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jcorb5 (Electrical) (OP)
9 Jun 03 16:13
I am looking into the failure (shorting across) of an MR760 diode that was installed in parallel with a solenoid valve to suppress back emf.  The solenoid coil is 35.1 watt, 465 ohms, 35.1 watt and 22 Henries.  What I am trying to do is find out how to size a diode for this application.  The MR760 diode has shorted across in 2 different instances when being used in this application.  What critical parameters should be used to  select the right diode.  Also should I be using something other than a diode to suppress the back EMF?  Maybe a transorb?

The contact opening time of the Struthers Dunn relay that controls the solenoid is on the order of 1-2 milliseconds.  The calculated EMF for the solenoid is  –6206 VDC for 1 ms contact open time, -3102 VDC for 2 ms time contact open time, -2069 VDC for 3 ms contact open time, and 1551 VDC for 4 ms contact open time.  

The MR760 diode is rated 1000 volts peak repetitive reverse voltage, working peak reverse voltage, DC blocking voltage.  It is rated 1200 volts non-repetitive peak reverse voltage (half wave, single phase, 60 HZ peak, and 700 volts RMS reverse voltage.
It is rated 22Amps for average rectified forward current, 400Amps non repetitive peak surge current for 1 cycle,

Any thoughts on properly sizing a diode for this application?  

37pw56gy (Electrical)
9 Jun 03 16:25
A resistor having a value between 1-5x the coil resistance might be a good alternative to the diode snub.  

Advantages: Quicker, more positive release of the solenoid; simple resistor less susceptible to sudden failure caused by surges; no tendancy to rectify stray AC energy.

Disadvantages: Resistor size, power dissipation much greater than a diode.  

MOV's are also a good, low-cost alternative.  Do not consider use of a capacitor snub in this type of application.
moturcu (Electrical)
9 Jun 03 17:17
37pw56gy:
Depending on the size of the coil resistance, inductive effect of the coil resistance will create another di/dt effect. Some applications may require another clamping diode accross resistance.
So bottom line, well sized diode would be better solution.
electricpete (Electrical)
9 Jun 03 18:03
I don't think I have ever seen a diode by itself... always a diode in series with a resistor connected accross a dc coil.

Without that resistor you at least one problem  - full current flowing through the diode for long period of time until it dissipates in the coil resistance.  Also the resulting slow dropout time may not be good for some applicaitons
electricpete (Electrical)
9 Jun 03 18:10
Wow, I didn't see that is a hefty diode.  22A continuous rating?  It must be a big one.
electricpete (Electrical)
9 Jun 03 18:13
Your normal current is on the order of SQRT(35w/465ohms) ~0.3A?  Are you sure there is a 22A diode in there?
Helpful Member!  cbarn24050 (Industrial)
9 Jun 03 18:39
Hi, the diode only needs to be rated at the supply voltage and coil current.
jcorb5 (Electrical) (OP)
9 Jun 03 19:28
Electripete,
thanks for the reply.
you are correct, the coil typically draws .28 amps (calculated). the diode is connected in parallel to the coil
The diode normally is reversed biased when the coil is energized (125 VDC) and then the maximum reverse current is between 25 microamps  (junction @25 degrees C) and 1 milliamp (jxn at 100 degrees C).
The maximum forward current is 22 amps.
When the contact opens and the coil collapses the voltage is reversed and the diode is forward biased, at this time it can handle 22 amps to "bleed off" the back EMF.
To repeat myself, the only time the diode is forward biased at 22 amps is when the upstream contact is opening and the coil field is collapsing and bleeding through the diode.  
does that make more sense?  do you have any advice on sizing the diode and resistor for this particular coil??
  
electricpete (Electrical)
9 Jun 03 21:41
You have exceeded my knowledge.  I have always seen the resistor in series but since you have huge margin in your diode continuous current rating it doesn't seem necessary.  
moturcu (Electrical)
10 Jun 03 0:06
jcorb5

First of all, you should determine what kills the diode. is it over current or over voltage. From the number you have given, both are possible. Over current and related over temperature kills diode quitely but over voltage kills it with an explosion. This might be the clue.

