Discussion in 'The Projects Forum' started by fender7802, Apr 7, 2013.

  1. fender7802

    Thread Starter Member

    Apr 7, 2012

    I am attempting to design a DC power supply which will be regulated, dual-output, and adjustable. Can you look at my schematic and see if the circuit makes sense?

    My main questions are:

    Will the Zener regulators work correctly to offset the voltage of 1.25V that the LM3x7's put out (in order to bring the output down to 0V)? I am concerned because the references are not from an isolated source. Will the zeners "hold" their voltage? The PN# for that reference diode is LM285.

    Will the voltmeter panels work or will it degrade the performance of the supply? I plan on using two 'Lascar V1' panels. I believe I have them connected in common ground configuration. Does this look correct?


    Thank you for your help.

    PS: I couldn't create the thread without underscores in the title. Bug in the forum?

  2. crutschow


    Mar 14, 2008
    Yes, the reference voltages you show should offset the internal 1.25V reference and allow you to adjust to essentially 0V output (within the tolerance difference of the two references). You don't need isolated references as long as everything goes to a common ground.

    The voltmeters won't degrade the power supply since they have a high input impedance.

    Do those voltmeters allow operation with a common ground? Many don't.
  3. fender7802

    Thread Starter Member

    Apr 7, 2012
    That sounds good, thank you.

    The panel does have a common ground mode. I found the datasheet to be a bit confusing, specifically where it says the following:


    So in other words, I can't measure anything below 5V? If that is true what does "200mV d.c. full scale reading" mean?

    Also, about where it says that L1 connects COM to IN LO. I would think, for common ground mode, L1 would be shorted. However it says:


    Does it make sense to have Link 1 opened?
  4. ifixit

    Distinguished Member

    Nov 20, 2008
    Hi Fender,

    The regulator reference current is 6mA, so the LM285 current should be at least that. Say 8mA, so the 10KΩ res can be changed to 470Ω. This will ensure that the references will regulate properly.

    How much current do you expect to get from this power supply?

  5. fender7802

    Thread Starter Member

    Apr 7, 2012
    Good point, ifixit. You're talking about the required minimum output current of the 7805, correct?

    Since the maximum current of the transformer is specified as 1A rms, I expect to get up to either 1.4A from each output or 0.7A from each output. I haven't figured out which it will actually be. Since the max current in the secondary is rated at 1A rms, and the transformer is rated at 44VA (22-0-22)V rms, does that mean I can get 30V @1.4A maximum for each output? That would be 22VA + 22VA = 44VA rms. My logic is that even though I am getting 2A rms total current, there is never a point in the secondary winding where the current exceeds 1A rms. Therefore, it is still within the rated current. I have no idea if this line of thinking is correct, I more or less was just going to wait and find out. But if you know, that would really clear things up! Thank you so much! :D
  6. #12


    Nov 30, 2010
    You will be lucky to get to 28 volts output because of losses in the rectifiers and the regulator chips.

    You can figure VA from the peak capacitor voltage times the current. Thus, the output current can not be more than Irated/1.414

    This handout from Hammond Transformers says you can't get more that 62% because the peak current to average current ratio blah blah blah...
  7. ifixit

    Distinguished Member

    Nov 20, 2008
    No. That is not what I was referring to, although you should ensure that requirement is met. I was referring to the LM285 load current of 6mA which comes through the pot from the 208Ω resistor. This current is higher than the 10K can supply and still maintain the 1.2V reference. So change the 10K to 470Ω.

    You need to be concerned about power dissipation in the LM3x7. The [voltage across it] times [the current through it] = watts. Assuming you are using LM3x7T then you are limited to 20 Watts and/or 1.5 Amps. AND you must ensure the case temperature stays within its limits.

    θjc is 4 °C/W. So with Tj limited to 100°C for resonable live expectancy and Tc limited to 50°C, then allowable Watts is (100-50)/4 = 12.5W.

    What are you using for a heatsink? Any fans?

  8. THE_RB

    AAC Fanatic!

    Feb 11, 2008
    Re the meters and your info in post #3, those meters require an ISOLATED power source (usually a 9v battery) that is not connected to the voltages being measured.

    You can get better meters that allow a common ground bewteen their power supply and the DC voltage they are measuring.
    #12 likes this.
  9. BillB3857

    Senior Member

    Feb 28, 2009
    When using LCD meters that require isolated power, I have used this circuit before. It works well. The voltage source to drive the opto-isolators can be pulled from almost anywhere within your design. You just need to adjust the value of the series resistor to stay within the current limits of the opto's LEDs.
  10. tubeguy

    Well-Known Member

    Nov 3, 2012
    You would use a resistor voltage divider on the input to scale the voltage reading.

    Hint: To display 5 volts, you would divide the voltage so 5 volts in provides 50 mv to the meter input. And then set your decimal point so the display reads 5.00.

    Edit: (200mv d.c. full-scale actually means the meter will display up to 199.9mv directly)
    Last edited: Apr 8, 2013
  11. fender7802

    Thread Starter Member

    Apr 7, 2012
    I think I am starting to see what you are saying. If the 6mA were to come entirely from the 7x05 through the 10k, the 10k would have to drop 60V which it can't do, but when it tries to it will take voltage away from the LM285? I think I was assuming that 6mA will be pulled up the the junction between the 10k and the LM285 no matter what. Is this correct now? I'd like to do some calculations to prove where the current through the potentiometer is actually coming from. Any tips?

