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Constant current power supply

Discussion in 'Power Electronics' started by Lamma, Jul 27, 2014.

  1. Lamma

    Lamma

    11
    1
    Jul 27, 2014
    Hi everyone,

    I was wondering if any of you could help with this circuit , in making the output current variable. Is it as simple as changing the Zener diode , Z1 , for a potentiometer and careful selection of R3 and R4 resistors ? I need a power supply that outputs a constant current with variable voltage , as this is the best method for anodising. It would however be preferable if that current could be adjusted up or down depending on the surface area of the anode. Thus far I have done some math and found as follows :

    R3 = 22000Ω
    R4 = 0.85Ω ( 0.25 + 0.25 + 0.25 + 0.1 , all 5W)
    Z1 replaced with 5k Potentiometer

    When POT is at 0Ω , collector current = 0A due to no voltage at Q1 base. When POT is at max 5000Ω , collector current = 3.839A , and Q1 base voltage is 3.963V. If my theory is correct , this increased voltage at Q1 base causes an increase in Q1 collector current , but more importantly that current is fixed and can thus be controlled by the potentiometer. Any ideas anyone?
     

    Attached Files:

  2. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Hello and welcome to Electronics Point :)

    It would be helpful to know the operating voltage of the circuit. If you don't know, can you tell us the secondary voltage for the transformer.

    The output (anodising) current is determined roughly by the emitter voltage and the emitter resistance using Ohm's Law, I = V / R. (This neglects the base current, which can be as much as 10% of the emitter current, possibly more!) The emitter voltage is equal to the base voltage minus the base-emitter junction forward voltage VBE, which for a power transistor, is roughly 0.8~1.0 volts, depending on base and collector currents. By setting the base voltage, you set the emitter voltage to a voltage that's roughly 0.8~1.0V lower.

    The usual way of making this simple current source variable is to use a fixed base voltage and vary the emitter resistor (R4). For example, putting 1.9V on the base gives you about 1V on the emitter, and therefore 1V across the emitter resistance. Neglecting base current, the anodising current is I = V / R where V = 1, so I = 1 / R.

    But with such low resistance values, and high currents, you can't just use a standard carbon potentiometer for the emitter resistor, because of the need for fairly accurate, low resistance, and the power dissipation, which (for the emitter resistance) can be calculated as P = V × I where V is the voltage across the emitter resistor, in volts, and I is the anodising current, in amps.

    Have a look at these options for the emitter resistor:

    www.digikey.com/product-search/en/potentiometers-variable-resistors/adjustable-power-resistor/262971?stock=1
    Adjustable power resistors. Not easy to change the setting. But you could use several of these, with a rotary switch to select between them. Make sure you use a good quality rotary switch with low contact resistance!

    http://www.digikey.com/product-search/en?pv1=1345&pv1=1025&pv1=250&pv1=382&FV=fff40004,fff80123,400fa,4017e,40401,40404,40433,4045a,40460,4048e,404b0,40523,40541,40546,4060f&stock=1&quantity=1
    Low-resistance rheostats (variable resistors). This would give you a rotary control for your anodising current.

    The circuit you have there is pretty simple and only gives rough control of the anodising current. The transistor's base-emitter voltage varies over a significant range, and with such a low emitter voltage determining the current, this variation affects the current significantly.

    You also need to ensure that the base drive circuit can supply significant current. Large transistors that can pass several amps have low current gain figures, and this means you need to supply quite a lot of base current to get the desired collector current to flow. R3 = 22k is far too high; it will limit the base current to less than 1 mA, and therefore will limit the collector current to around 10 mA (0.01 amps)!

    You also need to consider power dissipation in the transistor, which could be dozens of watts. You will definitely need a heatsink.

    There are some much better options available for this circuit. Have a look at http://www.google.com/search?q=op-amp current sink&tbm=isch or let us know the operating voltage, and the output current range you want, and I will design something for you.
     
  3. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    No, changing the zener diode for a potentiometer will not produce an effective constant current source.

    There are possibly ways of doing this in a more complex circuit, but a zener is still required and the load on the potentiometer has to be carefully managed (i.e. minimised) so that it is able to act as a voltage reference.

    One option is switching different resistors as R4. Beware that the switches need to carry the full current. You could have several resistors chosen for (say) 0.25A, 0.5A, 1A, and 2A that can be independently switched in and out. You would then be able to set the current anywhere between 0.25A and 3.75A in steps of 0.25A by carefully choosing which switches to close.
     
