Clamping diode not quite working ??? - Page 1
Forum section: electronics / beginners

author: Mike Jung
author_rank: Contributor
timestamp: March 23, 2021, 09:05:46 pm
content: Hi guys ! I am new to this but want to learn.I have an electromagnet powered with 24V and makes a spark each time I press the on/off switch. I read I must place a 1N4007 diode across it. I did that and the spark has become quite a bit smaller. But it did not go away completely.What must I do to make it go away completely ?

author: bob91343
author_rank: Super Contributor
timestamp: March 23, 2021, 09:58:46 pm
content: Do you have the diode polarity correct? As for complete elimination of a spark, you must bring the current down gradually. A capacitor may help.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 23, 2021, 10:05:09 pm
content: How much wiring length is between the switch, coil and supply?And, what is the solenoid rating?24V is in the regime where it may spark even if the current (or voltage peak) is relatively small. This may be normal, acceptable behavior for a switch, but it depends on the switch type, and any spark is still evidence of electromagnetic interference, which may or may not be irritating.Tim

author: Mike Jung
author_rank: Contributor
timestamp: March 24, 2021, 06:58:39 am
content: The solenoid ( clutch actuator for a machine tool ) is 24V , 1A . The total length of wiring is 1m i.e. 2 wires 50cm each. I power the solenoid from an ordinary adj. voltage regulator - I think it's a 317(?) . The regulator failed a few times and was told is the "back emf" doing the deed.

author: Psi
author_rank: Super Contributor
timestamp: March 24, 2021, 07:06:01 am
content: With the diode added, are you getting a spark when switching it off? or just when you switch it on?

author: srb1954
author_rank: Super Contributor
timestamp: March 24, 2021, 11:16:09 am
content: Quote from: Mike Jung on March 23, 2021, 09:05:46 pmAdding the diode across the electromagnet only partially protects the switch in that it clamps the inductive flyback voltage and limits the peak voltage across the switch to a little bit more than the 24V supply voltage. The switch can still arc across while the voltage across it is increasing from zero to the 24V clamp voltage.To more fully protect the switch against arcing you also need a resistor-capacitor-diode network across the switch contacts. The correct design of this network is not trivial and I refer you to chapter 7 of the the Henry Ott book "Noise reduction techniques in electronic systems" for details. I recommend this book as it gives a fairly comprehensive and practical treatment of the problems and solutions for contact protection along with many other protection and interference problems.

author: srb1954
author_rank: Super Contributor
timestamp: March 24, 2021, 12:01:31 pm
content: Quote from: Mike Jung on March 24, 2021, 06:58:39 amLM317s are vulnerable to damage from the output voltage rising above the input voltage, which might happen from the transient voltage from an inductive load being switched.The standard fix is place a diode across the input and output of the regulator with the diode cathode to the input side. Refer to the application notes for the LM317 for full details.

author: Mike Jung
author_rank: Contributor
timestamp: March 24, 2021, 02:49:07 pm
content: Quote from: srb1954 on March 24, 2021, 12:01:31 pmI did that and they still failed.

author: Mike Jung
author_rank: Contributor
timestamp: March 24, 2021, 02:51:03 pm
content: Quote from: Psi on March 24, 2021, 07:06:01 amWhen switching OFF. The spark is minute with the diode installed but I thought it shouldn't be any spark at all because the diode "opens" at 0.6V

author: CaptDon
author_rank: Super Contributor
timestamp: March 24, 2021, 02:53:37 pm
content: If the diode was backward it would appear as a short across the power supply. It must be in the correct way.

author: Mike Jung
author_rank: Contributor
timestamp: March 24, 2021, 02:55:21 pm
content: Quote from: srb1954 on March 24, 2021, 11:16:09 am1. I have to confess I do not understand what you explained. I thought that the inductor "wants" to hold on the current and increases voltage. The diode should then open quench this voltage. Is there something else going on ?2. Thank you. I will try to find that book and learn more. Thank you again.

author: Terry Bites
author_rank: Super Contributor
timestamp: March 24, 2021, 05:09:16 pm
content: Assuming the diode is the right way round- the next step is to put an RC snubber across your switch. Seehttps://sound-au.com/articles/relays2.htmsection 4.1Make sure to include the reverse protection diode on the LM317.Back EMF may only be part of your problem. Is your switch rated to carry 1A at DC? It might happily break a 1A AC circuit only to fail miserably at DC.

author: Mike Jung
author_rank: Contributor
timestamp: March 24, 2021, 06:19:48 pm
content: Quote from: Terry Bites on March 24, 2021, 05:09:16 pmThank you. I will do that. The switch is rated 5A but it does not say AC or DC. Thank you very much for the link - very nice article !

