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MPP revE need help

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  • dbanner
    replied
    The mosfet drain is connected to the heatsink tab in order to dissipate power(heat).
    When you touch the heatsink, it is if you are connected to the drain. The capacitative effects of the hand is now part of the circuit. There is a heatsink wafer to go between the mosfet and heatsink, which provides electrical isolation.

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  • dbanner
    replied
    Tick the ch2 box on the software. Hahahaha.It's not uncommon when getting familiar with new software.Change the time base on ch1.

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  • Albdelafar
    replied
    I did a test with another concentric coil and result is the same :'(
    But I see an improvement when i touch the heatsink of IRF840.....
    Could someone help me please to find the issue
    Thanks
    Alban
    Click image for larger version

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    EDIT: i have no signal on TX pin.... i will investigate on this side.

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  • Albdelafar
    replied
    Hi surfdetector in the first pick the timings are set up to 40 us by division. I forgot to mention it.

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  • surfdetector
    replied
    The time/Div on your scope is set to 400us. Looks like your TX pulse is ~1000us

    Why don't you look at TP3 with a lower time/div? that way you can see what it looks like.
    Cheers
    Last edited by surfdetector; 05-02-2020, 11:34 AM. Reason: correct spelling

    Leave a comment:


  • Albdelafar
    replied
    Strange,
    I think the link was working the fisrt time i'm tried it....

    For my MPP rev-E i'm still waiting for my 2W resistors but i received my new oscope and somme other stuff.
    I build the dampin resistor tool but i wait my 100X probe to test it I fear to burn my new oscope if I fail the voltage divider (I have only 1/4 w chinese resistors I am not confident with that).
    I change all my connection pin and I solder RX and TX pin. I use the damping tool at 700ohm to replace the damping resistor.
    I changed my pulse to 500hz with a pulse width of 200us and a pulse periode of 2ms but now i have my signal in tp3 like that :
    CH1 TP3
    CH2 TP2
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    What's going wrong for you ? Thanks

    Leave a comment:


  • 6666
    replied
    Same for me both of the links in post #3 go to coil calculator

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  • Qiaozhi
    replied
    Originally posted by Matt_Rowe View Post
    your link here goes to the coil calulator, not your book. Inside the METAL DETECTOR - Second Edition - Published 2015
    Hmmm ...
    I wonder if that's a forum problem, as there were two links in that post. The first one was for the coil calculator.
    As the second link was a cut-and-paste, it should have pointed to the book.
    Like this -> Inside the METAL DETECTOR - Second Edition - Published 2015

    Leave a comment:


  • Matt_Rowe
    replied
    Originally posted by Qiaozhi View Post
    There is a coil calculator on the forum here -> Coil Calculator

    A metal detector is not the easiest project to start with if you're trying to learn electronics.
    In order to gain a better understanding of how metal detectors work, you will find this useful -> Inside the METAL DETECTOR - Second Edition - Published 2015
    your link here goes to the coil calulator, not your book. Inside the METAL DETECTOR - Second Edition - Published 2015

    Leave a comment:


  • green
    replied
    Originally posted by waltr View Post
    I have used general 1/4W leaded resistors for the flyback Voltage divider with out issues.

    There has been many posts and theories about flyback voltage and MOSFET's going into Avalanche.
    Efficiency shouldn't change since this is NOT drawing battery power (if definition of Efficiency is power out / power in).
    However, it can cause longer recovery times which leads to longer first sample delay and can cause premature MOSFET failure so it is desirable to avoid.
    I was thinking avalanche current does discharge the battery

    Leave a comment:


  • waltr
    replied
    Originally posted by Albdelafar View Post
    Qiaozhi I'll try to find the data sheet of my Chinese resistance, but I'm not sure I will find it!!
    OK waltr thanks for the information that improves my understanding!!
    Do you think a high FLYBACK voltage could significantly affect efficiency by putting the MOSFET into avalanche mode?
    I have used general 1/4W leaded resistors for the flyback Voltage divider with out issues.

    There has been many posts and theories about flyback voltage and MOSFET's going into Avalanche.
    Efficiency shouldn't change since this is NOT drawing battery power (if definition of Efficiency is power out / power in).
    However, it can cause longer recovery times which leads to longer first sample delay and can cause premature MOSFET failure so it is desirable to avoid.

    Leave a comment:


  • Albdelafar
    replied
    Qiaozhi I'll try to find the data sheet of my Chinese resistance, but I'm not sure I will find it!!
    OK waltr thanks for the information that improves my understanding!!
    Do you think a high FLYBACK voltage could significantly affect efficiency by putting the MOSFET into avalanche mode?

    Leave a comment:


  • waltr
    replied
    Originally posted by Albdelafar View Post
    I see this on your sheme now so i need to mesue signal on B it is right ?



    I learn with U instead of E in france.
    So if I take this calcul you take a 500V flyback voltage for 2 parrallel resistor of 100k and 1k. I obtain 101kohm serial resistance.
    My duty cycle is 100usec in a 1msec period so i get 0.1 duty cycle (10%)
    So for total wattage i get 0.1*(500*500/101000) = 0.25w
    Yes, U = potential in Europe and I learned U as potential in Physics but E in electronics for Electrical Potential.

    It is the Duty cycle of the FLYBACK pulse not the TX pulse that produces the high Voltage. That is where I got 10usec.

    The 100k and 1k are in Series with the coil as shown in the schematic 6666 posted.

    Good info of resistor Q.
    Thanks.

    Leave a comment:


  • Qiaozhi
    replied
    You need to be aware that resistors also have a operating voltage rating. Although resistors are simple passive components, the datasheet specs can still be confusing.
    Here's some very useful information:

    Attached Files

    Leave a comment:


  • Albdelafar
    replied
    Originally posted by 6666 View Post
    its just a voltage divider across the coil, like what waltr was talking about
    I see this on your sheme now so i need to mesue signal on B it is right ?

    I little bit of math will tell you what Wattage resistor is needed.

    Ohm's Law: E = I*R or I = E/R
    Power = I*E = E*E/R

    Say you expect 300V so 500*500/10100 = 2.47W peak.

    But the flyback is low duty cycle. Say 10usec in a 2msec period (500Hz pulse rate). so 0.00001/0.002 = 0.005% duty.
    So total Wattage is duty * peak = 2.47W * 0.005 = 0.012 W average.
    I learn with U instead of E in france.
    So if I take this calcul you take a 500V flyback voltage for 2 parrallel resistor of 100k and 1k. I obtain 101kohm serial resistance.
    My duty cycle is 100usec in a 1msec period so i get 0.1 duty cycle (10%)
    So for total wattage i get 0.1*(500*500/101000) = 0.25w

    Leave a comment:

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