  4. Lamma

    Lamma

    11
    1
    Jul 27, 2014
    Thanks for your input fellas . The transformer outputs 26.2V , 6.3A max. That's written on it. That AC is rectified by a RS807 full wave bridge rectifier and smoothed by a big 10000uF 63V cap as per the schematic. See this link to find out where my info comes from http://www.shadowguarddev.com/alex/anodizing/ . Now you'll see that Alex uses a Zener diode of 4.3V for the circuit that eventually gives him a fixed output current of 1.8A . To anodise that way is fine because it makes it possible to calculate your anodising time using the low current density method. But when the items become bigger , that anodising time can be hours instead of minutes. To keep the anodising time at less than an hour I want to adjust the output current . I have mentioned this before , just want to explain where it all comes from. That put me onto this site http://pcbheaven.com/drcalculus/index.php?calc=trans_single_zener_const_current . With this calculator you can input your data and it works out the outputs. As long as your parts' specs fall within the parameters I guess your fine.

    Now if you input the following ( the parts I have on hand):

    VDD = approx. 20V ( RS807 drops 4.8V at max 8A)
    RB = 325Ω 2W
    RE = 2Ω ( 2 x 1Ω 5W)
    DZ = 5.1V
    Transistor = 2N3773 ( hfe min = 15 , max = 60)

    Then you get as follows :

    Vb = 5.1V
    Ve = 4.4V
    Ie = 2.2A
    Ic = 2.178A
    Ib = 0.145A
    Vce = 6.6V

    Power dissipated is within spec of parts , with some headroom at max. So with some further reading I come upon this page http://www.pcbheaven.com/userpages/LED_driving_and_controlling_methods/?topic=worklog&p=4 . Here the author explains how to dim a LED using a potentiometer instead of the voltage divider method , thus changing from a fixed current output to a variable current output , and that's exactly what I want . You hook up a voltmeter across the load and I know the voltage across my anodising bath , and an ammeter in line that shows the current flowing through the anodising bath . To an electronics idiot/newbie like me that sounds very logical.

    [​IMG]

    I quote : "R2 form the voltage divider can simply be replaced with a potentiometer and a limiting resistor in series. This way the ration of the voltage divider can vary, and therefore the base voltage can vary as well. The result is a variable emitter voltage which eventually leads to a variable current set for the LED. This is by far the simplest method of all and most straight-forward to understand."

    When the potentiometer is at lowest position, the expected voltage is:
    R1 = 4700 Ohms
    R2 = 0
    VB = VDD x R2 / (R1+R2) = 0
    VE = 0
    IC = 0

    And when the potentiometer is at maximum:
    R1 = 4700 Ohms
    R2 = 1000
    VB = VDD x R2 / (R1+R2) = 2.1 volts
    VE = 2.1 - 0.7 = 1.4
    IC = VE / RE = 1.4 / 47 = 29.7 mA

    Now if I implement his formulas into an excel sheet I get the following at max :

    Vdd = 20V
    R1 = 22000Ω
    R2 = 5K Pot
    RE = 0.85Ω
    Vbe = 0.7V
    hfe min = 15

    Then :

    Vb = 3.704V
    Ve = 3.004V
    Ie = 3.53A
    Pre = 10.61W ( I use 4 x 5W resistors in series )
    Vce = 5V
    Pt = 17.66W ( The transistor is mounted on a huge heatsink)
    Ib min = 0.236A

    I'm sure that I' missing something here , or not understanding a couple of things , so please help me out . I will appreciate a schematic if possible , just keep it simple please. I'm new to all this and must say very fascinated by it all. Can't believe I never thought of electronics in this way , and I'm almost 40 !

    Thanks again.
     
  5. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    The current will vary with temperature and significantly with Vdd
     
  6. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    That's a very rough "current source". Only slightly better than a big resistor.

    What is the maximum anodising current you want?
     

    Attached Files:

  7. Lamma

    Lamma

    11
    1
    Jul 27, 2014
    0-5A will do fine. It should cover all I will ever need.
     
  8. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    OK, here's my suggested circuit.

    269564.001.GIF
    The AC voltage from the mains transformer is bridge rectified by BR1 and smoothed by C1, which should be at least 10,000 µF. The thick lines represent connections that should be made with thick wire. For these wires, follow the topology shown in the diagram to ensure stable operation of the circuit.

