Varistor+Diodes as transient suppressor

Discussion in 'General Electronics Chat' started by abuhafss, Jan 13, 2015.

  1. abuhafss

    Thread Starter Active Member

    Aug 17, 2010
    154
    2
    Hi

    In a simple 555 based pump driver circuit two smd diodes PJ5N (in parallel) are connected across the pump (12V). But after some time of operation the diodes are blown.

    I have been replacing the two damaged diodes with a varistor successfully. I have a lot more of those unused pumps (with undamaged diodes) and I am thinking to add a varistor as a precautionary measure. My question is, is it okay to use diodes and varistor together as transient suppressor? Actually, I want to save time for the removal of the diodes.
     
  2. #12

    Expert

    Nov 30, 2010
    16,343
    6,827
    Probably work if you put them all in parallel, but the, "right" way is to install diodes the correct size.
     
  3. MaxHeadRoom

    Expert

    Jul 18, 2013
    10,565
    2,379
    What is the nature of the circuit? is this a single ended PWM controller?
    You could also use a R/C snubber.
    Max.
     
  4. abuhafss

    Thread Starter Active Member

    Aug 17, 2010
    154
    2
    Here is the circuit:
    Screenshot 2015-01-09 12.16.26.png
     
    Last edited: Jan 13, 2015
  5. #12

    Expert

    Nov 30, 2010
    16,343
    6,827
    Want to get at the math? That should tell why the diodes are dying.
    Is there a timing capacitor missing from that 555 timer circuit or are you using a reverse biased diode as a capacitor?
    That would be a pretty fast oscillator!

    Anyway, Energy = (1/2) LI^2
    A milliamp for a millisecond is a micro-watt-second.
    A million microwatt seconds per second is a watt second per second, usually referred to as a watt among people that are familiar with these things.
    For whatever amount of current is flowing in the 80 uH inductor at the moment of shut off, that energy has to go through the diodes and be dissipated as heat in the diodes and the resistance of the inductor.
    The initial current through the diodes is limited by the resistance (ohms) of the inductor. That's your Isurge, max.
    I have a feeling that the pump has a lot of energy stored as inertia. That might be where the estimation of energy gets hard to describe, and therefore, the reason the diodes got chosen as the wrong size.
    So, 1/2 x 80 uH x I squared x events per second is one answer for the watts that have to be dissipated.

    I'm getting nothing for PJ5N datasheets.

    So, measure the ohms in the alleged 80 uH to find your Isurge, max as 12Volts/ohms = amps
    Find the current the pump uses to do: Energy = 1/2 x 80 e-6 x Isquared
    Find out how many times per second this all happens.
    Measure what size package (that contains a diode) you can fit on the circuit board.
    Go looking for a good Isurge, max diode.

    http://www.mouser.com/Semiconductor...w7zmxZ1yw7z62Z1yw7zoxZ1yw7y3uZ1yuqh8oZ1z0z7pt

    Some of these might work. None of them might work. You can ignore me and I will not feel offended.
     
    abuhafss likes this.
  6. abuhafss

    Thread Starter Active Member

    Aug 17, 2010
    154
    2
    I'd posted the wrong schematic first, which I updated during your typing the response.

    L = 1200µH
    R = 2 Ohms
    I = 3A
    V = 14V
    Freq = 16Hz

    So the initial current is 14 / 2 = 7A.

    Energy = 0.5 x 1200µH x 3^2 x 16 = 72 x 1.2 / 1000 = 0.0864 Watts ......................Correct ?

    Sorry those are S1 diodes. http://www.vishay.com/docs/88711/s1.pdf

    Just for reference, here is photo of the PCB.

    555 pcb.jpg
     
  7. #12

    Expert

    Nov 30, 2010
    16,343
    6,827
    OK. Here's where we are. I have accomplished getting useful numbers from you. I can imagine several ways to approach this, but I'm over my head. I'm going to compile some options and ask the best people on the site for help.