If over voltage kills, based on your back emf calculation you have to go with a higher voltage rated diode. If you you do not like it you should look for other snubber solutions adapted from power electronics suchs RC accross the switch etc.

IF the over current is killer, after switch opens current changes as
I=Io(1-exp-(R/L)t).

By integrating this equation over a  5L/R distance, you can get average current by which your diodes needs to be sized.

I guess surge current spec is irrelevant for your case because inductor current will always change slowly and it is only 0,28A rated.
cbarn24050 (Industrial)
10 Jun 03 15:25
hi jcorb5, why do you think 22 amps flows through the diode, and where do you think it comes from?
nbucska (Electrical)
10 Jun 03 15:56
When the switch opens the current tries to remain unchanged
i.e. the parallel diode will have one forward VTG ( ~.7V)
accross and the coil's R will dissipate the power.

<nbucska@pcperipherals.com>

jcorb5 (Electrical) (OP)
10 Jun 03 15:56
Cbarn24050.
Thanks for the reply. My posting above was poorly written.
the 22 amps is the manufacturers rated value for forward current. (so are the other values) I believe when the coil field of a few thousands volts is collapsing that the 22 amps rating may be exceeded.  When the coil field collapses the back emf, which is opposite polarity of applied voltage,  forward biases the diode causing alot of current to flow.
The below hyper link which was provided by BrianR contains manufacturer technical data sheets for the failed diode. http://www.glencoe.com/ps/ee/bsee/electricity/6e/student/data_sheet_pdfs/MR750.pdf
you may need to copy the address into your address box
  






 http://www.glencoe.com/ps/ee/bsee/electricity/6e/student/data_sheet_pdfs/MR750.pdf
cbarn24050 (Industrial)
10 Jun 03 20:43
hi jcorb5, so now youve switched from an imaginary 22amps to an imaginary few thousand volts! Where do you think that comes from? This is all very basic stuff, allmost any diode will work in this application.
Helpful Member!  RickOzone (Electrical)
10 Jun 03 21:08
It seems to me that the diode is certainly capable of passing the surge current since the solenoid current is only 0.28 Amps.  Therefore it seems more likely the problem is an over voltage condition.  Perhaps the recovery time of the diode is an issue. Maybe a faster diode would work better. Also, the contact opening may be chattering causing multiple transients.  

A high voltage zener in parallel might help such as the 1N5388 (200 V). An RC snubber across the switch contact would probably help too.  
Helpful Member!(3)  Warpspeed (Automotive)
10 Jun 03 21:14
There seems to be a lot of confusion here folks !

What happens is that with the solenoid energized there is 125 volts applied, and about 290mA flowing through the solenoid. The diode sees 125v across it in the reverse direction.

Now when power is removed, the 290mA flows through the diode, which has about an 0.6v drop. So in an ideal world you would only need a 125volt 290mA diode to do the job.

The stored energy in the solenoid is 0.5LI^0.5
0.5 x 22 Henries x 290mA squared or 0.925 Joules.

So 0.925 watt seconds is dissipated each time the solenoid releases into the dc resistance of the solenoid. This heating energy does not go into the diode.

So the ratings of the specified diode should be far in excess of what is actually required.

BUT there is one other factor not included in the above.

It is assumed that the 125 volt supply is only 125 volts. What if there are very narrow high voltage spikes in excess of 1Kv on the 125 volt rail ?

When the solenoid is energized these spikes could easily puncture the diode. Spikes like this can easily be generated by series inductance in the power supply, and the fast disconnection of a heavy load elsewhere in the system.

