How to calculate Base for a PNP Darlington

Discussion in 'Homework Help' started by Geoffr67, Dec 6, 2011.

  1. Geoffr67

    Thread Starter Member

    Dec 5, 2011
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    I am new to transistors and have been reading what i can about them i just do not understand it. Could someone please explain this to me like a dummy.

    For example if i have a coil that is operated at 5 volts and has a coil resistance of 3Ω, Being triggerd by the pin of a pic chip, how do i calculate the base resistance to fully saturate the transistor when the coil requires 1.6 amps?

    Thank you so much!
    Geoff
     
  2. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    Using a transistor as a switch, usually at least 1/10th the Collector current is used for base drive.

    In your case, the base would need 160mA, which means you'll need another transistor between the PIC and the transistor in order to provide enough current (uC Pins can only sink/source 25mA)

    --ETA: Please post a schematic, It sounds like you are charging the coil through the transistor, remember to put a diode across the Collector and Emitter of the diode for a path of a reverse voltage spike to go through to prevent damage to transistor.
     
  3. Geoffr67

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    Dec 5, 2011
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    i understand about the flyback diode to prevent back emf from damaging the transistor, but how would i go about switching this thing? are there better alternatives it has to be a hiside switch. As the coil is naturally grounded.
     
  4. thatoneguy

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    Feb 19, 2009
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    A MOSFET would also work, but it would need a high side driver as well.

    What are you trying to accomplish? There may be a simpler solution to your problem than finding a solution for your solution.

    If you post a schematic of the overall "what needs to happen", that's language most here will understand, and you will get a lot more help. Right now, it's rather vague, a highside driver for a coil driven by logic level, but we don't know the high side voltage, only the current, etc.
     
  5. Geoffr67

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    Dec 5, 2011
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    ok when i get to work tomorrow i will make a schemetic, but let me Try and elaborate. The coil is a 5volt coil, is 3 ohms. I calculate 1.67 amps current draw MAX. My pic outputs 0 or 5 volts. my supply rail is going to be 5 volts with up to 3 amps total current. (LM323T) the coil is grounded thru the body. so i need to switch the + to 5 volts which will activate the coil. i hope ive filled in some blanks but i will make a schematic tomorrow. Thank you so much for you help so far!
    Geoff
     
  6. thatoneguy

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    Feb 19, 2009
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    With the darlington, you may be able to get by with 20mA drive, though after a base resistor was added to keep the PIC from overload death, I'm not sure that can be accomplished.

    Have you tried it with direct drive from the pin? You should have a 220Ω or greater resistor on the PIC pin to the base drive to prevent over 20mA going out of the pin.

    What is the part number of your PNP Darlington? You do realize you will lose 1.2V-2V across the transistor, even when it is on, correct?

    I'd suggest an N-Channel MOSFET with a high side driver for this.
     
    Last edited: Dec 6, 2011
  7. thatoneguy

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    Feb 19, 2009
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    ITS410E2 High Side MOSFET may do the trick. It's called a "PROFET", it is a MOSFET w/High Side driver built in, and needs at least 4.9V to run, so your 5V supply needs to be stable. It will take a logic input and switch it with only a few millivolts lost at 1.6A, far better response than you will ever get with a BJT or Darlington.

    Go through the datasheet to ensure it meets all the specifications you are looking for.
     
  8. crutschow

    Expert

    Mar 14, 2008
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    Another solution is a NPN driving a power PNP.

    The base of the PNP has the emitter to +5V and the collector to the coil. The base is connected through a resistor to the collector of the NPN. The resistor value would be (5V-0.7V) / (1.7A/10) = 25 ohms.

    The NPN's emitter is connected to ground and the base to the PIC output through another resistor. The resistor value is (5-0.7V) / (.17A/10) = 250 ohms.

    The circuit signal polarity is non-inverted, thus a high on the PIC output provides current to the coil.
     
  9. Adjuster

    Well-Known Member

    Dec 26, 2010
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    An NMOS plus high-side driver may perhaps be a little advanced for this homework application, particularly if this is a school project limited to parts available in class. A PMOS like the one suggested may be easier, again if OP can get one, although a NMOS could have less RDS(ON).

    If a Darlington is all that can be obtained, it should be rated for at least several times the 1.6A requirement: the saturated voltage should then typically be not much over 1V, though it is true that several volts is common as a maximum at full ratings. For a 5V application this is a serious disadvantage.

    The saturated current gain of such a device may however not necessarily be as low as 80, so 20mA drive may be more than adequate: the datasheet for this Fairchild TIP105 device specifies Vce(sat) at Ic = 500Ib.

    http://www.fairchildsemi.com/ds/TI/TIP105.pdf

    In the end it depends what parts you can get hold of. If all else fails, a straight PNP power transistor could be driven by a smaller intermediate PNP transistor, with its collector returned to the supply via a resistor set to give about 160mA drive. Not elegant, but it should work. Edit: Warning, this gives a different sense of operation to that obtained with an NPN pre-driver as in the previous post.
     
  10. Geoffr67

    Thread Starter Member

    Dec 5, 2011
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    Ok, I like this solution. Am i correct to assume this would be all i need to actuate the solenoid coil? Run one of the pins of my Pic chip directly to Pin 2 (Input, activates the power switch in case of logical high signal) Connect pin 1 to the commen ground and pin 3 to the positive 5V rail, Pin 5 directly to the Load. What should i do with pin 4? (st) diagnostic feedback?

    I can obtain a few of these and would like to if this would reduce my parts count and provide me with good realiability. It looks more simple to set up as well. As this is going to be driving a inductive load should i be putting Flyback diodes on this as well similar to a transistor?

    Geoff
     
  11. thatoneguy

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    Yes, a diode would go between ground and V+.

    The device family is called a "PROFET", which is a TO-220 package with a high side driver built in, basically. If you search for that term, you should be able to find them from a local distributor. I know DigiKey carries them.

    The extra pin is a fault indicator, which you can use or ignore.
     
    Last edited: Dec 6, 2011
  12. Geoffr67

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    Dec 5, 2011
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    okay so in my previous post i am correct in my pin assignments then, just to clairify? And when you say i should install the flyback diode, that would go between the Load output and ground correct? The extra pin for the fault indicator should i leave this NC or tie it to +5v or gnd?

    If i were to drive this with a PWM at say 35hz or so will this still work well for me?

    Sorry for all the newbie questions.
    Geoff
     
  13. thatoneguy

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    Take a look at the datasheet I linked above, it shows several circuit diagrams. The flyback diode goes from ground to V+, rather than load to V+ on the one I looked at, but it could have been for something else.

    The datasheet has several schematics and parameters that need to be met by outside devices (not many, but some). Make sure you give it a look through to be positive it will do what you need it to.
     
  14. Geoffr67

    Thread Starter Member

    Dec 5, 2011
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    It looks like it will fit my needs, my problem is implementing it. have you ever worked with these before thatoneguy? I am self teaching ATM and from time to time need my hand held :p. Forgive my ignorance.
    Geoff
     
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