Floating.pull-up resistor/Open Collector

Discussion in 'General Electronics Chat' started by Scottishness, Jan 24, 2013.

  1. Scottishness

    Thread Starter New Member

    Jan 24, 2013
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    Hi guys, never really posted on an Electronics site before but wondered if you could help me here. I've been given a project for college (scottish college not american college), Our project is to drive a switch reluctance motor from a 5 volt, 12 volt, 1 bit ring counter (100, 010, 001), ive peiced together a little MSP of what its ment to look like but theres a few gaps im not to sure about.

    [​IMG]


    my question is Mainly surrounding What is happening in this particular part of the circuit im confused with, it involves ''floating'' - which im not sure about at all. also ''pull up resistors'' im not really comfortable with them in this either exactly.


    Before i found this part out i had no idea i needed the 12v supply with a pull up resistor and earth there, i assumed that current would just flow into the gate opening it allowing current to flow through the coil. an opto coupler being a 3 prong device i belive?.

    So basically why wouldn't it work without the Pull up resistor and floating earth?, Whats the pull-up resistor doing exactly?. Ive tried looking some of this stuff up but its hard to pin down the answers to my questions.


    I will have missed out some details to the circut maybe. ill fill in the blanks best i can if it will help you answer, Please and thank yous :).
     
  2. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    What is your interpretation of "floating"? In electronics, floating means, not connected to anything else.
    Ground is the other terminal of a power supply, unless that terminal is explicitly shown on a schematic. In this case, ground is the return (negative terminal) of the 12V supply.
    The 5V supply may have a separate "ground", since the optoisolator does not require a common ground.
    In your circuit, when current is flowing through the LED in the optoisolator, the phototransistor will be turned on. This will turn off the IGBT. When no current is flowing through the LED, the internal phototransistor will be off, and the resistor to +12V will provide 12V of gate drive to the IGBT, turning it on, which will in turn energize the motor winding.
     
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  3. WBahn

    Moderator

    Mar 31, 2012
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    Your mixing up some of the terminology a bit, and that might be a part of your problem. Maybe.

    The first thing to note is that "ground" does not mean "the dirt outside the house". In this case, the 12V supply that is powering the right hand side of the circuit is connected between the node labeled Vcc and the node with the "ground" symbol. So for the coil to work, current has to have a path from the Vcc symbol to the ground symbol.

    I'm trying to think of a useful analogy for a floating input and the role of a pull-up. Imagine this: You have a trap door in the ceiling that is hanging vertically from some hinges so that it can go left or right. If it goes right, then a light turns on. If it goes left, then the light stays off. Now you connect a rope to the bottom edge of the door and hook it up to some mechanism that is off to the right that can pull on the rope in order to turn the light on. But since you can't push on a rope, that mechanism has no way to turn the light off because when it relaxes the rope the door goes to this "floating" condition in which it is neither on nor off. You could add a second mechanism with a second rope that pull it to the left and then link the mechanisms in such a way that one of them is always pulling on its rope when the other is not. In electronics, such an arrangement on the output of a circuit is often called a "push-pull" output or a "fully active" output. But other outputs, such as the optocoupler in your circuit, can only drive the output one direction. These are often called "open collector" outputs (for reasons having to do with the prototypical implementation of such and output) even when it isn't technically accurate for that circuit. Going back to the trap door analogy, if you can't build the second mechanism to the left, you might attach a spring from the door to the framework that's off to the left and tension it so that when the mechanism on the right isn't pulling its rope, the spring can pull the door far enough to the left to shut off the light. When the mechanism to the right wants to turn on the light, it simply pulls its rope like always. It will have to pull a bit harder than before, but unless the spring is tensioned too much, it will be able to "overdrive" the spring and pull the door to the right.

    The spring is acting like the pull-up resistor (or a pull-left resistor). When the optocoupler output is turned off, the node on its collector (the node with the pullup resistor) is pulled up toward Vcc by the pullup resistor. This then provides a path from the Vcc, through the resistor, to the IGBT control node. When the optocoupler output is turned on, this node is pulled down to a low voltage by the optocoupler.

    Hope that helps somewhat.
     
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  4. SPQR

    Member

    Nov 4, 2011
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    Analogies such as this are always helpful.:)
    ------------------------------------------------
    From one newbie to another, the concept of a pullup resistor is important and very common.
    Wiki has a nice piece on it too.
    But there is also something called a "pulldown" resistor that does the opposite.

    The general concept is if you want a particular part of a circuit to be at a particular voltage level - HI (Vcc) or LOW (Vdd) - calculate the value of the resistor based on Ohm's Law, and connect it to either Vcc or Vdd. That point in your circuit will be at the voltage you want, until acted upon by another part of the circuit.

    Pullups and pulldowns can also be used more flexibly - I did it last night.
    I wanted to press a button and turn ON an LED (with an IC in between).
    All I had was an inverting IC so in its "natural state" the LED would be ON and when I pressed a button, the LED would go off - not what I wanted.

    I connected one end of the switch to a pullup resistor, and to the input pin of the IC.
    The other end of the switch was connected to ground.

    The pullup resistor kept input to the IC HIGH (inverting IC - input HIGH, ouput LOW) and the LED OFF.
    But when you hit the switch, it pulls the pin to ground, and the LED comes on (inverting IC - input LOW, output HIGH).

    They are very important, very handy, and worth learning about.
    And I thank all the experts here for teaching me about them.
     
  5. shortbus

    AAC Fanatic!

    Sep 30, 2009
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    While not related to your question, it is related to the SRM. Driving a SRM by a logic circuit, your "one bit ring counter", will probably make it rotate, it won't make it rotate a "load". Without rotor position sensors, any resistance from the load will make the stator "lose" its place with reference to the stator poles.

    Srm's have to be tied to rotor position feed-back. Either by hall effect sensors, optical sensors or the newest way by sensor-less means. The sensor-less is the hardest, it means injecting an AC voltage into the stator coils and measuring the reluctance to choose which phase to turn on.

    Aberdeen University is one of the leading schools in the SRM technology advancement. Could I ask what stator/rotor configuration and phase number you are using in the motor? I've been working toward making a 12/8 three phase motor myself is why I ask. Good luck with your project.
     
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