Current Limiting in Series Regulators

Discussion in 'Homework Help' started by SkiBum326, Jun 18, 2014.

  1. SkiBum326

    Thread Starter Member

    May 16, 2014
    Hey Guys,

    I have another question regarding material from Electronic Principles (page 960). I attached an image to better illustrate.

    In discussing current limiting with respect to a series regulator, the book states that the load current has to pass through R4. Is this correct, or just that most of the load current passes through R4?

    The book goes on to say that R5 is added to increase the output impedance of the op amp. Also, the value of R5 is selected to be high enough to produce voltage gain in the current-sensing transistor, but not so high that it prevents the op amp from driving the pass transistor.

    It makes sense to me that with any changes in current through Q1, a larger R5 will create a greater and therefore faster/more sensitive voltage response. However, I don’t understand why this would impact the op amp’s ability to drive the pass transistor.

    I really appreciate everyone’s help.

  2. SkiBum326

    Thread Starter Member

    May 16, 2014
    I was again unable to submit the proper title. Could an admin change it to "Current Limiting in Series Regulators." Again, thanks for the help.
  3. SkiBum326

    Thread Starter Member

    May 16, 2014
    Here's the attachment.
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
    All the load current is flowing through R4 resistor. Also notice that R1,R2 voltage divider current also flow through R4. So IR4 = I1 + IL.


    As for R5 try assume that Vin = 10V; Vz = 2.5V; R1 = R2 = 100K; RL = 50Ω and Vbe2 = 0.6V; β2 = 49.
    So the Vout = 5V and voltage at Q2 emitter is equal to
    Ve = 5.1V. The voltage at Q2 base is equal to 5.7V.
    So now if you use R5 = 1.15KΩ the voltage drop across R5 is equal to:
    VR5 = Ib2*R5 = IL/(β + 1) * R5 = 100mA/50 * 1KΩ = 2.3V.
    So this mens that Op amp output voltage is equal to 5.7V + 2.3V = 8V.
    But what if we increase load current from 100mA to 200mA. What op amp output voltage we need to get 5V at output?
    To get this 5V at 200mA we need Q2 base current to be equal to 200mA/50 = 4mA . What will be the voltage drop across R5?
    Well VR5 = 4mA * 1.15KΩ = 4.6V. So the op amp output voltage need to be equal to 5.8V + 4.6V = 10.4V But op amp cannot give such a high output voltage, because op amp is supply from Vin = 10V.
    So as you can see op amp can give 10V max at his output. So Vout will drop from 5V to 4.79V. Because the op amp’s is unable to drive the pass transistor to get 5V at the output.
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  5. daviddeakin

    Active Member

    Aug 6, 2009
    Because base current flows through R5, causing a voltage drop across it. The opamp output can only swing rail-to-rail (less in the real world), so if there is too much voltage drop across R5 due to base current, the opamp won't be able to swing its output high enough to drive the transistor over the full desired range.

    (In reality you might not bother to include R5, because most opamps have their own output current limit. You would simply let Q1 suck current directly out of the opamp, causing it to shut down during an output overload.)
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