    I plan on using large heat sinks and having some ventilation holes, no fan. I will consider this more after I can be sure about the circuit.

    You are absolutely correct. Thank you for saving my ass on this one - looks like I misunderstood the data sheet.
  12. ifixit

    Distinguished Member

    Nov 20, 2008
    The LM285 is a shunt regulator, it doesn't supply any current.
  13. fender7802

    Thread Starter Member

    Apr 7, 2012
    Both schematics make great sense. I see no reason not to abandon the LM285. I appreciate your help.

    I am looking into the thermal aspect a bit more now. According to the datasheet, the maximum allowable power dissipation is PD(max) = (Tj - Ta) / θjc = (125°C - 25°C) / (4°C/W) = 25W. Now like you say, for reasonable life expectancy, 125°C comes down to 100°C and the result is 18.75W. I suppose if Tc is the temperature of the box, then the ambient temperature will increase and the answer will be closer to the 12.5W you calculated earlier.

    From each output, a max of 12.5W isn't unreasonable for me. I would like to use a heat sink and try to get that up to around 20W (or more) if possible. As a rough estimation, would you say a 2.5"x1.6"x1" (specified thermal resistance of 3°C/W) heat sink can do about that?

    Also, what is the difference between an LM3x7T and LM3x7K? Is it the temperature difference between the die and the case?
  14. ifixit

    Distinguished Member

    Nov 20, 2008
    Moving heat from a silicon junction to ambient air is like a step by step process. Temperatures are highest at the heat source (junction), but the physical volume, or surface areas are smallest so temperatures rise quickly. Devices are engineered to take this heat away to larger surface areas as efficiently as possible.

    Temperature definitions review:
    • T junction (Tj) is the maximum device junction temperature allowed. This is 125C. The LM3x7 will protect itself with a thermal limit circuit.
    • T case (Tc) is the device case temperature.
    • T heatsink (Ts) is the heatsink temperature, hottest near the heat source.
    • T enclosure (Te) is the ambient temperature inside the enclosure, or chassis, or box, or whatever. This is an optional term.
    • T ambient (Ta) is the air temperature in the room. 25C, or whatever you think is reasonable.
    Thermal resistance definitions:
    • θjc Junction to case thermal resistance. I your case this is 2.3C/W for the TO3 case style, or 4C/W for the TO220 case style. The lower the number, the better.
    • θcs Case to heatsink thermal resistance. With careful mounting and the use of a good thermal compound, you should be able to get the resistance down to .5C/W.
    • θse Heatsink to ambient inside the enclosure resistance. The air inside your box is somewhat unknown. You will have to make tests and measurements to be sure. It depends on what else is going on in there and how efficiently the heat in the box can get out. Ensure you have good vent holes near the bottom of the heatsink to let cool room air in, and at the top to let the heated air out. This term can be left out if the heatsink is not in a box.
    • θea Enclosure to room ambient resistance. Let’s assume for now the average air temperature around the heatsink is 25C, the same as room ambient.
    The first two items are well defined and easy to deal with, however, control of items 3 and 4 have a lot of options on how to deal with them and therefore are more difficult to deal with.

    Calculating the heatsink thermal resistance required

    Put whatever Tj and power values you want into the first formula below. You can juggle the other numbers to get the second formula to balance. Whatever you do in the first formula will have consequences in the second formula.

    • θsa = (Tj [max] – Ta) / power. (This is where the heat is created)
    • E.g. (125 – 25) / 20 = 5C/W
    • E.g. (125 – 25) / 12.5 = 8C/W
    • θsa - θjc - θcs - θse - θea = 0. (This is how you dissipate the heat to the air)
    • E.g. (1) 5C/W - 2.3C/W - 0.5C/W - 2.2C/W - 0C/W = 0
    • E.g. (2) 8C/W - 2.3C/W - 0.5C/W - 5.2C/W - 0C/W = 0
    • E.g. (3) 8C/W - 2.3C/W - 0.5C/W – 3.7C/W – 1.5C/W = 0
    • Let’s look at actual temperatures for example (3)…
    • E.g. (3a) Case Temp <= 125C - 2.3C/W * 20W = 79C
    • E.g. (3b) Sink Temp <= 79C – 0.5C/W * 20W = 69C
    • In the next step set the box temp to what we think it will be (40C) & calculate the box-2-room resistance…
    • E.g. (3c) Box Temp = 69 – 1.45C/W * 20W = 40C
    • The 1.5C/W is inserted into step 7 and the heatsink resistance changed to balance the formula.
    • The room ambient is arbitrarily set to 20 to 30C. Your choice.
    • You can juggle away until you find a situation that works for you.

    The manufacture spec says the part should last 1000 hours (42 days) at max junction temp, so if this is a lab bench power supply then maybe that’s not so bad if you push the limits of the device.

    On the other hand, if this is a device that runs 24/7 at max output, then reduce the max junction temp to 100C or so.

    Have fun,
    fender7802 likes this.