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Help with pulsed DC power circuit

Discussion in 'General Electronics Discussion' started by Gregory Miller, Mar 15, 2013.

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  1. Gregory Miller

    Gregory Miller

    7
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    Mar 15, 2013
    Hi,

    I am working on a project for a power supply - pulsed DC to a load. A plus 5V square wave is fed to H11D1 opto-coupler which is supposed to turn on a C38M SCR. This circuit does not work.

    I have attached a basic drawing.

    Could anyone please tell me the error in this circuit?
     

    Attached Files:

  2. Gregory Miller

    Gregory Miller

    7
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    Mar 15, 2013
    Additional information:
    1) the opto-coupler works fine if I supply a positive voltage on the collector (pin 5) and provide the emitter with a load to ground (pin 4).
    2) the SCR works fine (and powers the load) if I put the 100 ohm resistor from the SCR anode to the gate.
    3) when the two circuits are combined, the O-scope sees that the SCR attempts to turn on but can't do it.

    I don't think that I should apply any more current through the NPN in the opto-coupler because I might fry it.

    Anybody have ideas?
     
  3. duke37

    duke37

    5,364
    771
    Jan 9, 2011
    Upside down circuits annoy me since I have to stand on my head to understand them.

    I do not see what the 100 ohm resistor does, at 100V input it will allow 1A through, enough to kill the SCR and the optocoupler.
    There may not be enough current fed to the SCR to make it latch, show us the technical details of the optocoupler and the SCR so that we can suggest what can be done to make it work.
     
  4. john monks

    john monks

    693
    2
    Mar 9, 2012
    What are you driving the LED in the opto with?
    I suspect your current may be low and I suspect that the current required by the SCR may be a little on the high side.
     
  5. Gregory Miller

    Gregory Miller

    7
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    Mar 15, 2013
    This is actually not my circuit. I have the board in front of me with no schematic and some basic instructions. Trying to get it to work for a friend.

    I agree that 100 ohms is not enough to protect the NPN in the opto-coupler ( which looks to be .6 A max if I read the spec sheet correctly). The peak gate current is 2 A - so at least the 100 ohm resistor should keep the amps below that limit. I am attaching what I have in terms of the SCR and opto-coupler datasheets. This sight will not allow me to attach the document so I posted the URL to the document instead.

    I am thinking about redesigning the circuit. Maybe eliminate the SCR and replace with a MOSFET?

    http://www.jameco.com/Jameco/Products/ProdDS/14859.pdf
    http://www.datasheetcatalog.org/datasheet/vishay/83611.pdf

    Thanks for your help,
    Greg
     
  6. john monks

    john monks

    693
    2
    Mar 9, 2012
    My thinking is that you have everything hooked up correctly and that the parts polarences are such that the SCR just won't fire. So it takes at worst case 80ma to turn on the SCR and the opto puts out 100ma. And the opto requires 10ma minimum. So I'm suggesting trying more current to drive the opto like 50ma.
    By the way a 100 ohm resistor In the collector may be fine because when the SCR does turn on there is essentially no current feeding the opto and the trigger.
    From my experience with Mosfets they are much less forgiving than SCR's although it should work.
    So before ripping everything apart just try more current for the LED. Maybe a 9 volt battery and a 150 ohm resistor.
     
    Last edited: Mar 15, 2013
  7. Gregory Miller

    Gregory Miller

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    Mar 15, 2013
    You got me thinking here. The Current Transfer Ration (CTR) is based on the current through the LED. There is currently a 220 ohm resistor on the anode side of the LED inside the coupler. This might need to be reduced to get a better current flow. This should help the transistor to conduct better. This will apply more current to the gate of the SCR.

    Now to figure the correct amount of resistance. I could just put another 220 in parallel with the one already there - that would give 110 ohm to the anode of the LED. I need to make sure that it is not to low.

    Thank You,
    Sincerely,
    Greg
     
  8. john monks

    john monks

    693
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    Mar 9, 2012
    If I understand right you have a 5 volt pulse driving the LED through a 220 ohm resistor. This give you around 18ma of current. Using two 220 ohm resistors in parallel in this case is find because this would give you around 35ma. The maximum is 60ma. You should be fine.
     
  9. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,496
    2,837
    Jan 21, 2010
    Also, get a sensitive gate SCR. Something that requires only a couple of mA mat be better.

