Inverted input? (not inverting)

Discussion in 'General Electronics Chat' started by shortbus, Apr 25, 2013.

  1. shortbus

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    Ran across a mosfet driver that may be what I've been looking for, for my project. If "inverted input" means that it is on when the signal is a logic low.
    Would you guys look at the data sheet and tell me if thats what an inverted input is? The part number is IRS2183. Thanks.
     
  2. #12

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    Page 6 Fig 1. Hi in = Hout High = hi.
     
    Last edited: Apr 25, 2013
  3. GopherT

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    It looks like The high side driver is, high in => high out

    The low side driver is "not low" in => low out.



    Not
     
  4. praondevou

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    Pros:
    - you can use a single input signal if you tie /LIN and HIN together
    - you don't need to worry about deadtime because it is built-in

    Cons:
    - When both inputs are active (/LIN = low, HIN = high) then both outputs are off. Not necessarily a Con but strange. Other chips give priority to one channel.
    - deadtime is fixed for the 2183. If this is a project in progress I would prefer being able to adjust it. You can do it before the chip though if you use two separate input signals.
     
  5. Ron H

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    If you tie the two inputs together, and drive a half bridge with the outputs, the half bridge output will be low when the inputs are low, and vice-versa.
     
  6. Ron H

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    Are you sure? My interpretation is thatwhen /LIN is low and Hin is high, LO will be high and HO will be high, creating a short in a half bridge which is attached to the outputs.
     
  7. GopherT

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    I agree, Ron. If tied together, then a high input will be:
    > hi for hi side
    > low for low side
     
  8. praondevou

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    Which means there can't be a short.

    Figure 1:

    1. Input HIN = High, Input /LIN = High will turn ON HO and turn OFF LO.
    2. Input HIN = Low, Input /LIN = Low will turn OFF HO and turn ON LO.

    Which would mean I can tie them together without ever having a short (both HO and LO are ON) at the output. I assume we are talking about driving two NFETs.

    3. Then with Both inputs active (but not tied together), i.e. HIN = high and /LIN = Low, both outputs are OFF.

    Am I missing something?

    [​IMG]
     
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  9. GopherT

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    Correct.

    incorrect there. A low input to /LIN will turn ON LO


    Also, the dt pin (dead time) can be configured with a resistor (see DATASHEET) to set the delay between turning turning one off and turning the other side on to protect your MOSFETs.
     
  10. praondevou

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    I was talking about HIN = High and /LIN = Low, i.e. both inputs are active.
    According to figure 1: H on HIN AND L on /LIN turns both outputs OFF.

    Shortbus mentioned the IR2183 not the IR21834.
    Datasheet first page: 400ns internal fixed for IR2183 (or 500ns according to other datasheets) and adjustable for the IR21834.

    Looking at figure 1 there can never be both outputs ON at the same time.
     
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  11. Ron H

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    Now I get it. I failed to look at the timing diagram (Fig. 1). It makes sense that internal logic would prevent both outputs from being high at the same time.
     
  12. shortbus

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    The discussion here is the same as I was having with myself :). Praondevou's timing chart in post #8 is what I was thinking was correct.

    My reason for wanting to use this chip is eliminating the use of an inverter chip in my logic circuit.

    Maybe I should explain a little more of what I'm doing. I always put that off because its hard for me to put things in words and not screw up the technical terms, since I'm not an EE. This is for a long running project, the basics were given to me by an EE I used to work with, but he wouldn't help past doing a block diagram for me.

    1. Two capacitor banks, that are alternately charged and discharged. All fet switches are high side.

    2. In order to prevent having a charge pump to recharge the boot caps, and since a driver is needed for the fet's, I found the IRS2183. Since the cap banks prevent the source of the fet's from going to common to recharge the boot caps, I was going to use a second small value fet just to provide a path to common to recharge the boot caps. The only thing the small fet does is the recharge of the boot cap. There will be diodes to prevent drawing the charge out of the cap bank while the recharge of the boot caps is going on.

    3. This is a hobby level version of an industrial machine tool, a pulse EDM, electrical discharge machine. While it doesn't make sense to do this to most people, there is a lot of people working toward this goal in the model engineering world.

    I know this probably how most of you guys would approach doing this, its what I've come up with. But I'm open to ideas and critique. Thanks for all the help you guys have given.
     
  13. Ron H

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    To what voltage are you charging the caps?
     
  14. praondevou

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    Can we see a circuit? :)
     
  15. shortbus

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    Ron, the open circuit/charge voltage is ~100VDC.

    Praondevou, I'll draw up a better one than the "napkin" sketch I have now and post.
     
  16. shortbus

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    OK, here is one leg of the charge - discharge part of the circuit. There will be two such circuits. When the charge switch of one is on the, the discharge of the other is on.

    The two circuits will alternate back and forth, at between 5kHz to 50kHz. The frequency is chosen to suit the job at hand, to make the best cut depending on conditions. Once the best frequency is found,it is not changed, but left at that setting.
     
    Last edited: Apr 26, 2013
  17. praondevou

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    Did you already make a choice for the power FETs?
    Will there be a controlled current limiting for the charge/discharge?
    What size are the capacitors? What current are we talking about?
     
  18. shortbus

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    Yes, I have the power fet's, they are STE53NC50. They are isotop style mosfets like used in inverter style welders. The low sides I haven't bought yet but some thing in a TO220/TO247 package will be used.

    For the power supply it is a 70V, 20A transformer, a 400V, 50A bridge rectifier, a large cap that can't remember the specs on off hand. This is all from an old servo motor drive I bought at auction. The there is two 250W power resistors in parallel that I can't remember the ohm's of, but they were figured to only allow 16A to flow in case of a short in the electrode. All of this stuff is in storage but if needed I'll dig it out.

    The caps vary in value, from 1μF to 4.7μF, but all are 450V polypropylene pulse caps, the box style. Rifa PHE 450 and Kemet MKP. They are switched as needed in 1μF, 2.2μF, 3.3μF, 4μF, 6μF and 8μF value steps.

    The current is 16A max due to the power resistor. But it is also adjustable by using the PWM part of the logic circuit.

    The logic also uses comparators from the cap bank. One won't allow the pulse if the voltage on the caps isn't at least a set value. The second will shuts the pulse off if the voltage falls past a set value. This is done on commercial machines, its known as 'iso-pulse'. Its done to keep each machining pulse the same energy level. None of the other DIY machines out there have incorporated iso-pulse that I know of.
     
  19. shortbus

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    Anybody have an idea if this will work? And if not what needs to be changed?
    PLZ I NED HLP. :)
     
  20. praondevou

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    The resistor is in series with the 100V power supply I assume?

    You say it is also adjustable by using the PWM part. That means there is an inductor in series somewhere? If we assume wires to be ideal components with zero inductance there would be nothing to limit the current to the capacitors apart from the power supply output impedance and capacitor impedance.

    How is the output current limited?
     
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