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How a current buffer transistor cancels the early effect (Cascode)

  1. Apr 17, 2015 #1
    In the given circuit Q2 is a current buffer tranistor which maintains constant current voltage for Q1 just to cancel the harms of early effect...how does it do so??

    Attached Files:

  2. jcsd
  3. Apr 17, 2015 #2
    Simply Q2 keeps Vc_Q1 fairly constant at 1.75V. And since Vce_Q1 is not changing, no early effect (base width modulation) occurs.
    Without Q2 load variations causes Vce_Q1 to change with load. For example for RL=1kΩ we have Vc = 9V and Vce = 8V; But for RL = 5kΩ we have Vc = 5V and Vce = 4V.
    And this change in Vce will also change collector current due to transistor β changes (early effect).
    So by adding Q1 we make Vce_Q1 fairly constant. And therefore Ic_Q1 ≈ I_load is also constant. Vce_Q2 will vary with the load but this variations will have almost no effect on load current. Because Q2 work here as a current buffer (common base), and this is why changes in Vce_Q2 due to load variations will have no effect on load current.
    I_load = IcQ1* β_Q2/(β_Q2 + 1)
    I_load = 1mA * 50/51 = 0.9803mA
    I_koad = 1mA * 100/101 = 0.9900mA
  4. Apr 18, 2015 #3
    If i for any reason want to make Vce_2 constant then do i have to add another transistor which would isolate Q2 from load and acting as a buffer for Q2..?
  5. Apr 19, 2015 #4
    Yes, you need to add another transistor on top off Q2.
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