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HELP ( BJT as switch )

  1. Dec 2, 2009 #1
    hello everyone ....I'm new here and I have so many questions need to be answered ! so I will be really really really thankful if you help :)

    what should be the out put voltage waveform shape on the Oscilloscope "Vce" , when the 1Kohm resistor inserted and when it's not ....
    the BJT as switch.doc (37.0 KB)

    I made this experiment but I'm not sure about what I came up with !
    can anyone explain to me what is the effect of the 15v DC & the function generator on the output signal & should I consider 1 Micro Farad capacitor to be an open circuit ??!!!!

    The square wave function generator gives input signal with Vpp = 2v & 1KHZ

    need your help as soon as possible
    thanx in advance :)
     

    Attached Files:

  2. jcsd
  3. Dec 2, 2009 #2

    vk6kro

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    Science Advisor

    can anyone explain to me what is the effect of the 15v DC
    This is a pretty sure signal that you have not yet studied transistors.
    The transistor can draw current from this 15 volt supply, or it can stop current flowing.
    Do you know what controls this current?

    The device on the output is an oscilloscope, which lets you watch the output as a waveform on a screen. In this case, it is a dual trace oscilloscope, so it lets you watch the input and the output at the same time.

    You can work out the REACTANCE of a capacitor at any frequency. In this case, you know the frequency and the capacitance, so you can estimate the effect such a capacitor would have at that frequency.
    Have a look here :
    http://en.wikipedia.org/wiki/Electrical_reactance
    d1093caebdb7b6d3f6adfd94d00d03ec.png
    Beware that the C is in Farads.
     
  4. Dec 3, 2009 #3
    well , unfortunately I studied transistors before but it's my first time for being in Electronic Lab & I realized that I don't have that much information , almost nothing !


    I'm a little bit confused , I know that DC supply is for biasing the transistor to work either in Active , cut-off , or Saturation mode....by locating the Q-point on the DC load line

    but is this the only aim for using it !? and what's it's effect on the output voltage ?!

    for example , I got these shapes for the output voltage ....and I'm not really sure about why they shaped on this way !

    can you help me with this please ?!
     

    Attached Files:

  5. Dec 3, 2009 #4

    vk6kro

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    Science Advisor

    I can't see anything in the second image except the room lights and a flat line.

    When they said to remove the 1 K resistor, maybe they meant to replace it with a piece of wire.

    It is best to regard the transistor as a device whose resistance can be changed. So, if you put another resistor in series with it from the power supply, the two will form a voltage divider.

    If the transistor is behaving like a low resistance, most of the supply voltage (your 15 volts) will appear across the series resistor. If it is behaving like a high resistance, most of the supply voltage will appear across the transistor.

    So, if the power supply wasn't there, none of this could happen. The transistor would always have zero volts across it.
    If you break the circuit, no current could flow, so none of this could happen.

    Voltage gain is output signal voltage / input signal voltage.

    For a given transistor, the voltage gain depends on the load resistance. The higher the load resistor, the higher the gain.
    So, if you have a pair of resistors in series as the load resistance, and measure the gain, the output will reduce if you put a capacitor across one of them.

    This effect will be different at different frequencies.
    At higher frequencies, the capacitor will have more effect. That is why you have to work out the reactance of the capacitor and compare it with the resistance of the resistor you are putting it across.
    Then you can estimate what effect this capacitor will have on the overall frequency response.
     
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