I might suggest you either fit a transorb instead of the diode, or a capacitor right across the 125v supply, or fix the overvoltage transients at the source. There is nothing wrong with the solenoid, or the diode. The problem is external.

 
jbartos (Electrical)
11 Jun 03 22:15
Suggestion: Reference:
John Markus "Electronic Circuits Manual", McGraw-Hill, 1971 includes several circuits where a diode is applied to suppress the voltage spikes of a switched inductance and to dissipate the inductive load stored inductive energy.
cbarn24050 (Industrial)
12 Jun 03 13:24
Hi jbartos, I think you have misread the reference you cited. Warpspeed's explaination is 100% correct.
OperaHouse (Electrical)
15 Jun 03 16:30
This is an example of taking a bad idea and beating it to death.  If this diode is failing it is because of something else in the circuit besides this solenoid coil.  I've found that these higher voltage DC control systems are usually found in old crane systems and the power is extreamly noisy.  Most likely cause is a voltage spike that breaks down the diode when the solenoid is engaged.  I would suggest that you place a 47 ohm resistor in series with the power to this solenoind with a .47 uf cap at the solenoid end to the common.  This will basically filter the DC supply to the coil removing any spikes.  A 1A 1000V diode is more than sufficient in this application.  The company I worked for sold tens of thousands of these potted in epoxy.  Adding a resistor in series with the diode is optional and is only used to increase speed.  A MOV might work but it might have to be sized excessivly large to take the spikes from whatever else is on the line producing the noise.
buzzp (Electrical)
16 Jun 03 17:06
The higher votlages at switching can be attributed to the voltage equation for inductance V=L di/dt. Since the current rate of change is at its maximum (limited by circuit parameters such as stray capacitance and inductance) when the switch is open, the voltage can be a lot higher than the applied voltage. However, it does not reach this value in an instant, it takes time due to parameters (circuit capacitance and inductance). Often the switching time of the diode is adequate to 'limit' (energy disipated in the diode and the coil before reaching maximum voltage of inductor) the voltage (before gets to peak value) before any damage is done. The standard diode series (1N400x) are generally more than adequate for obtaining the results you want. There is something being overlooked (other transients,other built-in flyback protection in the solenoid, etc) that has not been communicated in this forum. I would suggest some measurements with a scope or if possible, put the diode in the circuit without de-energizing the load to see if failure occurs. This test is highly unlikely unless the diodes are failing rather quickly.
Using strictly diodes in this configuration is very common for DC coils. You may want to visit some mfg sites to check through there app notes such as www.onsemi.com, or phillips, etc. The solenoid mfg is probably the first place to check to make sure you don't have something unique that requires special considerations. Good luck and please post your results.  
jbartos (Electrical)
16 Jun 03 19:55
Suggestion: Visit
http://www.hvca.com/advprodsearch2.jsp
for 600V 25A diode and add a series resistor, or
find a diode manufactured for the higher voltage and current, e.g.
http://us.st.com/stonline/books/ascii/docs/3179.htm

OperaHouse (Electrical)
17 Jun 03 10:37
The problem is not with the diode or the solenoid circuit.  The spike which causes failure is coming from somewhere else.  A small resistor might delay the eventual failure of the diode junction from breakdown.  Three diodes in series might also give you enough reverse breakdown to survive.  A capacitor from the 125VDC to common would also help but the capacitor would also be destroyed unless you use a high quality snubber capacitor.  These have specs that say something like 5A @ 20KHZ and are double wound (two capacitors in series) to avoid corona effects. Standard metalized film capacitors will suffer end weld failures and punctured dielectric.  The circuit I described in the previous post with the resistor and capacitor will protect a standard cap from destruction and filter out the large spikes from the solenoid circuit.  The relay will operate on a slightly lower voltage which is just fine.  The capacitor will supply enough additional current at initiation for reliable operation.   Most electro-mechanical devices have a pull in voltage to hold ratio of 5:1 or better.   Definitely there will be no change in performance with a 10% drop. Of course, I assume this is not being rapidly cycled.  Dropping the solenoid voltage after activation is a common practice in the gluing industry to prevent hot coils from changing the viscosity at the glue head.
jbartos (Electrical)
21 Jun 03 20:04
Suggestions to Warpspeed (Automotive) Jun 10, 2003 marked ///\\\
There seems to be a lot of confusion here folks !
///Please, be specific about a lot of confusion here.\\\
What happens is that with the solenoid energized there is 125 volts applied, and about 290mA flowing through the solenoid. The diode sees 125v across it in the reverse direction.
///True, however, very elementary.\\\
Now when power is removed, the 290mA flows through the diode, which has about an 0.6v drop. So in an ideal world you would only need a 125volt 290mA diode to do the job.
///Apparently, what is meant here is a 125V 290mA diode circuit, perhaps with a voltage drop resistor.\\\
The stored energy in the solenoid is 0.5LI^0.5
///The equation needs a clarification.\\\
0.5 x 22 Henries x 290mA squared or 0.925 Joules.
So 0.925 watt seconds is dissipated each time the solenoid releases into the dc resistance of the solenoid. This heating energy does not go into the diode.
///True, especially, if there is a resistor in series with the diode.\\\
So the ratings of the specified diode should be far in excess of what is actually required.
///Please, clarify.\\\
BUT there is one other factor not included in the above.

