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Misc Electronic Questions

Discussion in 'General Electronics Chat' started by Billy Mayo, Apr 4, 2013.

  1. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    In general, yes. The idea is that below a certain value on the diode setting you get a beep. This is generally used to check continuity of wires without looking at the meter.

    The best way to find a shorted or degraded transistor is to use the diode function. You just have to watch the effects of transformers and inductors in a circuit.
     
  2. Billy Mayo

    Billy Mayo Member

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    Degraded transistor means?

    Yes I use the diode mode only when find shorted diodes or transistors
     
  3. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    Degraded might mean leaky. Say you measured 0.3 V for a transistor, you might suspect it's bad. Just like a pregnancy test can tell you a female is pregnant, but it cannot tell you she is not with 100% confidence.

    As a reminder many tests are designed to detect bad components, but not good ones. I may have told you about the 3AG fuse problem I saw twice in my life. It tested fine with an ohmmeter out of circuit. In circuit, it would open, The voltage across the fuse was 120 VAC. Transistors and diodes could still be bad. Only a curve tracer knows for sure. A transistor checker - no.
     
    Last edited: May 12, 2013
  4. dave

    Dave New Member

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  5. Billy Mayo

    Billy Mayo Member

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    At work we had to changed out 12 volt regulators with another 12 volt regulator because of a IN RUSH current problem

    I don't understand what's the difference between IN RUSH current and a LOAD , because the current goes up and the voltage goes down

    If you load the regulator to much the current goes up and the voltage goes down

    If the IN RUSH current it high on the output of the regulator the voltage goes down

    So the IN RUSH current is loading the regulator? but how is it different than a LOAD?

    It's a 12 volt regulator at 1 amp, they switched it with a different company that is the same 12 volt at 1 amp, my manager said that the internally the regulator must of had a bad feedback loop internally because the regulator was current limiting and not the new one which is rated the same voltage and current output

    What is it internally in the regulator that can cause current limiting? both regulators have the same voltage and current output but one current limits and drops the voltage in half

    It goes from 12volts to 6 volts as more current is draw by the regulator
     
  6. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    You actually could have been a victim of counterfeiting which is a BIG problem for semiconductor manufacturers.

    Do you have the part numbers of both of the regulators that were switched? I can think of a series which is 1A from one manufacturer and from another it's 1.5A, but it's not the same regulator. This guy http://www.electro-tech-online.com/custompdfs/2013/05/329.pdf is probably similar to the 7812, but is 1.5A instead of 1A.
     
    Last edited: May 13, 2013
  7. Billy Mayo

    Billy Mayo Member

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    What u mean counter fitting?

    I will get the part numbers tomorrow , they keep changing them

    How does In rush current LOAD a regulator differently than a LOAD on the output of a regulator?
     
  8. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    Like counterfeit money. Fake. Manufactured by someone else, but has the markings of the real thing.
     
  9. Billy Mayo

    Billy Mayo Member

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    oh ok so the datasheets are fake date I know what u mean know

    Yes true would have to try it out in circuit to really know how its operating
     
  10. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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  11. Billy Mayo

    Billy Mayo Member

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    lets say they are both not counterfeit or fake

    both are 12 volts at 1 amp regulators

    What internally would cause the regulator to lower the voltage when there is a HIGH current draw ?

    There is a feeback look inside a regulator and a comparator , what else is there inside a regulator?
     
  12. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    Well allowable "inrush" could be related to just how fast a regulator responds. A slower responding regulator would allow a higher inrush current than one that responds quickly. A side effect of responding too quickly is the tendency to oscillate. So, there is a tradeoff one has to make.
     
  13. Billy Mayo

    Billy Mayo Member

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    So how long does a Inrush current last? Nano seconds or microseconds?

    What is Inrush current caused from?
     
  14. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    Well, it comes down to the following rules:
    The voltage across a capacitor cannot change instantaneously and the current through an inductor cannot change instantaneously.

