Calculating Power Dissipation of LM138

Discussion in 'General Electronics Chat' started by maw455, Jul 26, 2016.

  1. maw455

    Thread Starter Member

    Jul 20, 2016
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    I'm an amateur at this, and I'm planning on building my first bench power supply and need some help checking my homework. The design I'm working on is for an adjustable low voltage and current supply (1.5-12VDC, 0-1.0A). Before getting into the finer details of the circuit design, I first want to make sure the voltage regulators I've chosen can handle the power dissipation and won't go into thermal shutdown under max conditions since the circuit is built around them.

    I'm planning on using an 18V transformer (9-0-9) rated for 2A max, and I want to have adjustable, regulated DC output (after full-wave rectification and smoothing) of 1.5-12VDC and adjustable current of 0-1A max.

    Two questions:

    1. I have chosen to use the LM138 in the TO-3 pkg. due to its better heat dissipation properties over a LM317 and will be using 2, one for current regulation and one for voltage. I calculate that at either max current output of 1.0A or min voltage output of 1.5V I will need to dissipate between 16.9 and 22.7 Watts in either or both regulators (depending on the voltage and current settings), and depending on whether I’m using the rated transformer voltage of 18V or peak voltage of 23.8V. Planning for worst case, I used the peak voltage of 23.8V in the following calculations:

    As a current regulator:

    P = (Vmax-Vref) * (max current + 10mA) {10mA being sunk current necessary for stable regulation}

    (23.8 – 1.25) * 1.01 = 22.77W

    As voltage regulator:

    P = (Vmax – Vmin) * (max current + 10mA)

    (23.8 – 1.5) * 1.01 = 22.52W

    Have I done this correctly?

    2. To determine the heat the regulators would be dissipating with a heat sink on each one under the above conditions, I attempted to calculate the junction temperature for the 2 LM138’s under my max current and minimum voltage conditions, using information from the LM138 data sheet http://www.ti.com.cn/cn/lit/ds/symlink/lm338.pdf as follows:

    Rjc (thermal resistance junction to case) = 1.0 deg. C/W (TO-3 pkg.)

    Rch (thermal resistance case to heat sink) = .25 deg. C/W (based upon published avg. thermal resistance properties of various thermal paste materials)

    Rha (thermal resistance of heat sink) = 2.70 deg C/W (from data sheet of heat sink)

    Ta (Ambient temp) = 25 deg. C

    P = 22W

    Tj = P(Rjc+Rch+Rha) + Ta

    = 22(1.0+.25+2.7) + 25

    = 111.9 deg. C

    While this is hot enough to boil water, it shows the Tj temp to be under the 150 deg. C max operating temperature of the LM138. I plan to add a cooling fan to pull air through the chassis case to improve the cooling efficiency and keep the temps down.

    Have I calculated this correctly and does anyone see any problems with what I’m proposing to do? What is not clear to me from the data sheet (and I freely admit I barely understand them) is what effect input-output voltage differentials above 15V will have on power dissipation per note 1 at the bottom of page 2 of the data sheet. Any thoughts here would be appreciated as well.

    Thanks.
     
  2. crutschow

    Expert

    Mar 14, 2008
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    For 12V output, 18Vrms from the transformer will give a higher voltage than you need (about 23Vdc from the bridge rectifier and filer) and thus higher dissipation in the regulators.
    You should need no more that 15Vrms from the transformer which will give about 19Vdc.
    That still gives 7V of headroom for the regulator and current limiter.

    If you used a current limiter with a lower drop, than you could likely even get by with a 12Vrms transformer giving about 14.8Vdc output from the bridge/filter.
     
  3. maw455

    Thread Starter Member

    Jul 20, 2016
    30
    1
    Thanks. I struggled with the transformer selection quite a bit. I trust your answer, but I'd like to understand it a bit more. Here is where I get confused: With a 15Vrms 2A transformer the 19VDC is peak, with no load. With a load at the transformer's max of 2A, the voltage drops back down to 15A. So, what happens to the voltage when the load is half that, at 1A? How far below the 19V peak does it drop, and what does that do to my headroom, or am I totally confused about this?
     
  4. #12

    Expert

    Nov 30, 2010
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    No it doesn't. The peak voltage of a 15 VAC sine wave is still 21.2 volts. With a load at the transformers max of 2A, the transformer delivers what it's label says it will deliver.
     
