Help analyse the circuit with transistor

Discussion in 'Homework Help' started by anhnha, Jun 8, 2012.

  1. anhnha

    Thread Starter Active Member

    Apr 19, 2012
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    Hi all,
    Can anyone help me analyse this circuit.This is a circuit inside RS485 but I modified serval things.
    I assumed that the base of Q1 and Q4 is connected to 5V power supply and the base of Q2 and Q3
    is connected to ground.
    According to the book(Serial PortComplete, Second Edition by Jan Axelson),with the circuit above,
    "Q1 and Q4 to switch
    on and Q2 and Q3 to switch off. The voltage on line A causes Q6 to switch on.
    Current flows into Q6 and returns to the driver via the ground wire. In a similar way,
    the low voltage on line B causes Q7 to switch on, and current flows from Q7 into Q4,
    returning to the receiver via the ground wire.
    "
    I had try to explain this but I fail.In this circuit, I see that the volage between
    base and emitter of Q2 and Q4.With Q2 the voltage is 0V therefore it is switch off
    and the voltage between
    base and emitter of Q4 is 5V and therefore it is switch on.
    But why Q1 is switched on?
    I think that if Q2 is off then the voltage at emitter pole of Q1 is 5V thus the voltage between base and
    emitter of Q1 is 0V and it must should be off.
    And can you help me explain this statement:
    "Q1 and Q4 to switch
    on and Q2 and Q3 to switch off. The voltage on line A causes Q6 to switch on.
    "
    May be that I do not understand how transistor really work, therefore I feel that it is
    difficult fo me to analyse the circuit.
    Could you help me the steps to analyse or how to know what transistor is on or off in a circuit?
    Many thanks in advance.
     
  2. panic mode

    Senior Member

    Oct 10, 2011
    1,321
    304
    book is wrong, Q4 does not switch on, it explodes when you connect 5V to BE junction.
    to prevent that you need to add base resistor (for Q4). otherwise book explanation is correct.

    when Q1 and Q4 are on, voltage Vce across both of them is small (0.2V or so, we can consider this to be zero).

    when Q1 is on, it's Vce is low so line A has voltage 5V - Vce1 = 5V (approx). This means Q5 is shorted and hence off (it's base is at +5V because of base resistor, it's emitter is at +5V because of voltage at line A, difference between base and emitter is less than 0.7V so transistor Q5 turns off).

    Emitter of Q6 is again at line A potential (5V) and R6 pulls base of Q6 down so we get 0.7V across it's BE junction and Q6 is on.

    Opposite happens on line B. Q3 is off because Vbe is negative (base is on lower potential than emitter so being NPN transistor, it turns off). Q4 (assuming it has base resistor) turns on and because of low Vce when on, line B is near ground potential.

    This shorts Q8 which turns off. Q8 turns on. the explanation was oversimplified to point out what happens so it neglects couple of things (like 16.8k resistors)
     
    Last edited: Jun 9, 2012
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  3. anhnha

    Thread Starter Active Member

    Apr 19, 2012
    774
    48
    Thanks.
    Sorry for my poor English.
    I am sorry.It is my mistake,the book has base resistor.
    Could you give a detailed explaination about this?Assumming that Q1 is on, the line A has 5V, but why you can say for sure that difference between base and emitter is less than 0.7V so transistor Q5 turns off? I assume that the base current is very small and then in the circuit I will have VA=3.2V and VB=1.8V with the voltage drop in each diode is 0.7V.But I can not to figure out the votage between base and emitter of each tranistor.The 16.8K and two 960ohm resistors make me difficult to calculate it?
    Why you know that the result above is right when you neglect these resistors?
    And could you help me explain why Q1 is on? As for me
    I see that the voltage between base and emitter of Q1 is 0V and I think that it have to off.
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,963
    1,098
    The voltage between the bases of a Q5 and Q6 transistors is equal 3.2V - 1.8V = 1.4V.
    And if we assume Vbe = 0.6V then voltage drop across 960R resistor is equal 0.1V.
    So the voltage at point where 16.8K and two 960ohm are connect is equal to 2.5V.
    If R5 = R6.
    So when Q1 is on the voltage at point where 16.8K and two 960ohm are connect the voltage are going to increase slightly. So Q5 will be in cut-off.
     