    U1 is a TL431CLP voltage reference. These are more accurate and stable than zeners. VR1 taps off a variable voltage, and op-amp U2 controls the bias on QD and QP to produce an equal voltage at RF, i.e. an equal voltage across RE1 and RE2. Therefore, the voltage at VR1's wiper determines the emitter current in QP. QD and QP are connected as a Darlington transistor, which has a current gain of at least 1000, so its base current is negligible. Therefore, by regulating the emitter current, the circuit regulates the collector current.

    It is possible to make a good constant current circuit without using an op-amp but the op-amp gives predictable operation and requires no offset adjustment. The other op-amp in the package is not used.

    RB limits the current into QD and QP so the op-amp does not try to provide the full 5A when the collector circuit cannot pass that current. RF protects the op-amp inputs against high differential voltage during transients.

    QD and QP will both require heatsinking as noted on the schematic. RE1 and RE2 may also require heatsinking. The amount of heatsinking needed on QP depends largely on the characteristics of the load (the anodising current path) and I don't know how this will behave. Worst case power dissipation in QP will occur at maximum anodising current and minimum voltage drop across the anodising circuit.

    The small components should be assembled on stripboard. Use point-to-point wiring for the high-current connections.

    All components are available from Digi-Key (http://www.digikey.com).
     
    chopnhack likes this.
  9. Lamma

    Lamma

    11
    1
    Jul 27, 2014
    Thanks a billion for your effort Kris. When I say I'm a novice , I mean NOVICE. I don't understand most of what you said in your post but I will give it a go nonetheless . I'll make up a PCB schematic and post it for your approval , but I must ask if there is really no easier way to do this?
     
  10. Miguel Lopez

    Miguel Lopez

    246
    54
    Jan 25, 2012
    Use a 7805 or an LM317 with a variable resistance between output and adjustment pins. Take output from the adjustment pin. This will provide a useful and simple constant current source. Surely you will need a TO-3 to stand 0-5A, and a big heatsink.

    [​IMG]
     
  11. chopnhack

    chopnhack

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    Apr 28, 2014
    An elegant and simple design, very nice Kris. Why are pins 5,6 of the unused op-amp connected?
     
  12. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Thanks chopnhack :) Really it's pretty conventional. I grounded the pins of the unused op-amp just to stop them picking up noise. For CMOS ICs it is important to connect unused inputs to something, but not really for op-amps. It does avoid one track cut on the stripboard though (between pin 4 and pin 5). (Yes, I do think about those things when doing a circuit design.)
    Well, it's the simplest way that I can think of, given the design requirements that I used, which include using standard components, not requiring any setup adjustment, having a linear current adjustment from 0A to 5A, being fairly consistent from one run to the next, and being comfortably over-engineered so it's unlikely to fail.

    Miguel has suggested a simple alternative to the regulator section in post #10 and it meets most of those criteria. The problem with that design is R1. For the LM317 current regulator, the R1 resistance for a given output current can be calculated as 1.24 / IOUT. This gives the following values for typical output currents:

    IOUT = 0.1A --> R1 = 12.4Ω
    IOUT = 0.5A --> R1 = 2.48Ω
    IOUT = 1.0A --> R1 = 1.24Ω
    IOUT = 1.5A --> R1 = 0.83Ω
    IOUT = 2.0A --> R1 = 0.62Ω
    IOUT = 2.5A --> R1 = 0.5Ω
    IOUT = 3.0A --> R1 = 0.41Ω
    IOUT = 4.0A --> R1 = 0.31Ω
    IOUT = 5.0A --> R1 = 0.248Ω

    So there are several practical problems with this approach.

    If you need low anodising currents, you need a fairly high value variable resistor, and almost all of the useful adjustment will be at one end of the travel. You could use a logarithmic variable resistor, if you can find one with such a low resistance (I doubt it), but then the high current end would be the fully anticlockwise end.

    Variable resistors with such low resistance values are pretty specialised components. Digi-Key are probably the biggest component supplier; look at their range and pricing: http://www.digikey.com/product-sear...=1028&FV=fff40004,fff80123&stock=1&quantity=1

    Power dissipation in R1 will be over 10W at some settings.

    You could use a switch, or several switches, to select various fixed resistors, as mentioned by me and Steve in posts #2 and #3.

