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linear amplifier with voltage-divider bias

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  1. #1
    mockingbird mockingbird is offline

    linear amplifier with voltage-divider bias

    My lab homework is to use Multisim to design a linear amplifier with voltage-divider bias. We are using an example in the textbook as a template. Values for Vcc and the first four resistors were given from an example we did in the previous chapter. It looks like it should be easy, but my Vc isn't right. Here is a screenshot:

    120416.jpg

    If I let it run long enough, Vc actually starts to approach 0. Whatever I did wrong is beyond my current level of comprehension. Will anyone please help me troubleshoot this?

  2. #2
    ronsimpson ronsimpson is offline
    Problem-1
    Your input signal is 10 volts peak, or 20 volts p-p.
    Your supply for the amplifier is 10 volts.
    The gain is about 2.
    10 volts in 20 volts out peak with only a 10 volts supply.

    Change the input signal to 1 volt pk and see what happens.
    1V in pk, 2V p-p in
    gain=2
    output = 2V pk or 4V p-p.

    Problem-2
    The DC model shows almost 0 volts across the transistor, C to E.

    You need to change the bias point.
    Cut R2 in half. (2200 approximately). You need voltage across C to E.
    0

  3. #3
    Roff Roff is online now
    Nodes 2 and 3 need to be connected together.
    0
    Ron


  4. Thread Starter #4
    mockingbird mockingbird is offline
    Thank you both for the responses. I have made (I think) the recommended alterations to my circuit, with unsuccessful results. How do I establish VCE? In my notes, I have VC=VCC-ICRC. Are you saying the transistor is operating in saturation?

    120419.jpg
    0

  5. #5
    Roff Roff is online now
    It still looks like you need to attach the voltage divider to the base. See attachment.
    Attached Images
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    Ron


  6. #6
    JimB JimB is offline
    Quote Originally Posted by mockingbird View Post
    Whatever I did wrong is beyond my current level of comprehension. Will anyone please help me troubleshoot this?
    At the risk of appearing a bit harsh, I am not surprised with that schematic.
    How is any one supposed to read that!

    OK, bear with me, you seem to be making mistakes which are common in beginners by having wires un-necessarly crossing all over the place, and components which do not line up, leading to dog-legs in the wires.
    All this contributes to a lack of clarity and understanding.

    Please look at my redraw of your simulator schematic, I hope that you find it easier to understand than your original.

    JimB
    Attached Images
    0
    Experience is directly proportional to the value of the equipment ruined.

  7. #7
    audioguru audioguru is offline
    It would be nice if you turn off the dots on the schematic that are all over the place.
    0
    Uncle $crooge

  8. Thread Starter #8
    mockingbird mockingbird is offline
    Not harsh at all JimB, for my level of experience. Your redraw is easier to understand. I find it easier to draw these by hand. If I was better with drawing in Multisim, I might have caught on to what Roff was saying earlier.

    I have to be at work in 7 hours and I am up way past my bedtime studying circuits...

    120421.jpg

    I have added yet another dot to the circuit, this one connecting the base to the voltage divider, as was kindly suggested earlier. I haven't checked the math, but the output looks like what I expected from the beginning.
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  9. #9
    JimB JimB is offline
    Still a few odd unnecessary kinks in the wires, but much better and more understandable. Keep it up.

    JimB
    0
    Experience is directly proportional to the value of the equipment ruined.

  10. #10
    Winterstone Winterstone is offline
    Quote Originally Posted by mockingbird View Post
    Thank you both for the responses. I have made (I think) the recommended alterations to my circuit, with unsuccessful results. How do I establish VCE? In my notes, I have VC=VCC-ICRC. Are you saying the transistor is operating in saturation?
    120419.jpg
    I am afraid you forgot the dc drop Ic*Re across the emitter resistor, didn't you?
    0

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