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  5. anhnha

    Thread Starter Active Member

    Apr 19, 2012
    774
    48
    Thanks.
    As you said, I determined the voltage in base-emitter of Q5 when Q1 on.
    When Q1 is on, it's Vce is low so line A has voltage 5V - Vce1 = 5V (approx).
    The current through the 16.8K is (5-2.5) /16.8K = 0.1488 mA
    And the voltage in 960 ohm of Q6 will increse amount:
    0.1488 x 0.960=0.1428V
    Therefore the voltage between base of Q5 and the point where 16.8K and two 960ohm are connected will decrease 0.1428V and equivalent to 0.7 -0.1428 = 0.5572.So Q5 will be in cut-off.
    Am i understanding what you mean correctly?
    And could you please tell me about the ways to know weather a transistor is on or off?
    Is therre any special way to know state of it?
     
  6. panic mode

    Senior Member

    Oct 10, 2011
    1,321
    304
    you are to make assumptions unless when you have exact data for given transistor. for example some assumptions are:
    - all semiconductors are made from Si

    - hence all forward voltage drops are (about) 0.7V, except for Schottky diode (~0.3V)

    - Vce of transistor that is on is about 0.2V

    - therefore, when 5V is at base of Q1, A is at 5V-0.7V=4.3V; and when Q2 is on, voltage at A is about 0.3V+0.2V=0.5V;same goes for B line

    - R5 and R6 are same, and D5 and D6 are same so voltage between them is exactly 2.5V

    - Q5 and Q6 are matched so their emitter currents are same when there is no current through 16.8k and in that case voltage between 960 ohm resistors is at 2.5V and entire circuit on the right side is in balance (voltage at point between two 960 Ohm resistors is at 2.5V, just like point between D5 and D6).

    trick: if the voltage at any two points is the same, nothing will change if we short those two points, but this will usually offer some new insight or at least circuit simplification (a great thing when doing analisys).

    result is very common circuit known as constant current circuit or so called Widlar circuit. important thing to note is that emitter current will be very small because of fairly large value of 960 Ohm resistor (assumed 0.7V forward drop was good approximation but in reality there are some differences that depend on current).

    suppose collector of Q5 is not connected and that D5 and b-e of Q5 have exactly the same characteristic.
    also suppose emitter resistor is zero ohm. then Q5 and D5 are in parallel so current through Q5 and D5 will always be same, for any value of R5.

    now let's put back that resistor of 960 ohm. current through Q5 is much smaller because b-e junction of Q5 and D5 are no longer in parallel, that resistor seriously reduces current through Q5.

    now lets connect back the collector of Q5 to positive rail so circuit is again complete. even if the base current of Q5 is small, emitter current should be bigger because of transistor gain. but bigger current through 960 ohm means voltage drop across 960 Ohm is bigger, so Vbe must be reducing (because 0.7V across D5 is fixed and equal to 0.7V; this is same as sum of Vbe and drop across 960 ohm, if drop across resistor increases, then Vbe must be decreasing). this means that emitter current is tiny... but if the emitter current is tiny, what is the base current? it is even smaller because of transistor gain. in fact it is so small that transistor is near tripping point.

    even slight decrease in emitter current and transistor shuts off. slight increase in emitter current and transistor is on. this is exactly what happens when A voltage changes. when line A is high , current through 16.8k causes imbalance in Q5 and Q6 such that Q6 turns on and Q5 shuts off. When A is low, emitter current of Q5 increases (and Q5 is on), while Q6 turns off.
     
    Last edited: Jun 10, 2012
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