What would you use?

Discussion in 'General Electronics Chat' started by marshallf3, May 16, 2011.

  1. marshallf3

    Thread Starter Well-Known Member

    Jul 26, 2010
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    2.5 mA Active low input drives +2A or +3A out at 10V - 12V?

    Supply is usually +13.6V to common however it varies due to the battery state of charge and whether it's getting charged.

    I realize the gains would be at least 800 or 1,200 but all I've come up with is power hungry.
     
  2. THE_RB

    AAC Fanatic!

    Feb 11, 2008
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    Why not use a PNP darlington TO220 package (as a high side driver) on a small heatsink. At 3A it will dissipate about 3W.

    2.5mA should be plenty to drive it on.
     
  3. marshallf3

    Thread Starter Well-Known Member

    Jul 26, 2010
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    That sounds better already. I've got two problem here.

    The controlling voltage is from an IC that operates from a regulated 5V
    2) I need an output in reverse of the controlling logic state capable of driving raw, unregulated 12V at 2A or 3A such that the light it's hooked to comes on when the input drops to logic level low.

    It would certainly be preferable for the output to be +12 so the other end of the bulb just needs to see common..

    I apologize for the stupid question I should know the answer for but I'm on some heavy medication that severely. impairs my thinking and I need to get the 200 PC boards ordered such that I can start building these around the end of the month.

    http://www.innoengr.com/CX500/fan_controller/FAULT_option_1.jpg

    See what I mean as stupidity? R1 would be putting 200 mA though it and Q1 most of the time. as that light would rarely come on.
     
    Last edited: May 16, 2011
  4. marshallf3

    Thread Starter Well-Known Member

    Jul 26, 2010
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    So you see my problem, I like the TIP126 but if I've got +12V going to t emitter than even fully on logic of 4.5V is going to keep it on.

    Put an 8V zener in series with the base?
     
  5. retched

    AAC Fanatic!

    Dec 5, 2009
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    Can I get some of that medication? :D
     
  6. THE_RB

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    I would use a PNP high side driver with a NPN low side stage connected to your logic level 0v/5v pin.

    That requires 5v to turn ON.

    If you need 0v to turn ON, you need another logic gate to invert the signal, or another logic level NPN to invert the logic signal.

    Of course if it is a microcontroller you can invert the pin in software. :)
     
  7. SgtWookie

    Expert

    Jul 17, 2007
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    Well, he's going to need an open-collector drive for the PNP high side, so simply inverting the logic in a uC itself wouldn't help much; he'll still need to get the base (or MOSFET gate) up to whatever the battery is putting out.

    I'd use a P-ch power MOSFET instead of a Darlington, as a Darlington will have a rather high saturation voltage (somewhere between 0.7v and 1.6v, depending on the Darlington and the load current). No sense in throwing away power in a Darlington when a MOSFET could be used.
     
  8. marshallf3

    Thread Starter Well-Known Member

    Jul 26, 2010
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    Well, here's one simple circuit I've come up with, does it appear to be valid?

    Remember I've got an open collector active low output from a 5V logic level device and I'm trying to drive a 12V high side switch from it.

    Input 5V (from pullup resistor) when not active, IND1 off

    Input 0.3V, 2.5 mA max current sink when active, IND1 on
     
    Last edited: Jun 10, 2011
  9. THE_RB

    AAC Fanatic!

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    What's the max VHIGH voltage rating on the open collector gate output?
     
  10. marshallf3

    Thread Starter Well-Known Member

    Jul 26, 2010
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    Unfortunately 5V thus the reason for the 10V Zener. I figure (simplified) that with +12V on the cathode and +5V on the anode the Zener isn't going to conduct thus not trip the Darlington pair on. I had to go with something that would guarantee me a minimum gain of 800 so I can switch 2A from 2.5 mA.

    I've got some wonderful high-side MOSFET switches but can't use them as this is a product that's going into production and I may need thousands of that part.
     
  11. shortbus

    AAC Fanatic!

    Sep 30, 2009
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    Probably a dumb idea(still learning this stuff), but why not use a low side mosfet to switch the load? A logic level mosfet could be used with the 5V output of the 'OC' gate.
     
  12. marshallf3

    Thread Starter Well-Known Member

    Jul 26, 2010
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    That setup wouldn't invert. The logic output is active low (as all OC would be considered to be) and I'd really like to keep it simple as in the indicator lamp can have one switched hot line and a plain chassis ground.

    BTW: The +5V is regulated, the +12V is not as it's the + end of a motorcycle electrical system thus it would vary from around +12V to +14.4V depending on if it's charging. Figure on around 13.4V nominal.

    No, this is not in any way part of any emissions control, the bike is vintage and has no electronic control at all. It is a very lean burn Honda engine though, probably meet today's California emissions tests as the 31 year old thing sits.
     
  13. marshallf3

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    It got real quiet, am I to assume that means what I finally came up with should work?
     
  14. nigelwright7557

    Senior Member

    May 10, 2008
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    Look at the tc4420 datasheet.
     
  15. marshallf3

    Thread Starter Well-Known Member

    Jul 26, 2010
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    That's an interesting IC but I see several potential problems with it.

    1) It was primarily designed to drive a lot of capacitive load such as a hugh handful of MOSFETs or CMOS ICs.

    2) The High-State Output Resistance vs Supply Voltage graph looks like it would have trouble driving an incandescent lamp at 12V / 2A.

    3) It appears to be an older design that's falling off of production, that could cause me problems down the road if demand for the product remains steadily popular.
     
  16. SgtWookie

    Expert

    Jul 17, 2007
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    Won't the open collector output go as high as your supply voltage?

    A P-ch MOSFET would get rid of the power dissipation problem in the Darlington.
     
  17. Wendy

    Moderator

    Mar 24, 2008
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    With an open collector output R1 is not needed. When the output of the comparator is low the transistors will turn on, when it is off R3 would turn the transistors off. R2 probably needs to be larger, but it is all that is needed, you don't need the zener. Transistors are current devices.

    [​IMG]
     
  18. SgtWookie

    Expert

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    What Bill said; 6.2k would be about right for R2. You can also use that same value for R3. 100k is really too high.
     
  19. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Here is another way, using a common base transistor as a level shifter.
     
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