General transistor question

Discussion in 'General Electronics Chat' started by mbohuntr, Mar 31, 2010.

  1. mbohuntr

    Thread Starter Active Member

    Apr 6, 2009
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    I have been expierimenting with several transistor circuits, and have a question. If you bias the gate with say 5v, Vdd is 12v, source is 12v. On all three types of transistors (bjt,mosfet, and fet) is it reasonable to expect the drain voltage to be Vgate - Vthreshold? I was using a 339 comparator circuit to ground the gate of a PNP 3906, and the drain of the 3906 to bias the gate of a TIP 3055 to switch on a 12v fan. Poor circuit design led me to get 6v at the gate of the 3055, and the fan was only getting 5v. In order to properly drive the 3055, will I need to provide Vdd - Vthrshold at the gate?
     
  2. SgtWookie

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    Jul 17, 2007
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    BJT's (bipolar junction transistors) have an emitter, a base, and a collector.
    FETs have a drain, a gate and a source.

    BJT's are current controlled devices; current through the base controls a larger current through the collector.

    FETs are voltage controlled devices; Vgs controls the state of the FET; on, off, or partially conducting.

    FETs and BJTs are quite different devices, but very rough correlations are:
    source = emitter
    gate = base
    drain = collector

    Please post your schematic and explain what you are seeing.
     
  3. mbohuntr

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    Apr 6, 2009
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    Is there an easier way than multisim??
     
  4. SgtWookie

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    To draw a schematic? Or to simulate a circuit?

    I don't have Multisim. I have LTSpice installed. It's a very good (and free) PSPICE schematic capture/simulation software tool.

    But whatever schematic capture software you are using, just bring up the schematic, do a screen print by pressing Ctrl+PrintScreen, then start MSPaint, and paste the image into it.
    Save the image as type .png, and attach/upload the image to the thread using the Go Advanced and Manage Attachments buttons.
     
  5. mbohuntr

    Thread Starter Active Member

    Apr 6, 2009
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    Here is another attempt since I have dis-assembled the origional circuit. I used a PNP only because I had a 339 comparator handy. I guess my question is this... Why cant I get approx 11v at the motor? (It only needs 800mA, the TIP3055 is rated higher. R3 is to limit the PNP current as well as protect the 339 as they both are limited to 40mA. R2 is just to measure the base of the PNP. I think I am not using my resistors correctly. I am learning to set-up and bias the transistors correctly, before I advance further. The circuit worked ok, except that the NPN was not turning off fully, and the fan was still pulling current. I then changed the NPN to a reed relay to get more power to the fan. any help finding my mistake(s) or better design would be great. Thanks.
     
  6. SgtWookie

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    Jul 17, 2007
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    You are using the NPN transistor as a voltage follower instead of a saturated switch.

    Your resistor values are way off.

    Try it like this instead:

    [​IMG]

    R1 limits the maximum current through the base of Q1 and the 339. It's really more current than the 339 is rated for, but you might get away with it.

    R2 ensures that Q1 turns off when the 339 is not sinking current.

    R3 limits the maximum current through Q1 and the base of Q2 to about 100mA. This is overkill, as with an 800mA load, Q2 really needs just 80mA base current. 140 Ohms would be just about right. R3 should be rated for 2 Watts.

    R4 ensures that Q2 turns off when Q1 is not sourcing current.

    D1 is important, as when Q2 turns off, the motor will produce a reverse-EMF current. Without D1, Q2 may be destroyed due to this reverse-EMF.
     
  7. mbohuntr

    Thread Starter Active Member

    Apr 6, 2009
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    Thanks, When I get some more time, I'll ask how you are determining the currents, I was just limiting the total at 12V. Thanks again!!
     
  8. SgtWookie

    Expert

    Jul 17, 2007
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    When you are using a transistor as a saturated switch, the basic rule of thumb is that the base current needs to be 1/10 of the desired collector current.

    A generalize formula that you can use is:
    Rbase = (Vsupply - Vbe)/(Ic/10)
    where:
    Rbase is the resistance in Ohms needed to limit current flowing through the transistor's base.
    Vsupply is the voltage available to source current to the base.
    Vbe is the voltage difference from the base to the emitter when the transistor is saturated. For most purposes, you can use 0.7v to 0.8v, unless you are operating the transistor near it's design limits. In the latter case, you are better off to change to a more capable transistor.
    Ic is the desired collector current.
    In your case, you have a 12v supply, and you want 800mA collector current.
    So:
    Rbase = (Vsupply - Vbe)/(Ic/10)
    Rbase = (12v - 0.9V)/(0.8a/10)
    Rbase = 11.1/0.08
    Rbase = 138.75 Ohms
    140 Ohms is the closest standard value.

    For Q1, you need 80mA collector current to drive Q2.
    So that means you need around 8mA base current. LM339's have a pretty wimpy open-collector output stage; their Vce gets pretty high when you're trying to sink more than around 6mA, but let's try it anyway.
    Rbase = (Vsupply - Vbe)/(Ic/10)
    Rbase = (12v - 0.8v)/(.08a/10)
    Rbase = 11.2/0.008
    Rbase = 1.4k Ohms.
    That's pushing it a bit much for the LM339, so I bumped it up to 1.5k.
    11.2v/1.5k = 7.46mA, which should be enough to saturate the 2N3906. It's not that far off. Had it been 1/20 the desired collector current, you might run into difficulties.
     
    Last edited: Apr 1, 2010
  9. mbohuntr

    Thread Starter Active Member

    Apr 6, 2009
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    Thanks! Mabye i'll get a better handle on it now.
     
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