    The short answer is that that design is my best attempt at what I would use if I wanted to do what you want to do.
     
    chopnhack likes this.
  13. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    I think the main point is that assuming you don't want to purchase a very expensive high current (i.e. high power low resistance) potentiometer, then you need to either switch the resistors or have a more complex circuit to allow you to compare the voltage across a fixed resistor to a variable voltage reverence.
     
    KrisBlueNZ likes this.
  14. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Here is another option that falls between the two in complexity.

    vari.png

    In this circuit the current through D1 can be varied from about 700 mA to about 4.5A by replacing R2 with a 1k pot. R2 has a maximum of around 7mA flowing through it, so any small pot can be used.

    R1 will need to be rated at 25W (I'd get a 50W resistor) and Q1/Q3 need to be on a heatsink. (the 50W resistor will probably need a heatsink too).

    D1 is just what I used for simulation, you can replace it with your electroplating bath.

    If you switch R3 between 1 and 10 ohms, you could get a current down to about 70mA.

    You're not saving an awful lot in complexity or parts count though.

    The added disadvantage is that R1 might end up dropping almost 5V, and Q1 another 1.5V, leaving only 5.5V for your bath.
     
    chopnhack likes this.
  15. Miguel Lopez

    Miguel Lopez

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    Jan 25, 2012
    I think that a continous current adjustment won't be required (just my humble opinion). A step-selector will do the job at the exact value of the several required current. Fixed resistors can be selected to stand high power dissipation. This way, an ammeter is not needed.
     
  16. Lamma

    Lamma

    11
    1
    Jul 27, 2014
    I thank you all gentlemen for your input and help. Much appreciated. I have been away for a couple of days so only saw the new replies today. Before I make up the PCB , just a few questions though :

    1. Can I make a PCB on 1 oz / sq. ft board with the ironing method or is there a specific reason for making this on a strip board? If only strip board , please define "thick" wire.
    2. Can I use a 2N3054 at QD?
    3. Can I use a 2N3773 at QP?
    4. Will a plain TL431 work at U1?

    The reason I ask is because I have these components on hand already. These type of things are hard to find locally , so I'm scrounging off the rubbish dump at the scrap yard literally.:( My PCB layout attached. I will appreciate any advice on faults or perhaps better practice. I'm very new to all this remember.
     

    Attached Files:

  17. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Yes you can use 2N3054 and 2N3773. They are both a bit low on gain but that's OK. Yes any TL431 will work; the CLP suffix I specified means commercial temperature range and TO-92 suffix, so I think that's what you'll end up with anyway.

    No you don't need to use stripboard.

    Is that layout viewed from the top or bottom side? If it's a bottom view, the pin numbers on the op-amp are mirrored.

    The spacing for the op-amp looks wrong. Pins are spaced 0.1" apart, and the two rows are spaced 0.3" apart. Yours looks like the rows are spaced 0.2" apart.

    Ideally CD should be very close to the op-amp and the track to the op-amp and the track from the rest of the circuit should connect on opposite sides of the CD pad. Keep the loop enclosing the op-amp and CD as small as possible.

    You may need to heatsink BR1.

    I would avoid putting mains voltages on your PCB. It means the PCB is dangerous to work on. The fuse, switch, and incoming power connector can be easily wired using insulated wire with heatshrink to prevent shock hazards.

    I would keep all the high-current stuff in one area of the PCB. It means the high-current tracks can be nice and short. I don't like the long track from the negative of C1 to the rest of the circuit.

    Personally I would run some reinforcing along all the high-current tracks, unless you used 2 oz copper perhaps. You can use braid, or just thick copper wire - at least 0.5 mm diameter; I use about 0.8 mm.

    I don't see any other obvious problems.
     
  18. Lamma

    Lamma

    11
    1
    Jul 27, 2014
    OK Kris the spacing on the op amp has been fixed , it was 2.5mm. The view is from the top so the op amp is the right way round. BR1 is heatsinked . The reason I put the 230V lines on the board is because it was like that on the commercial power supply PCB from which I took the transformer , op amp and 2N3773 transistor. Didn't know better. I'll redo the layout to incorporate a shorter ground track and use 2 oz board instead as suggested. Thank a million for your help again. I'll post a couple of pictures of the completed PS once it's done. Cheers
     
    KrisBlueNZ likes this.
  19. chopnhack

    chopnhack

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    Apr 28, 2014
    Lamma, what software are you using for your board layout/schematic?
     
  20. Lamma

    Lamma

    11
    1
    Jul 27, 2014
    Designspark PCB 6.0
     
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