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Why 2 power transistors in my power supply ?

Discussion in 'Electronic Basics' started by Abbie, Nov 29, 2003.

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  1. Abbie

    Abbie Guest

    Hello everyone,
    I have a variable current/voltage power supply 15V/3A.
    At the back of the unit there is a large heatsink with to
    2N3055. I have looked at the datasheet of these and
    found that they can conduct up to 15A, hence I don't
    understand for what reason there are 2 of them ? Another
    question that I have is that I often see in power supplies
    designs that the power transistors have a resistor from
    e to b such that they operate only when current goes above
    ,lets say, 1A. Is there any particular reason why power
    transistors need to be controlled this way ?
  2. Chris Welsh

    Chris Welsh Guest

    If it is a Switching supply, there could be 4 or more power transistors,
    depending on the configuration....

  3. They are capable of handling the heat produced by the passage of 15
    amps only if they are saturated on (used as saturated switches). If
    they are passing current and dropping large voltages at the same time,
    the current has to be reduced to keep the heat within capability.
    This is called the safe operating area (area being the product of
    volts and amps, instead of length and width). Look at figure 2
    (Active Region Safe Operating Area) on the data sheet. The DC line is
    the one that applies to linear power supply regulators, and then, only
    with an adequate heat sink to keep the case temperature below 25 C.
    So a hot heat sink allows less than shown here. Then you have to
    derate the power by the curve shown in figure 1.
  4. Abbie

    Abbie Guest

    I can tell you that when this unit works at 3amp the heat sink is much
    higher than 25 C, in fact it is so hot that I cannot touch it with my hand
    for more than one second.
  5. It was probably a simple engineering trade off to use two transistors
    in parallel (with each pretty severely derated) but a smaller heat
    sink (for a higher peak temperature and worse derating) as a total
    cost compromise. The transistors are cheap compared to a triple sized
    heat sink and a bigger overall package. This is especially true when
    you consider the cost of shipping thousands of units.
  6. Bill Bowden

    Bill Bowden Guest

    They probably use 2 transistors to spread the heat more
    evenly on the heat sink. A fixed supply could use one
    2N3055 at 3 amps since most of the power is going to the
    load, but a variable supply will produce much more heat
    when the output is set low. When the output is low compared
    to the input, most of the input power will end up as heat
    on the transistor, so they use 2 of them to run cooler.
    Another problem is the gain of the 2N3055 drops off at higher
    current, so they probably have more gain at 1.5 amps than 3 amps.

  7. Abbie

    Abbie Guest

    I remember having a discussion with you regarding my power supply
    with the circuit that I had, did you manage to see how it works? I
    understand it now, it took me a long time, but I had component
    values and node voltages to make life easier. I drew a new diagram
    which simplifies the circuit by eliminating all components which are
    related to current limiting. These components can be ignored when
    we analyse how the voltage is regulted and can be studied later. So
    here is the diagram, see if you understand it, if not I will explain, it
    is very easy.
  8. You have made considerable progress in documenting this circuit,
    correctly. It makes much more sense than the original version, and
    corrects a couple errors on the second. Rather than going through an
    explanation, I would like to hear your take, so I have a data point on
    your level of understanding. That way, if there is something I
    disagree with, I will be in a better position to communicate with you,
    effectively. There is still a compensation network missing around the
    opamp, isn't there?

    Have you made a similar, simplified schematic, that documents the
    current limit system?
  9. Abbie

    Abbie Guest

    The fact that the minus input of the op-amp is
    connected to the ground confused me at first, but then I realised that since
    ground is also psu red o.p. means that there is a negative feedback from
    the o.p. of the op-amp, through all the transistors, and back to the minus
    i.p. of the op-amp. Therefore from op-amp basic principles both minus
    and plus inputs are equal, and are equal to gnd, i.e. 0V.

    This means that the voltage at VDIAL (top) is 0V. The voltage at Zener 431
    is 2.5V. Hence the voltage across the 910 ohms resistor is always 2.5V
    (assume the variable resistor below the 910 one is not there).
    Now, constant voltage, so we can know the current through the 910 resistor.
    This current must go through VDIAL, so the voltage of the negative rail will
    be VDIAL times this current. That's it. By varying VDIAL we vary our
    negative rail.

    This circuit was difficult for me to understand because all the circuits I have
    seen up to now regulate the positive rail, whereas here we are regulating
    the neative rail. It's the same thing, but it is like looking at something upside
    I don't think so, but tell me what you mean or why.
    Not yet, but I will.

    Please tell me what you think. I need to know if what I think is correct.
  10. That is one of the basic simplifying assumptions of linear opamp
    Right. The voltage divider between the negative output and the +2.5
    volt reference must always produce 0 volts. You adjust the negative
    output with respect to this internal zero volts by changing one of the
    divider resistors.
    Yes. you have to be flexible. :)
    An opamp has very high gain, and if the feedback loop has enough gain
    at the frequency where the net phase shift turns negative feedback
    into positive feedback is higher than 1, the loop will oscillate.

