Reverse voltage protection diode for relays - Page 1
Forum section: electronics / beginners

author: npelov
author_rank: Frequent Contributor
timestamp: September 30, 2013, 04:31:24 pm
content: Which diode is better for reverse voltage protection for relays: 1N4148 (faster) or 1N4007 (can handle more current). Also should I bother putting schottky diode instead? Most absolute maximum values in datasheet specify -0.3 voltage for pins, so maybe 0.6-0.7V of rectifier diode would be too much. And if I have to put schottky, what current should it be able to handle?

author: madires
author_rank: Super Contributor
timestamp: September 30, 2013, 04:41:12 pm
content: For standard relays the 1N4148 should be fine.

author: Bryan
author_rank: Frequent Contributor
timestamp: September 30, 2013, 06:30:45 pm
content: And the reason why a diode should be used. Rule of thumb is the diode current capacity should match the motor, so would assume the same applies for the relay.http://youtu.be/LXGtE3X2k7Y

author: JackOfVA
author_rank: Unknown
timestamp: September 30, 2013, 08:08:37 pm
content: A standard silicon power diode has a more than adequate turn on time for relay protection. The switching speed problem associated with power diodes such as the 1N400X series relates to turn off time where it may be much longer than in a high speed switching diode such as the 1N4148.I've written a detail web page on the topic athttp://www.cliftonlaboratories.com/diode_turn-on_time.htm

author: npelov
author_rank: Frequent Contributor
timestamp: October 01, 2013, 10:38:00 am
content: @JackOfVA That's a long study... It would help if you put title to the pictures. I read that you measured 136 mA current on the diode. 1N4148 is capable of handling 450mA if it's pulsed or 200mA contnuous. So it should be able to handle 30 - 70mA relay. Also I can't understand why diode would affect the release time of the relay. I thought that any RC snubbler would delay the release more than a diode.So can I use a rule of thumb that the reverse current is about double the switch current of the relay. So I put a diode that can handle minumum 3-4 times the switch current of the relay. I guess the DC resistance of the coil limits the reverse spike current too.

author: Chris Wilson
author_rank: Super Contributor
timestamp: October 01, 2013, 11:52:39 am
content: I can maybe help here, although I am a beginner. I came across diodes and their effect on relay release time for fast switching of a transceiver. here's Tyco Electronics sheet on the various suppression options, attached, hope it helps.

author: JackOfVA
author_rank: Unknown
timestamp: October 01, 2013, 12:13:55 pm
content: Quote from: npelov on October 01, 2013, 10:38:00 amYes, the maximum current through the protection diode is limited by the resistance of the relay winding. A typical small 12V relay will have a winding resistance around 1000 ohms.If you require the relay coil to de-energize quicker, you should allow the discharge voltage to build up and then have the diode clamp -- such as a Zener diode.

author: npelov
author_rank: Frequent Contributor
timestamp: October 01, 2013, 02:55:02 pm
content: And why having diode shortens the relay live (according to the table Chris attached)?Also - correct me if I'm wrong - when I use zenner clamp the zenner diode will limit the reversed voltage on the coil which will sum with supply voltage at the transistor. I must make sure that supply voltage + zenner voltage is < break voltage of the transistor.

author: Zero999
author_rank: Super Contributor
timestamp: October 01, 2013, 10:21:31 pm
content: Quote from: Chris Wilson on October 01, 2013, 11:52:39 amYou can achieve a fast turn off time by putting a resistor in series with the free-wheel diode. If you make the resistor value roughly equal to the coil resistance, the voltage spike won't exceed the power supply voltage.

author: JackOfVA
author_rank: Unknown
timestamp: October 01, 2013, 11:47:31 pm
content: Quote from: npelov on October 01, 2013, 02:55:02 pmYes, you do not want the transistor collector voltage during the transient to exceed the device rating. It may not fail the first time but it will eventually degrade and fail.Some MOSFET devices are rated for avalanche breakdown and in that case you can use the MOSFET as a dual purpose switching device and peak voltage clamp. If you do this, be certain the particular MOSFET is avalanche rated, however.