    That said, at the instant a capacitor sees voltage it's ALMOST a short.

    So, we could be back to a Time constant discussion. It takes about 5 time constants 5* (R*C) for the voltage to reach 99% of the final value. R in ohms and C in Farads; resulting units are seconds.
     
  15. Billy Mayo

    Billy Mayo Member

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    I talked with my manager today about this regulator issue

    He said that the filter capacitors on the input to the regulator gets discharged and lowers the voltage down when there is a in rush current or high current draw on the output of the regulator.

    Also there is a voltage drop across the regulator it self which has a resistance from input to output of the regulator

    So it's like a voltage divided with the LOAD

    They have been trying different regulators that are 12 volts at 1 amp , but each one behaves differently when there is a HIGH current draw on the output

    He was saying that when there is a HIGH current draw on the output that the LOAD is not constant, which i don't understand , I would think the load would be constant , he is saying the load is not constant

    I'm also not sure why a linear regulator discharges the filter capacitor

    When using a Switching Regulator , it doesn't discharge the filter capacitor on the input of the regulator, why is that? is it because a switching regulator is isolated?

    He was saying that a switching regulator using a PWM oscillator which has a "switching Mode LOSS" only in nanoseconds, that the leading and falling edges of the PWM signal will create a voltage drop on the output of the regulator

    A Switching Regulator doesn't have a voltage drop across it or a resistance from input and output , so there is no voltage loss , is this true?
     
  16. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    This question is rather hard to answer because of the buck, boost and buck-boost topologies. The latter, I'm not too familiar with. Buck, means the voltage has to be lower than the input, Boost means the output must be higher than the input. The question is "how high"?

    Ask you rmanager to look at this http://www.alliedelec.com/search/productdetail.aspx?SKU=70051995 as a possible replacement?
     
  17. Billy Mayo

    Billy Mayo Member

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    Thanks

    When transistor switches or IGBT that are used as a Push pull network , if the push or pull is displaced or offset what happens?

    Because the PUSHing transistor switch will turn OFF , but the PULL transistor switch will still be on if they are not synced up, what will this do?

    When using Two IGBT transistors as a push pull network, you have to use the same value Resistor GATE value

    We use 24 ohms as the Resistor GATE value for both IGBT transistors , they are configure for a push pull network

    If one of the Resistor GATE which is called RG , is lets say 22ohms or 20ohms or 15ohms on either the Push or pull switching Transistor or IGBT transistor it will cause a displacement and offset of timing, what will happen?

    example pushing switching transistor or IGBT RG is 24ohms
    pull switching transistor Or IGBT , RG value is 20 ohms

    What will happen?

    My manager said that one of the IGBT will have a different DT/VT , But if both pull pull IGBT have different DT/VT what does that mean?

    The push IGBT will shut while the pull IGBT will shut off later , causing a short?
     
  18. Billy Mayo

    Billy Mayo Member

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    I'm confused on DT/DV on push pull switching transistors or IGBT, mostly the values of the resistors are the same value to keep the DT/DV in sync with the on time and off times, but if the resistor values are slightly different between the push pull switching transistor pairs , what happens?
     
  19. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    Are you sure he doesn't mean
    ?

    I'm trying to figure out what DT and DV stands for? DT=Dead time?

    Vt= V (threshold)?
     
    Last edited: May 15, 2013
  20. Billy Mayo

    Billy Mayo Member

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    yes he means dv/dt Applied to the push pull PAIR , if the PAIR is slightly off DV/DT what happens?
     
  21. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    actually somewhat determines power dissipation. The faster it turns on and off, the better. The gate current does affect how hard it turns on.

    In most cases the designer will implement some sort of "dead-time" control, so both push-pull drivers cannot be on at the same time.

    In an audio amp, class AB required that one of the bipolar transistors in a push-pull stage to be partially on and thermally compensated. This eliminates "crossover distortion".
     

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