    Alternatively, have the optocoupler driving the base of another transistor to turn on the SCR. That transistor will need a significant Vceo (300V?)
     
  10. Electrobrains

    Electrobrains

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    5
    Jan 2, 2012
    Gregory's circuit can not work.

    The optocoupler (H11D1) has only CTR=20% (minimum).

    To work securely, the SCR (C38M) demands IGT=40mA (maximum).

    That means the optocoupler input (diode side) would need a minimum current of 200mA (40mA / 0.2)!
    But the specified max current of the optocoupler input, is only 60mA!

    So the two component simply do not match one another in this circuit.
     
  11. Gregory Miller

    Gregory Miller

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    Mar 15, 2013
    I will try the first idea (putting another 220 ohm in parallel). If Electrobrains is correct I will have to redesign. Maybe insert a darlington pair to drive the SCR. Can anybody suggest a good device - one that has 200ma current capacity and a Vceo of 300V?

    Thanks,
    Greg
     
  12. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    I haven't looked, but does your optocoupler have an output with a Vceo of 300V as well?

    I'd just use some regular transistor, you don't need a darlington. There are lots with ratings this high. I recommend you see what your supplier has. Pick something capable of 500mA of Ic or thereabouts.

    edit: Yeah the opto is OK for 300V
     
  13. Electrobrains

    Electrobrains

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    5
    Jan 2, 2012
    As Steve writes, a normal transistor like MJE340 would be ok. That transistor has hFE minimum 30. It would give you a minimum total amplification of 6 (0.2*30), which is fine.

    Suggestion: put a 180 Ohm resistor as current limiter on the diode side (gives approx. 20mA input).

    I am quite sure the resistor R in your circuit can be reduced (I have made similar circuits without that resistor). It would allow you to switch on the SCR at an earlier point of the sine wave.
    The avalanche characteristic of the SCR is faster than the signal transfer speed of the optocoupler/transistor, which will not allow the transistor current to raise too high (as John points out).

    A problem could occur though, if the optocoupler is in the switched on state and suddenly the AC power is applied. Even worse if your load would have a built-in capacitor. Suggestion: R=47 Ohm /0.6W.
     

    Attached Files:

  14. Gregory Miller

    Gregory Miller

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    Mar 15, 2013
    Thanks for redrawing the circuit. The plan makes sense. I do have a few questions. If the 100 ohm resistor is changed to 47 ohms - the amps thru the gate would be a function of the resistor, the MJE340 "on" resistance and the input resistance of the SCR gate - I will need to find out those values to determine the current thru the resistor. If for example, the amps turned out to be 1A through the SCR gate side of this circuit, then the resistor would need to be 47 Watts? Need I be concerned about this?
     
  15. Electrobrains

    Electrobrains

    259
    5
    Jan 2, 2012
    You wouldn't speak about resistance of the transistors and SCRs. The transistor is a current source and the gate of the SCR (as far as remember) is to be seen as a diode (or zener diode).

    The resistor is only there to protect the transistor/SCR at abnormal transient occasions, as I already mentioned.

    If the transistor is on by the optocoupler and the SCR is off because no voltage is present, a problem could occur if you suddenly connect the AC power at the peak of the sine wave. Then it will be a race between the avalanche function of the SCR and the rising edge of the current through the transistor/SCR (di/dt). That is a very fast process. The quicker the SCR switches on, the less current the transistor/resistor have to take.

    Often Snubbers and serial Inductors are used with SCR's and Triacs. They help removing disturbance and prevent self firing, but also reduce di/dt (du/dt).

    In the normal case (AC voltage present when the optocoupler switches), the resistor would not be needed! That's because the transistor will switch on rather slowly. Thus, when the transistor/resistor/gate current reaches max 40mA, the SCR will "take over" all current and practically protect the transistor circuit.

    I suggested a 0.6W (metal film) resistor. It should be well able to handle the high current for the very short (micro seconds), transient moment. Just don't use "fusing resistors"!

    In the normal case, you have the max transient, power in a 100 Ohm resistor of P=40mA*40mA*100=160mW.
    Of course the RMS power is even much less!
     
  16. Gregory Miller

    Gregory Miller

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    Mar 15, 2013
    Thanks for the great explanation.

    Sincerely,
    Greg
     
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