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The use of diode in this circuit...

  1. Sep 12, 2015 #1
    please help me guys

    My inference to above circuit.....
    The negative cycle of the output signal from the driver is entirely clipped off before getting to switching transistor..this however don't have any effect on output because the transistor won't conduct for negative cycles but when due to any circumstance the reverse bias voltage on the base of the transistor increases the excess current would flow from emitter to base and then to ground via the diode thus protecting the transistor...am i right..?

    Does this circuit always sink current...?
    What would be the output when the diode is not present..?
    In order to make this circuit source current do i require a negative supply..?

    Attached Files:

  2. jcsd
  3. Sep 12, 2015 #2


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    You are right.

    The purpose of the diode is to protect the emitter-base junction from "reverse overvoltage".

    Absolute maximum VEB = 7V


    But the current will go to VEE ( negative power-supply, not shown ).
  4. Sep 12, 2015 #3
    For this circuit the voltage at base cannot be lower than -0.7V because the diode do not allow to this to happen.
    No wrong. For negative driver voltage larger than -0.7V the diode will start to conduct a current. And this current will flow from GND through diode--->1kΩ resistor---> driver output--->Vee---->GND.
    For the positive voltage BJT is in saturation.
    But for the negative voltage larger than 5V the base-emitter junction is in reverse biased and current will flow through it, but no current will flow through collector.

    Which part of a circuit?
    Last edited: Sep 12, 2015
  5. Sep 12, 2015 #4


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    As seen from the load: Yes.
    I assume the output is a squarewave. If you remove the diode, you will have to supply the op-amp by VCC and ground, thereby making the upper corners in the squarewave more soft: instead of .

    It requires something like a complementary output stage.
  6. Sep 12, 2015 #5
    Are you sure about this ? How can a current flow from base to ground via forward biased diode for negative voltage at the driver?
  7. Sep 12, 2015 #6
    But the base emitter junction could be reversed bias for any bias voltage larger than -0.7V..so why you said that it should be more than -5V in order to reverse bias it..
    Are you taking about maximum reverse breakdown voltage....provided by the driver ( in extreme cases) ??
  8. Sep 12, 2015 #7
    Also as you said that the voltage at the base cannot be greater that -0.7 V ....then how can the voltage be -5V..??
  9. Sep 12, 2015 #8
    But i don't think so that the emitter-base junction would go into reverse overvoltage ..since the diode wont let it do ..as Jony 130 said..
  10. Sep 12, 2015 #9


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    Say that the diode is removed and that the opamp is supplied by ±12V ( Vcc/Vee ), then the opamp could pull down the base of the transistor to say -11V while the emitter voltage is 0V ( ground ).

    It would be quite normal to supply the circuit by these voltages.

    That's why I think your assumption is correct in #1.
  11. Sep 13, 2015 #10
    First you ask :

    What would be the output when the diode is not present..?

    And my answer is :
    For the positive voltage at the driver output BJT is in saturation, so base voltage is 0.7V and Vce is = 0.2V.
    For the negative voltage at the driver output, voltage larger than -5V can breakdown transistor base-emitter junction, so the base-emitter is in reverse biased and if driver voltage is larger than -5V the b-e junction breaksdown and current will start flow through it, but no current will flow through collector. So Vce = Vcc.
  12. Sep 13, 2015 #11
    Actually i am sorry for the typo in post 7...instead it should be "cannot be less than"...after that now my confusion is solved..(self perpetuating vortex)...since the diode strictly hold that voltage to -0.7V...
  13. Sep 13, 2015 #12
    So when the diode is present all of the current which flows across the b-e junction(earlier in diode's absence) now flows through diode as the reverse bias ain't increases to larger levels (-5V)..hence the junction gets protected from breakdown ....am i right?
  14. Sep 14, 2015 #13
    Simple please notice that in this circuit we have two diodes connected in parallel but with their polarities reversed (anti-parallel connection). And as you should know from the basics that if we have two component connected in parallel with each other there will be the same voltage drop across them.
    So when one of a diode is in forward biased the other diode will see the "forward" voltage of the conducting diode.
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