I designed an external device for Android phones, but i am in trouble with temperature.

Discussion in 'The Projects Forum' started by Bipsiljo, Apr 16, 2016.

  1. Bipsiljo

    Thread Starter New Member

    Apr 16, 2016

    I designed an external device for Android phones, but i am in trouble with temperature. I am getting power from phone's OTG output voltage (5V)
    My device needs only 3.0 volts and 0.4 Amper maximum. I used many 5V to 3V adjustible regulators but i am not happy with temperature of device, it is about 60-65 degrees celsius.

    I used 3A, 5A and 7A regulators, but i can't solve temperature problem from 5V to 3V. Than i add 2 pieces serial Diodes to front side of device. 2 pieces diodes droped the voltage to 5V > 4.3V and than 4.3V > 3.6V.
    Regulator has better temperature after the 2 diodes, but now 2 diodes are hot now. I am using 600V 16A and 400V 30A diodes. Both are TO263AC and D2PAK package types.
    Anybody know a good solution for for 5V to 3V and 0.4A requirement without temperature problem. I ordered hundred type component from Digikey but i didn't find suitable solution already.
    Many thanks for your comments...
  2. DickCappels


    Aug 21, 2008
    Silicon semiconductors can easily operate at over the temperature of boiling water. Of course if your enclosure is melting or there is a chance for painful contact to one's skin, then you have something to worry about.

    I assume that the device for you are concerned about the temperature is the voltage regulator. If you are using a linear regulator (7805m, etc.) the input voltage is always the same and the load current is always the same, the power dissipation will always be the same (nearly) regardless of the current rating of the regulator.

    Two options come to mind: 1) Add a heatsink or 2) Change to a switching regulator.

    If the device for which you are concerned about temperature is the entire circuit, then you need to supply adequate heatsinking and/or ventilation to handle up to 1.2 watts overall without anything getting too hot.

    If you post your circuits you might get advice that is more specific.
  3. R!f@@

    AAC Fanatic!

    Apr 2, 2009
    Welcome to AAC

    What do you mean by temperature of device?
    What device ? Phone or the device that you use ? or the diodes ?
  4. Bipsiljo

    Thread Starter New Member

    Apr 16, 2016

    I used linear regulators before, i just searched in internet, there are some comments about switching regulator, are they produce less temperature? Can i use the switching regulator like linear?
    My external device has 42 mm x 26 mm dimensions, so there is no enough place for aluminium heat sink or ventilation :-(

    Nice to meet you.

    I meant " high temperature of my external device" My mobile phone's temperature is normal ( about 30 degrees celsius)
    There are USB module, AV Rx module, 5V to 3V regulator and 2 pieces diodes on my external device. When regulator and diodes become hot, my external device is discomfort for to hold with hand.
  5. bertus


    Apr 5, 2008

    Do you have decoupling capacitors on the regulator?
    Without them the regulator might oscillate and get hotter as with the capacitors.
    Decoupling or Bypass Capacitors, Why?

  6. Bipsiljo

    Thread Starter New Member

    Apr 16, 2016
  7. DickCappels


    Aug 21, 2008
    You can probably buy or build the switching regulator your need and cut your power losses and along with that your temperatures.

    Take a look at the LM2956 Simple Switcher from Texas Instruments www.ti.com/lit/ds/symlink/lm2596.pd

    There are several similar chips and assemblies available.
  8. wayneh


    Sep 9, 2010
    Yes. A DC-DC converter can be 90+% efficient, compared to 60% efficiency of dropping 5V to 3V by any linear method. That difference in efficiency will dramatically reduce heat production.
    Yes and no. They are a little bit more complex to wire but not bad. The simplest solution is often to buy a pre-made module that meets your specifications. They are inexpensive and you cannot build your own for less. But I'm guessing you would want to incorporate the circuit onto your own board with your other components?
  9. Alec_t


    Sep 17, 2013
    Dropping from 5V to 3V at 0.4A current will dissipate 2 x 0.4 = 0.8W (and hence make things hot), whether you use a linear regulator, diodes, resistors or any combination thereof. A switching regulator is the only way to avoid that.
  10. Bipsiljo

    Thread Starter New Member

    Apr 16, 2016
    Friends, many thanks for recommendations.

    I bought few different type (all adjustible output voltage) Switching Regulator-Evaluation boards from Texas Instruments. I just tested them quickly for temperature of regulators, my first impressions were good. I will finish the detailed tests within 2-3 days, i will write the results here.

    But i didn't understand something, can you explain please?

    I have a digital Power supply (0-30V , 3A with display screen) I set the output voltage as 3V of Power supply, than i connected my external device to power supply, the current drawn was 0.4A. It worked.

    Secondly I set the output voltage as 5V of Power supply, than I connected the switching regulator board to power supply, I adjusted the output voltage of regulator board as 3V (5V input and 3V output). At the end, i connected my external device to switching regulator, the current drawn was 0.22A. It worked again.

    My question is : Why current consumptions are different? One is 0.4A and other is 0.22A? Is switching regulator getting less current from Power supply?

    Last edited: Apr 27, 2016
  11. ronv

    AAC Fanatic!

    Nov 12, 2008
    The current is different, but the power is the same.

    3X.4 ~= 5X.22
  12. AnalogKid

    AAC Fanatic!

    Aug 1, 2013
    When people say a switching power supply is more efficient, that has real world energy consequences. A switching buck regulator reduces the input voltage and "amplifies" the input current.

    Direct: 3 V x 0.4 A = 0.8 W device power
    Switching: 5.0 V x 0.22 A = 1.1 W total system power
    0.8 / 1.1 = 0.727

    So the switching regulator is 73% efficient. The output current is greater than the input current, but not by an amount that exceeds how much the output voltage is *less* than the input voltage.

    So it looks like the switcher is not that great a deal, since the total power drawn is0.3 W more than before. But that's not apples to apples. If you run a linear regulator on 5 V, the input current is the same as (actually a tiny bit more than) the output current.

    Linear: 5 V x 0.4 A = 2 W total system power

    Now the switcher is looking pretty good.

  13. Alec_t


    Sep 17, 2013
    Where were you measuring current? At the input side of the regulator or the output side? (From the measurement it looks like the input side).