author: geraldjhg
author_rank: Regular Contributor
timestamp: October 14, 2013, 08:16:02 pm
content: shottkey dont have the votage rating4007 are a bit slow but the work4148 cant take the poweruse 1n4936 1n4937 and you cant go wrongit is to protect the transistor from inductive kickwhen it turns off

author: scott216
author_rank: Regular Contributor
timestamp: October 20, 2013, 11:40:03 pm
content: Here's a really good video that talks about reverse voltage protection.

author: madires
author_rank: Super Contributor
timestamp: October 21, 2013, 01:04:10 pm
content: Quote from: scott216 on October 20, 2013, 11:40:03 pmThe video is about reverse voltage protection for the power supply but the topic of this thread is about how to deal with the reversed voltage of the self inductance of a relay when switched off.

author: ee.jmlp
author_rank: Contributor
timestamp: October 21, 2013, 01:47:05 pm
content: well... there are many ways to discharge a relay coil.If you use theantiparallel diode(the one you were asking for) you could use 1N400X diodes, or a diode that supports the transient current for a while (at least).Keep in mind the current flows in the coil until the energy stored as a magnetic field is dissipated or stored again.Normally you loose this energy in the coil resistance and the diode forward voltage/resistance.The antiparallel methodincreases the "turn off"time because you need to wait until the current falls below the holding value.There is asecond methoddischarges the coil without increasing the turn off time and you can control the discharge time by modifiying a value. That value is the breakdown voltage of azenerit's in antiparallel with the switch that feeds the coil of your relay.The restriction is to put a zener (or a TVS) with a Vz higher than supply voltage and lower than switch (mosfet of bjt) breakdown voltage. As you approach to the switch breakdown voltage thedischarge time will decrease.This method has higher EMI than others because the magnetic field is collapsed very fast (you can always snub it in the zener). Normally the EMI is very weak so you don't have to care about it.The third methodis to use asnubber(R and a C) in parallel with the coil. This is far more complex but gives you agood relation between EMI/discharge_time.Other methods could involve re-storage of the energy in the coil.So you can now decide how to properly cancel your kickback transient.

author: Zero999
author_rank: Super Contributor
timestamp: October 21, 2013, 03:58:06 pm
content: The good thing about the resistor and diode is not only does it give a fast off time, but it also won't be damaged by reverse polarity, as long as the resistor can safely dissipate the heat.

author: ee.jmlp
author_rank: Contributor
timestamp: October 21, 2013, 11:00:43 pm
content: Quote from: Hero999 on October 21, 2013, 03:58:06 pmThat method is the snubber one without the capacitor, it could be as fast as the zener method but you need to calc your resistor, btw you don't need to care about the supply voltage.I you move that with a mosfet you must take care about the value of the resistor, if it's too large a high voltage spike will appear. For example, if your coil works with 80mA and you put a 10k resistor you will get a 800V spike like a champ. SO...To calc the voltage peak do this:coil_current*resistor+Vf_diode=Vpeak

author: Zero999
author_rank: Super Contributor
timestamp: October 21, 2013, 11:16:46 pm
content: Quote from: ee.jmlp on October 21, 2013, 11:00:43 pmSelect a resistor with the same value as the coil and you can't go wrong: the turn off time should be fast enough for most applications with the spike being no more than the supply voltage.

author: ee.jmlp
author_rank: Contributor
timestamp: October 22, 2013, 12:16:30 am
content: Quote from: Hero999 on October 21, 2013, 11:16:46 pmLet's see... 500 Ohm relay at 12V with it's 24mA could be between 40H and 50H, let's get the better case 40HSo if you put a 500Ohm with the diode in parallel:40H/(500 Ohm+500 Ohm)=40ms!For me it is not fast enough. But always the application will tell...