    The capacitor across VDIAL provides phase lead (increasing gain with
    increasing frequency), so it cannot be rolling the gain off at high
    frequencies to prevent oscillations. The opamp has an internal
    capacitor that rolls of the gain proportional to frequency that gets
    the gain below one at a lower frequency than the one that has 180
    degrees of phase shift internal to the opamp, but it does not account
    for the extra gain and phase shift of the three transistor amplifier
    chain hooked onto the output of the opamp. I would have expected some
    high frequency roll off or other phase compensation network somewhere
    in the loop to reduce the high frequency loop gain.

    By the way, this chain has a lot more phase shift than it could. The
    turn on bias for the transistor chain is provided by the 10k pull up
    resistor above the zener diode, and this is stolen by either the
    voltage regulating opamp or the current regulation opamp (whichever
    pulls down more) to turn off the transistors.

    The first current gain stage lowers this signal source impedance from
    10k to 1k, which is about right for a single stage of emitter
    follower, but the second emitter follower should drive a base to
    emitter resistance (across the pair of 2N3055) of something more like
    100 ohms, not 1k, to speed up the turn off of those big transistors
    when the upstream transistors turn off.
  11. Terri

    Terri Guest

    Is anyone still reading this thread ? Calling John Popelish ?
    I have some questions regarding your reply, see below

    Very good, thank you !!
    Which frequecy are you thinking about, 50 Hz?
    This is an electrolitic, it is there, I presume, to stabilize the voltage at that
    point, and for the same reason there is an electrolitic over the zener.
    Why will there be a phase shift to the chain of transistors, is it stray
    capacitance, or is it the time it takes the transistors to react to changes ?

    There are two capacitors, one is 10nF from pin 2 to pin 6 of the 741, i.e.
    from the minus input to the output, and another, 100pF, from pin 5 to
    pin 6. Pin 5 is one of two offset-null pins.
    There is a resistor from output plus to output minus of 2.2k so we have
    current also with nothing connected to the supply. What is R7 for, is it
    thermal compensation. Which component doesn't look right, is it the
    way in which the transistors are connected or the values of the e-b
    resistors in the 2nd and 3rd stages ?
  12. Probably low kilohertz or high hundred hertz. Especially with that
    high value of base to emitter resistor on the 2N3055s.
    If it is an electrolytic, you are probably right. Too big ot be part
    of the feedback stabilization.
    It is mostly the internal capacitance that limits how fast any
    transistor can react to an input signal. There is an inherent limit
    based on the charge carrier mobility, but this is a minor factor
    compared to the junction capacitances in a circuit like this.
    The one from pin 2 to 6 is the ordinary way to roll off the gain as
    frequency rises. The one from output to pin 5 is more unusual. but
    the zero offset pins certainly have a lot of gain to the output, so
    they are usable for frequency response shaping. But it looks sort of
    desperate. I suspect they had an unstable system, and experimented
    rather than analyzing the circuit to design a good solution. I am
    willing ot bet there is a better way to give good stability and also
    higher frequency response (ability to regulate during load
    The resistor across the output provides a minimum load so the
    regulation doesn't have to work all the way to zero current, where
    leakage through the 2N3055s would dominate. I would have lowered the
    resistor that connects base to emitter on the 2N3055, and maybe put
    one base to emitter on each.
  13. Abbie

    Abbie Guest

    I have come up with some sort of d.c. analysis for the transistors, see
    if you agree:

    1. lets assume for simplicity that the 12V supply is 10V
    Lets also assume that Vbe for our transistors is at 1V
    instead of 0.6V
    2. the 100 resistor from c1008 collector to 10V limits the current of
    the 2sd313 to 10V/100ohm = 100mA
    3. the 1k from b to e of 2sd313 limits the current through c1008 to
    10V/1k = 10mA . (it's actually 1k +100, but let's ignore it)
    4. the same resistor from (3) will make sure that c1008 has a minimum
    of 1mA for operation of the supply with very small currents.
    5. the 1k from b to e of 3055 supplies a minimum of 10mA to
    2sd313 (9 of which go through the collector of 2sd313 and 1mA
    through it's base to supply the minumum 1mA of c1008)
    6. the R from (5) has no impact on the large currents.

    What do you think ?
  14. You have to take the base to emitter drop of the 2SD313 into account.
    If this voltage is 1 volt, then the current through the 1k base to
    emitter resistor is limited to 1/1k = 1 ma. But the base ot emitter
    path can still carry what is left of the original 100 ma limit.
    Its purpose is to make sure there is some path to drain stored charge
    out of the base of the 2SD313 when it needs to turn off fast. Without
    the base to emitter resistor, when the C1008 is off, its leakage
    current will still be feeding a tiny current into the base of the
    2SD313, and the 2SD313 will still amplify that in addition to reaction
    to the base stored charge as it fades away.
    Again, it is not there to supply a load for the preceding stage (the
    base to emitter junction can conduct almost any current at a nearly
    constant voltage drop), but to help reduce the amplified leakage and
    give the stored charge a discharge path.
    Please add Q and R numbers, etc. to the schematic to make these
    discussions more efficient. Thanks.
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