author: srb1954
author_rank: Super Contributor
timestamp: March 24, 2021, 09:40:01 pm
content: Quote from: Mike Jung on March 24, 2021, 02:55:21 pmYes, the inductor opposes changes in the current flowing through it and releases the energy stored in its magnetic field to oppose that change of current. All real inductors have a finite resistance so the the voltage across the inductor is composed of two components: the voltage across the inductive component and the voltage drop across the resistive component of the inductor.Consider the situation where you have the inductor connected at the top end to your 24V supply and the bottom end is connected through the switch to ground. In the steady state a current will be flowing through the inductor determined by its resistance and the supply voltage, I = V/R. The current flowing through the inductor will have stored a certain amount of energy (E=1/2 LI2) in its magnetic field but the voltage across the inductive component is zero because the current in not changing.Upon opening the switch the current the inductor current will start to reduce but the inductor will convert some of that stored magnetic energy to produce a voltage across itself, V=L di/dt, where di/dt is the rate of change of current. Since the current change is negative the voltage across the inductor is negative and opposes the 24V drop across the resistive component and the voltage at the bottom end of the inductor will start to rise. The more rapid the decrease in current the greater the negative voltage across the inductor and if the current decreases rapidly enough the bottom end of the inductor can actually produce a spike "fly-back" voltage which exceeds the 24V supply by a considerable margin, maybe hundreds of volts.The rapidly increasing voltage across the opening switch creates an arc which is sustained by the stored energy in the inductor until such time as the switch contacts are too far apart for the the inductive spike voltage or all the stored energy in the inductor is dissipated in the switch arc and the resistive losses in the inductor. Thus the flyback voltage can produce a considerably larger arc and cause more damage to the switch contacts than what might be expected from the 24V supply alone.The addition of the diode across the inductor helps dissipate more of the stored energy in the resistance of the inductor rather than in the switch arc. As the voltage across the inductor flies back the diode will start conducting when the voltage at the bottom of the inductor reaches the 24V supply voltage plus 0.7V for the diode. When the diode starts conducting the current flows in a circular loop through it and the inductor until all the remaining stored energy is dissipated in the resistance of the inductor.The effect of the diode is to clamp the maximum voltage across the opening switch to 24.7V rather than the hundreds of volts that might be seen in the case of an unclamped inductor. There is still a switch arc produced during the period while the voltage across the switch is increasing from zero to 24.7V but this arc is more quickly extinguished than in the unprotected case. To fully extinguish this remaining smaller arc requires a more complex resistor-capacitor-diode network across the contacts as detailed in the Ott book.

author: Mike Jung
author_rank: Contributor
timestamp: March 25, 2021, 01:29:43 am
content: Quote from: srb1954 on March 24, 2021, 09:40:01 pm

author: srb1954
author_rank: Super Contributor
timestamp: March 25, 2021, 03:11:29 am
content: Quote from: Mike Jung on March 25, 2021, 01:29:43 am

author: Mike Jung
author_rank: Contributor
timestamp: March 25, 2021, 07:39:09 am
content: Now, that's what I'd an explanation ! Got it ! Thank you VERY VERY much !

author: Psi
author_rank: Super Contributor
timestamp: March 25, 2021, 08:56:29 am
content: try a schottky diode, maybe your current diode is super slow and cannot clamp it quick enough

author: Mike Jung
author_rank: Contributor
timestamp: March 25, 2021, 01:19:06 pm
content: Quote from: Psi on March 25, 2021, 08:56:29 amI know nothing about Shottky diodes. Could you please suggest one given my voltage and current ? I will see is I can buy it from RS Components here in RSA.

author: Psi
author_rank: Super Contributor
timestamp: April 19, 2021, 09:19:48 am
content: Quote from: Mike Jung on March 25, 2021, 01:19:06 pmhttps://za.rs-online.com/web/p/schottky-diodes-rectifiers/6882202/

author: MarkF
author_rank: Super Contributor
timestamp: April 19, 2021, 09:58:36 am
content: You could also try the UF4000 Series instead of the 1N4000 Series.They are faster.The UF4004 has a lower forward voltage rating but faster than the UF4007 (50ns vs 75ns).Also, a SA24CA transient suppressor.The 'CA' suffix is bi-directional vs the 'A' version which is not.

author: Mike Jung
author_rank: Contributor
timestamp: April 19, 2021, 06:47:26 pm
content: Thank you. What voltage should the supressor diode be ? I power with 24v

author: srb1954
author_rank: Super Contributor
timestamp: April 20, 2021, 05:47:53 am
content: Quote from: MarkF on April 19, 2021, 09:58:36 amIn this situation where the diode is placed directly across the inductor the transient suppressor will less effective, and a lot more expensive, than the ordinary diode. The peak clamp voltage will be at least several volts higher than what can be achieved by an ordinary clamping diode.The faster UF4004 will be slightly better than a 1N4007 but the benefit of the faster diode will be negligible when protecting a mechanical switch. A faster diode would mainly be of benefit when you are using a MOSFET or transistor switch, which are more sensitive to damage from rapid over-voltage transients than mechanical switches.In this application you don't really need to use a 1000V diode like a 1N4007 or UF4007. There is a common misconception that, because inductors produce high fly-back voltages, you need a high voltage diode to clamp that flyback voltage. In fact, the diode is in forward conduction when clamping the inductor fly-back so the requirements are only that it can carry the full inductor current and the forward voltage, particularly the forward recovery voltage, be as low as possible. Lower voltage diodes tend to perform a little better than higher voltage diodes in both these respects.When power is applied to the inductor the clamping diode is reverse biased so the requirement in that voltage rating has to be greater, plus a little safety margin, than the supply voltage to the inductor. In this case with a 24V supply a 1N4001 diode with a 50V reverse voltage rating would be adequate.

author: ledtester
author_rank: Super Contributor
timestamp: April 20, 2021, 07:20:57 am
content: Quote from: srb1954 on March 24, 2021, 11:16:09 amWhile looking into this I found that Ott published "Electromagnetic Compatibility Engineering" (2009) which may be considered an updated and expanded version of this text.