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Power supply: university project

Discussion in 'Power Electronics' started by walkingbeard, Mar 24, 2010.

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  1. walkingbeard


    Mar 24, 2010

    I'm doing first year Electronics at Queen Mary University of London.

    As part of our basic circuits course, we have to design a power supply, based on a rectifier, smoothing cap and voltage regulator.

    We can ignore for the moment, the transformation from UK mains voltage (253V maximum).

    The rectification and smoothing aren't a problem for me. What I am confused about is the voltage regulator. We have been shown three regulators, two of which we are allowed to use in our project.

    I attach the circuits underneath.

    As things stand, I have gone for the second one, on the basis that which the first one, you have difficulty in picking components, due to feedback and the fact that a real Zener diode requires a certain current (probably more than I would want at the base of the transistor), in order to reach a near-constant voltage.

    I don't think I really know what I'm doing though. Would someone be able to explain to me why I would or wouldn't use these circuits, and how I would go about picking actual components, given that I have to pick values before I even start looking for the components themselves?

    I'd be most grateful.

  2. walkingbeard


    Mar 24, 2010

  3. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    Jan 21, 2010
    The questions you should ask yourself (for both circuits) are:

    0) What is the voltage across C?

    1) What zener voltage is required?

    (for (a) only)
    2i) What base current is required (approx)?
    2ii) How is this base current regulated?

    3) What current must pass through R?

    4) What voltage will be across R?

    5) What power will be dissipated in R?

    6) What power (in the worst case) will be dissipated in Z?

    Tell us what you think and we'll tell you if you've got it right.
    Last edited: Mar 24, 2010
  4. walkingbeard


    Mar 24, 2010
    Right, well the voltage across C is going to be alternating between 0V and 23V. The cap is going to smooth that out to almost-DC.

    The Zener voltage will be about 9.1V in either case.

    Now the stuff for circuit (a) is where I'm having the problem.

    I can see that the transistor's base current should be 1.1mA (for the maximum output current), given a transistor with a gain of 100, which my text book says is reasonable for a general purpose transistor.

    So then the resistor will have a maximum voltage of 25.3-9.1=16.2V dropped across it, giving a resistance of 14.73kΩ.

    That's all fine, because even if the output voltage of the rectifier drops to its minimum of 22.08V, I can still get enough leeway from the capacitor to achieve the Zener voltage across the output terminals.

    The problem I'm having is that some of the current going through the resistor is going to go through the Zener diode, but I don't understand how much.

    I saw a tutorial on this exact circuit, but the guy added the current needed on the base of the transistor, to the current needed to achieve the Zener voltage over the diode, and this was the total current through the resistor. I don't understand that, because how is the transistor going to select just the smaller current to amplify?

    Anyway, the power dissipated in the resistor will be 10E-3 x 16.2 = 16.2mW and in the diode, 10E-3 x 9.1 = 9.1mW.

    I still don't understand why I would or wouldn't use either type of regulator though.

    Hope I'm making sense.


  5. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    Jan 21, 2010
    Wrong -- 23V? are you sure?

    And it's either alternating or almost DC. Which is it. Assuming zero load, at what voltage?

    hint: What units are used for the measurement of the AC voltage?

    Wrong or very wrong. Why 9.1V in the simple case?

    Shouldn't the transistor require a different zener voltage? Why? Higher or lower? By how much?

    With a transistor turned on, what is the relationship between the voltage at the base of a transistor and the voltage at the emitter?

  6. neon


    Oct 21, 2006
    the emitter follower is ok and the zener circuit are both ok. The question that must be unswered is the load the transistor follower cannot ever give you 8.4 volts it will be 8.4 v plus Vbe or . 7 volts added to it 8.4*.7=9.3v . the second circuit the zener must be able to donduct both the regulating current plus the load current . the previous circuit the follower the zener current need only to conduct the the load current and zeners current divided by the beta of the transistor. since beta varies greatly with Ic the design [ to make sure should be beta 10. ] finaly the design calls for a zener of a least 1 watt the series limiting resistor looks like two watts assuming that the caps are not charging up to peak volts. what you need is 5w resistor s 2 watts zener and a TO3 case trnsistor to keep this thing from thermal ruaway. good luck
    Last edited: Mar 25, 2010
  7. walkingbeard


    Mar 24, 2010

    Okay, even by my own understanding I made some mistakes there. :)

    Here are what I think the voltage looks like before and after rectification (before the smoothing).


    The peak on both is at (beg pardon) 25.3 * sqrt(2) = 35.8V. The maximum of 253V on the UK mains is RMS.

    In the transistor circuit, the Zener voltage will need to be 8.4+0.7 = 9.1V.

    In the diode/resistor circuit, the Zener voltage will need to be simply 8.4V.

    The voltage across C is going to be 35.8V in either case.

    The base current into the transistor still needs to be 1.1mA. Why will I need to recalculate this? I don't get it. That would mean that the resistance of the loan resistor in that circuit will be (35.8-9.1)/1.1E-3 = 24.3kΩ. The current is regulated by that resistor.

    In the resistor/diode circuit, the current that needs to flow through the resistor is the specified maximum current of 110mA, giving a resistance of (35.8-8.4)/110E-3 = 249Ω.

    The power dissipated:

    P(R) = V(R)xI(b) = (35.8-9.1) x 1.1E-3 = 0.02937W
    P(Z) = V(Z)xI(b) = 9.1 x 1.1E-3 = 0.01001W
    P(T) = V(T)xI(e) = (35.8-8.4) x 110E-3 = 3.014W

    P(R) = (35.8-8.4) x 110E-3 = 3.014W
    P(Z) = 8.4 x 110E-3 = 0.924W

    So are you saying that there is less power dissipated in the regulator by the transistor circuit, so if I want to maximise power transfer, I need to use that one?
  8. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    Jan 21, 2010
    Firstly, it may be best to ignore neon. Whilst some of what he said was correct, some was badly wrong. And I'm not going to tell you what part of his information was wrong :)

    Where did 25.3 come from? The mains voltage in the UK is not of any concern.

    No, but close enough for the moment.

    Look at the initial assumption you used to determine what the gain of the transistor would be. Later on, you may find that such a transistor will not be suitable and you'll have to recalculate for a different type of transistor. I don't get it.

    No, the current is not regulated by that resistor. Consider what would happen if you used a smaller value resistor.

    Close, but no cigar :) Will the voltage across the capacitor stay constant under load? what about when the mains voltage changes?

    close... but what current do you want to flow through the zener? remember that the voltage across the zener depends to some extent on the current flowing through it. Should you not allow more current to flow so that the zener voltage is stabilised?

    No, there's more to it than that. Consider the range of currents through the zener. What happens to the voltage across the zener when this happens? What types of regulators are these? Consider the power dissipated under a range of loads from zero to 110mA. Consider the forward voltage drop in the bridge rectifier.
    Last edited: Mar 26, 2010
  9. Laplace


    Apr 4, 2010
    One thing that has not been discussed, yet should probably be the first that is considered, is the ripple voltage on the capacitor. This is mostly dependent on the AC frequency, the capacitance value, and the DC current drain. Before designing any electronic regulation, just model the load as a constant current sink to find the ripple voltage. Another complication in high current applications is that the ripple voltage equates to a ripple current through the capacitor's equivalent series resistance which heats the capacitor, sometimes magnificently. That shouldn't be a problem here. Just design the regulation to accommodate the ripple voltage.
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