    I found a way to get 22% less power load on the suppressor diodes just by changing the part numbers, but I don't think that's enough. I'd like to see 50% to 67% improvement to get some reliability in this design.

    When I come back, I will post some options that need somebody to run the math on them.

    http://www.mouser.com/ProductDetail/Rectron/FM150-W/?qs=ZlFwIjeWdbHOI%2b8vCtdGCA==
     
  8. #12

    Expert

    Nov 30, 2010
    16,343
    6,827
    There is something odd about saying the normal coil current is 3 amps and the surge current is 7 amps.
    The only way this can happen is if the, "on" time of the coil is so small that the current never reaches 14V/2 ohms.
    The change in current (rate of increase) is limited by the inductance.

    I'm double checking my math...

    IL/V = time
    3 x 0.0012/14 = about 257 microseconds of "on" time.
    True?
     
    Last edited: Jan 14, 2015
  9. MikeML

    AAC Fanatic!

    Oct 2, 2009
    5,450
    1,066
    Which of the S1 suffix letters is the PJ5N?
     
  10. #12

    Expert

    Nov 30, 2010
    16,343
    6,827
    This is the research. Can somebody help choose the right approach?
    @MikeML It's S1A
     
  11. abuhafss

    Thread Starter Active Member

    Aug 17, 2010
    154
    2
    You can see the diodes in the photo shown in post #6.
     
  12. abuhafss

    Thread Starter Active Member

    Aug 17, 2010
    154
    2
    Just for your reference, these pumps are actually automotive fuel pumps to be operated with a 12V automotive battery. However, they work fine even with a 12V 2A power supply. I have been bench testing with a 12V 5A SLA battery.
     
  13. #12

    Expert

    Nov 30, 2010
    16,343
    6,827
    That still doesn't seem right for 3 amps through 2 ohms.

    While you're in there, you need to use a tooth brush on that board. It's getting really ugly with all that white powder on it!

    and do not say the fuel pumps will be in cars or you will get your conversation locked up. Cars are against the rules here.
     
  14. MikeML

    AAC Fanatic!

    Oct 2, 2009
    5,450
    1,066
    50piv for the diode is a bit low. I would have used the 400piv one...

    Two diodes in parallel do not work unless you add a low value (1Ω) resistor in series with each diode.

    The peak current through the diode(s) is ~7A. The average current is below 1A.

    269.gif
     
    #12 likes this.
  15. abuhafss

    Thread Starter Active Member

    Aug 17, 2010
    154
    2
    I already mentioned, I have lot of those pumps with me. It just happened that the PCB which I snapped had some dust on it otherwise, all other are neat and clean.

    For that particular reason I have been using the common noun "pump" earlier. By the way, the pump I am talking about is a universal type which can also be used for other purposes.
     
  16. #12

    Expert

    Nov 30, 2010
    16,343
    6,827
    It is good to know that. Bill Marsden gets irritated with me if I break the rules by answering automotive questions. :rolleyes:
     
  17. abuhafss

    Thread Starter Active Member

    Aug 17, 2010
    154
    2
    I think, we are making this discussion more and more complex. My question was:

    I have lot such pumps in working order. I am thinking just to add a varistor in parallel with the two free-wheeling diodes as a precautionary measure.
    Would it be okay to use a varistor with the diodes? Although I have already tested them with 5A battery, I just want to be sure it won't be a problem in actual run.
     
  18. MikeML

    AAC Fanatic!

    Oct 2, 2009
    5,450
    1,066
    What would the varistor do except waste power?
     
  19. #12

    Expert

    Nov 30, 2010
    16,343
    6,827
    It won't hurt anything, but if I remember correctly, I chose the TVS because it's better than a MOV at wasting the energy that was killing your original 1 amp diodes.

    Right now I'm hoping somebody as educated as MikeML will look at my research and say, "There it is. That's the best option".
     
  20. abuhafss

    Thread Starter Active Member

    Aug 17, 2010
    154
    2
    So, should I remove the diodes and only use varistor?
     
Loading...