It is assumed that the 125 volt supply is only 125 volts. What if there are very narrow high voltage spikes in excess of 1Kv on the 125 volt rail ?

When the solenoid is energized these spikes could easily puncture the diode.
///Depending on the diode reverse voltage spike rating.\\\
 Spikes like this can easily be generated by series inductance in the power supply, and the fast disconnection of a heavy load elsewhere in the system.

I might suggest you either fit a transorb instead of the diode, or a capacitor right across the 125v supply, or fix the overvoltage transients at the source. There is nothing wrong with the solenoid, or the diode. The problem is external.
///Not quite external since the original posting is concerned with the diode sizing. The diode experienced two shortings according to the original posting.\\\
Nandos (Electrical)
6 Jul 03 18:28
Jcorb5:

The Joules available is = L * I^2 = 22 H * .28^2 = 1.728 joules per pulse.

Having 125 volts at 0.28 amps a 1N4004 will do the job and is just one amp diode, since it will take 30 amps for 8 milliseconds.

A diode 200 volts, 1 amp should be suficient

You may have another problem, since the diode will clamp the voltage to 125.7 volts.

Regards

Nando
ScottyUK (Electrical)
8 Jul 03 13:09
Just a thought here, from the back-to-basic principles of passive components.

The coil inductance will, on removal of the supply voltage, attempt to keep the current of 290mA flowing through the coil resistance, and will do so by collapsing the magnetic field of the relay coil and thus releasing the stored energy within it. the current circultaes throught he forward-biased diode. The diode should therefore see a decaying current which is initially 290mA and falls exponentially toward zero. The forward voltdrop is approx. 0.7V.

The diode in the 'relay on' state will be reverse biased by the supply voltage, and for the 1N400* series the reverse leakage will be in sub-uA level.

At no point will the diode see 22A. It might however see a high transient voltage if the diode is a 'slow' type designed for low frequency rectification. It might be worth considering a faster type such as one of the UF540* series, although I am doubtful that response time is the issue as the coil doesn't store enough energy to exceed the I^2T rating of a 22A rectifier and cause damage.

The R-C type snubber networks mentioned earlier are typical for AC switching to reduce arcing; diode clamps are standard for DC coil switching.

I must ask, why was such a large rating (22A!) chosen for a clamp diode? Is this the manufacturer's design?

The only other questions I would ask are:

Where is the diode physically in relation to the coil? Ideally it should be directly connected across the coil.

Are there any large loads on the DC bus which can dump energy into the system, such as regenerating motors, any heavy inductive switching, etc? I would consider getting a fast 'scope onto the supply rail and set the single-shot trigger level at, say, 200V and see what is happening elsewhere on the supply, as the problem sounds as though it is external to the coil / diode.

jbartos (Electrical)
12 Jul 03 18:03
Suggestion: When it comes to basics, the solenoid valve consists of inductance and resistance (and some stray capacitance); the diode equivalent circuit includes resistance and capacitance, visit
http://www.tpub.com/neets/book11/45n.htm
for a diode approximate equivalent circuit

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