Voltage Regulator questions.

Discussion in 'Power Electronics' started by vandaycalta, Jun 12, 2016.

  1. vandaycalta

    Thread Starter Member

    Mar 22, 2016
    Can someone help me in answering questions I have involving voltage regs.
    1) If you have a vcc of 18v and you need two different voltages in your circuit of 5v and 3.3v, can you have the vcc feed the 5v reg and then the 5v reg feed the 3.3v reg OR is it better to run EACH reg from the VCC of 18v? I was thinking the first option would be better as going from 5-3.3v would create less heat and allow a smaller package.

    2) Can someone describe to me the proper way to choose a volt reg as the last time I did I did not take into consideration the junction temperature, ambient temp, theta ja, etc but only the input voltage, output voltage and the output current. Needless to say my regulator package heated up pretty quickly( lessons learned as my motto is becoming " fail and fail often" in my learning process.)and went into thermal shut down.

    3) Do I need to derate the voltage regulator any? In my circuit on the 5v side I used a 250mA reg and had a current useage of about 220mA, I thought this would be ok but I am not sure. As is on the 3.3v side I used a 3.3v regulator rated at 30Ma and needed only average of 17mA draw.

    4) I was using linear regulators for my 5v and 3.3v regulators. I am looking for the smallest package I can as the pcb is less than one inch in square and I put my voltage regs on the bottom layer to keep them away from my microcontroller and other ics that are on the top layer.

    Any direction would be appreciated. Thanks for your help on my learning as it is a vast topic to digest.

  2. bushrat


    Nov 29, 2014
    It would be easier to run each voltage regulator seperately.

    Are you looking for Thru-hole or SMT components?
  3. benta


    Dec 7, 2015
    No matter how you configure your regulators, you'll need to dissipate around 3.5 W of heat.
    That's your main concern. You'll not get away with that without heatsinking.

    ErnieM and #12 like this.
  4. vandaycalta

    Thread Starter Member

    Mar 22, 2016
    Thank you for your repsonse. I am looking for only SMT components.
  5. vandaycalta

    Thread Starter Member

    Mar 22, 2016
    Benta, thank you for your reply. Can you explain your outcome of 3.5w of heat so I can understand your logic?

  6. bushrat


    Nov 29, 2014
    Digikey has your part (not in stock currently): dual voltage output (5 and 3.3V)
  7. vandaycalta

    Thread Starter Member

    Mar 22, 2016
    I think the point here is being missed. I am looking for direction on how to determine if I need to derate my volt reg as well as what I need to look at in determining the proper heat dissapation
  8. jellytot


    May 20, 2014
    If you hook up EACH regulator to the 18v source:
    • If one regulator fails, the other will still work.
    • But I'm pretty sure this method generates more heat, compared to cascading the regulators together.
    If you hook up the 5v regulator to the 18v source, then hook up the 3.3v regulator to the 5v:
    • Pretty sure this generates less heat than the above.
    • Note: You can only hook up the 3.3v regulator to the 5v regulator IF the 3.3v regulator can take 5v (check the datasheet).
    Not sure what you mean by "help you choose one"? Well, if you are going to cascade them together, like I said above, make sure the 3.3v regulator can take a 5v input. If it's overheating, you need a heat sink on the overheating regulator (with thermal compound). Actually, hooking up a 18v source to 5v regulator will generate a LOT of heat. If you have a lower voltage source (e.g. 9v or 12v) the 5v regulator will generate less heat compared to hooking it up to a 18v source.

    I don't know what "derate" means. Sorry.

    I don't have experience with PCBs. But if your 5v regulator is burning off a lot of heat, having components cover it seems like a bad idea?
  9. #12


    Nov 30, 2010
    ErnieM likes this.
  10. ronv

    AAC Fanatic!

    Nov 12, 2008
    The problem you have is the 18 volt supply and the 5 volt 250 ma requirement. To get from 18 to 5 volts the regulator needs to drop 13 volts. Since P=IE the power in the regulator is .25X13 or 3.25 watts. That's quite a bit even for a large linear regulator. The heat sink will be about the same size as your board.
    Here is an example for a TO220 IC
    It's temperature rise is 16C per watt. so it would get to about 75C in your case. 25C for ambient and another 50 in generated heat.
    Do you need to start with 18 volts?
    Last edited: Jun 13, 2016
  11. WBahn


    Mar 31, 2012
    If you are using linear regulators, then you are going to be dissipating the same amount of power in either case, though the distribution between the two regulators will be different.

    You want to run them in parallel for a few reasons. First, running them in cascade will increase the power dissipation in the one that would already be dissipating, by far, the most when in parallel. Second, if cascaded then noise in the 5 V circuitry will be seen at the input of the 3.3 V regulator which will increase the noise coupling.

    Your total circuitry, given the numbers you provide, draws 270 mA from an 18 V source, meaning that it will dissipate 4.86 W. Of that, 1.2 W will be in the 5 V circuit, 99 mW will be in the 3.3 V circuit, and the rest, 3.56 W, will be in the regulators.

    If you run the regulators in parallel, then the 5 V regulator drawing 240 mA will have to dissipate

    P_5 = (18 V - 5 V)(240 mA) = 3.12 W

    The 3.3 V regulator drawing 30 mA (not Ma) will have to dissipate

    P_3 = (18 V - 3.3 V)(30 mA) = 441 mW

    If you run them in cascade, then you will have all 270 mA flowing in the 5 V regulator, so it will dissipate

    P_5 = (18 V - 5 V)(240 mA + 30 mA) = 3.51 W

    While the 3.3 V regulator will only have 1.7 V across it and will dissipate

    P_3 = (5 V - 3.3 V)(30 mA) = 51 mW

    To cascade them you also have to use a low-dropout (LDO) regulator because most non-LDO regulators can't work with just 1.7 V of overhead.

    When possible, it is better to minimize the heat dissipated in integrated circuits, so instead of putting big heat sink on the 5 V regulator (the 3.3 V regulator may or may not require one), you can shift the heat from the regulators to a resistor (or set of resistors) by putting resistors between the 18 V supply and the regulator input. You can size the resistors to drop just enough voltage to bring you near the minimum dropout voltage at just a bit above the max current you plan to use -- this also gives you pretty good short-circuit protection. For the 5 V circuit that means (roughly) you want to drop about 10 V when the current is, say, 250 mA meaning that you want a 40 Ω resistor. So you might use a 39 Ω resistor, but at your max current draw of 240 mA that resistor would be dissipating 2.25 W, meaning it is going to be a good size resistor (you would want to size it for about 5 W to give a safety margin). You can spread that across a resistor network if you want very easily. If you want to use 1/4 W resistors with a 20% safety margin, then you would need 11 or 12 resistors. Using 12 would be easy to size in a 3x4 (or 4x3) array of resistors that are all the same value.

    If you don't have room for this (or the necessary heat sink) then an alternative to look at is one of the many DC-DC converter ICs that are out there. Because 5 V and 3.3 V is so common, there are single ICs that provide both outputs. The are switch-mode regulators so they are very power efficient, typically above 70% (and sometimes above 90%). At 70% that means that the regulator IC would need to dissipate somewhere around 600 mW, which is pretty easy to manage. Now, switch-mode regulators produce a fair amount of noise, but unless you have analog circuitry this is probably not a big concern.
    absf, ErnieM and Techno Tronix like this.
  12. hp1729

    Well-Known Member

    Nov 23, 2015
    As long as you realize the 3.3 V current has to come from the 5 V regulator you should be okay. Total watts used is the same. Just more stress on the 5 V regulator your way.
    Last edited: Jul 15, 2016