  5. maw455

    Thread Starter Member

    Jul 20, 2016
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    If the label says "7.5V + 7.5V AC 2A", doesn't that mean 15Vrms at a load of 2A? If not, please help me out.
     
  6. #12

    Expert

    Nov 30, 2010
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    That's exactly what it means. Full rated voltage at full rated load.
     
  7. maw455

    Thread Starter Member

    Jul 20, 2016
    30
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    Right. But that same transformer with no load is putting out 19.6V after rectification, correct? I thought that is why you should de-rate a transformer, to account for the drop from peak to rms voltage at full load? Or, have I confused myself?
     
  8. #12

    Expert

    Nov 30, 2010
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    The rectifiers will still rectify to the peak voltage of the 15VRMS sine wave at full load. Transformers don't drop from peak voltage to RMS voltage. They produce sine waves which always have a peak to RMS ratio of 1.414 to 1.
     
  9. maw455

    Thread Starter Member

    Jul 20, 2016
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    Sorry - my ignorance of transformers is showing. If I understand you correctly, a transformer rated at 15V 2A AC is still putting out peak voltage of 21V whether under full load or zero load, so after rectification the voltage is 19.6VDC, even at full load?
     
  10. #12

    Expert

    Nov 30, 2010
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    An unloaded transformer will produce more than the rated output voltage. It loads down to the rated voltage at full load (if the power line is at the voltage rated for the primary winding). The difference depends on the quality of the transformer. A 15 volt transformer might produce 17 volts RMS unloaded or 19 volts RMS when there is no load. At my house, such a transformer will produce 15.625 VRMS at full load because the power line voltage here is 125 volts and transformers are usually rated at 120 VAC.

    In addition, transformers are not rated for DC output unless you're talking about old vacuum tube transformers. The DC current you can get out of a 2 amp rated transformer is less than 2 amps. How much less depends on which rectifier arrangement you use. I've been having trouble with my computer lately or I would post the conversion chart.
     
  11. maw455

    Thread Starter Member

    Jul 20, 2016
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    I think I just realized my mistake. I'm measuring the unloaded Vout of the transformer using a DMM that displays average Vrms, which for an unloaded 15V transformer with 120V AC input shows up as 21V - which is not actually peak. It just happens to be approximately the same as peak which threw me off. So, the rectified voltage is the peak voltage of the transformer at its max load, correct?
     
  12. #12

    Expert

    Nov 30, 2010
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    The rectified voltage is the peak of the sine wave minus a bit for the rectifiers. At 2 amps, you will lose about 7 or 8 tenths of a volt for each diode. The difference between unloaded peak at high line voltage and loaded peak at low power line voltage is probably the worst case of trying to deal with a range of uncertainty in a power supply design. When I was getting paid for that, I used 105 RMS to 125 RMS, from zero load to full load. Transistors and diodes must survive the highest voltage that can ever happen at no load and the highest current that can ever happen when the power line is at low voltage. In a 15 volt supply, you are working on the easy end of the problem. Meanwhile, there is a guy in the Power Electronics forum that wants to work with 30 VDC to 480 VAC as the input voltage. HE has a serious design problem.:D
     
  13. maw455

    Thread Starter Member

    Jul 20, 2016
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    Holy @#$%! What country's grid is 480VAC????
     
  14. #12

    Expert

    Nov 30, 2010
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    None of them, and I'm not allowed to call him a nut-case. I just have to guess what he has that needs that kind of range because he won't tell me.:rolleyes:
     
  15. maw455

    Thread Starter Member

    Jul 20, 2016
    30
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    Thanks for your help. Can you tell me if I used the right formulas in my original post?
     
  16. #12

    Expert

    Nov 30, 2010
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    I thought crutschow did that. I try not to read posts that are over 500 words.:(
     
  17. maw455

    Thread Starter Member

    Jul 20, 2016
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    He did not say yes or no, which MIGHT imply that I was correct. As for the 500 word post, I'm a lawyer by profession, so that post was relatively short by my standards. :D
     
    #12 likes this.
  18. #12

    Expert

    Nov 30, 2010
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    A quick look says you are right. A bit high with an 18 volt transformer, but crutschow covered that.
     
  19. maw455

    Thread Starter Member

    Jul 20, 2016
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    Got it, and you helped clarify why a 15V transformer would be better suited. Thanks a bunch.
     
  20. #12

    Expert

    Nov 30, 2010
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    A TO-3 can survive 23 watts, but attaching a heat sink can only make it more reliable.
     
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