes clip the waveform you present to them is because the "resistance" of the diodes changes with voltage, from near infinity to only a few ohms when the diodes begin their work and conduct.This clipping circuit is a voltage divider made of the series resistor from the signal generator to the diodes and the effective, voltage-dependent resistance of the diodes themselves. For the diodes in question, this sudden drop in resistance begins as the diodes begin to conduct and this is a characteristic of the silicon junction itself, this effect begins around .65V and the resistance drops immensely with each few millivolts of voltage above .65V.Ok, if you have two 100-ohm resistors in series and you apply two volts to them, there would be a voltage divider and the voltage at the junction of the two connected together would be 1/2 the applied voltage or 1-volt.Now if instead of the bottom resistor in this voltage divider you have a diode, the voltage across this same voltage divider will equal the input voltage until the applied voltage reaches the diode's conduction threshold voltage of approx. .65V. At this point the top resistor is still 100 ohms, but the bottom resistor(the diode) starts changing from open circuit to a conducting resistor in series with a dc bias voltage, which is.the conducting "on" voltage of the diode.Then as the voltage applied increases (as in the case of a sinewave signal changing in voltage with time) The voltage drop across the top 100-ohm resistor increases while the voltage drop across the bottom "resistor/the diode clipper) increases only slightly and the effect is a regulation of the voltage across the diode, and bingo you see clipping because the voltage waveform flattens out at the diode's conducting voltage because increases in current only result in small increases of voltage across it.

author: ferdieCX
author_rank: Regular Contributor
timestamp: April 18, 2018, 07:29:54 am
content: In the circuit that I have posted, the +V and – V supplies polarize the D9 trough R26 and R27 respectively. It has then always a voltage equal to Vz.In the positive semicycle, when the AC voltage reachs 0,7 V + Vz + 0,7 V, current flows from the signal generator to ground through D5, D9 and D8.In the negative semicycle, current flows from ground to the generator trough D7, D9 and D6.Compared with the clipper with 2 Zener diodes, by using the same Zener diode for both semicycles, you avoid the problem of the dispersion in Vz between different diodes of the same type and so you get a more symmetric clipping.Also, the Zener diodes don't switch suddenly from blocking to conducting, because they have a high internal capacitance. By keeping D9 always conducting, the switch is made by the rectifier diodes, who have a faster blocking to conducting characteristic.

author: Tomer
author_rank: Regular Contributor
timestamp: April 18, 2018, 10:18:35 pm
content: Hi PaulThank you very much for the explanation, sorry it took me time to response, i read it again and again and i think that i understand now.ferdieCXThank you for that additional info about the scheme !