Voltage regulator allowable "headroom"

Discussion in 'General Electronics Chat' started by Andreas, May 2, 2012.

  1. Andreas

    Thread Starter Active Member

    Jan 26, 2009
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    4
    Hi,

    I'm using a small +5V (TO-92) volatge regulator, namely a LE50CZ by ST electronics. I've calculated that my load devices should consume less than 100mA of current so this device should be OK. However the spec sheet says that it can handle a Vin = 20V (max) or 18V (max) !!! depending on what part of the spec sheet you look at. Now common knowledge suggests that the input to output should not exceed 2 to 3V else the thing will overheat and may shut down but they never seem to tell you this in the spec sheet other than a Test Condition that happens to be 7V.

    In any case I need to run my system (for only 2 hours) from a 12V 1.2Ah SLA battery so am proposing to use a basic volatge divider (on the front end before the input of the Vreg) using 47 Ohm (1 Watt) and 68 Ohm (1 Watt) resistor to ground. Tapping off in the middle will result in 7V (and give 104mA of current) arriving at the input of the Vreg.
    (If I use resistors of higher values then I will increase my overall battery run time and can use smaller Wattage resistors. But all at the cost of reducing my available output current).

    I realise that the voltage divider is necessary to drop the input voltage to a suitable level but wastes energy in the form of heat. Is this good standard practice or is there a method that is more efficient and better suited than this approach?

    Hope there's enough info here for a resonse.
    Thanks.
     
  2. #12

    Expert

    Nov 30, 2010
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    Your voltage divider is wrong. The regulator can't pull 100 ma off the side of a 104 ma string. Whatcha need is a series resistor at 10 volts per ohm eating up some supply voltage.

    The attached datasheet for a random TO-92 package shows 200C per watt so you have to stay below 1/2 watt (5 volts headroom) on the chip.

    about 39 ohms, 1 watt should do it.
    and it's fair game to glue a dime or a quarter to the chip as a heatsink.
     
    Last edited: May 2, 2012
  3. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    Look at its datasheet. The LE50 is a low-dropout 5V regulator. Its input must be at least 5.4V. It is tested with a 7V input because its output voltage changes a little when the input voltage changes. With a 7V input then the output is 5.00V. With a 6V input then the output might be 4.99V. With a 5.4V input then the output might be 4.90V.

    You DO NOT feed it from a voltage divider because the resistor to ground simply throws away battery power.
    Use only one series resistor to share the heat. If the load draws less current then the input voltage will rise and not do any harm.

    The datasheet does not show its thermal resistance but its maximum dissipation is probably 625mW like little TO-92 transistors.

    Its output capacitor MUST BE at least 2uF and is typically 10uF. It is tested with a 2.2uF output capacitor.
     
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  4. KJ6EAD

    Senior Member

    Apr 30, 2011
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    Instead of a big resistor and a little regulator, why not use an 8V regulator ahead of a 5V regulator (both TO-220's) so you can easily bond to a sink as needed.
     
  5. crutschow

    Expert

    Mar 14, 2008
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    Why would you use two regulators instead of one with this low current (if it is in a TO-220 package on a heat sink)? :confused:
     
  6. Andreas

    Thread Starter Active Member

    Jan 26, 2009
    68
    4
    Yes No.12, thinking about it, now that you've shed light on it. I have seen the light!
    I don't know why I thought that about the current. In any case, I see your point so thank you for your insight.

    P.S. Due to limited space in my box and the current demands of the load I decided that a T-220 style package was not necessary, hence the smaller TO-92 case.
     
  7. #12

    Expert

    Nov 30, 2010
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    Geez. I said that wrong. It's 10 ohms per volt.

    Meanwhile, is a 1 watt resistor plus a TO-92 chip going to be more convienent to the space requirements than a TO-220 package? You have to make a judgement call on that.
     
  8. nomurphy

    AAC Fanatic!

    Aug 8, 2005
    567
    12
    The way you go about this is to reason that you want 5V from 12V, that means 7V will be dropped across the regulator (12V-5V = 7V). You state the current is less than 100ma, but not the real current, so I will assume 100ma. 7V x 100mA = 700mW dissipated by the regulator. You can probably see that this is not practical with a TO-92 case.

    So, consider that you would prefer a minimum 7V input to produce 5V, that would be a 5V drop across an input resistor (12V-7V = 5V) and at 100mA would be 5V/100mA = 50 ohms. And, 5V x 100mA = 0.5W dissipated by the resistor. This would require a 50 ohm 1W (min) resistor. Placing a cap between the resistor and the regulator also creates a filter Fc = 1/(2pi*R*C).

    With this input resistor, the regulator would dissipate (7V-5V = 2V) and 2V x 100mA = 200mW.

    Once you know the max load current, and if it is significantly different than 100mA, you may want to recalculate and tweak the resistor value.
     
  9. KJ6EAD

    Senior Member

    Apr 30, 2011
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    Just to distribute the heat over two stages so with the heat sinking effect of the tabs an additional sink may not be needed. A big resistor is no smaller than a regulator and the project is a one off so cost is not a factor.
     
  10. crutschow

    Expert

    Mar 14, 2008
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    At 100mA current (perhaps 1W max.) I don't think there's any need to distribute the power over two tabs.
     
  11. #12

    Expert

    Nov 30, 2010
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    While there is no need to "double" regulate this one, I did one some years ago where the (2) LM317 regulators divided the excess voltage between themselves. As the output voltage was adjusted lower, the excess voltage was distributed evenly so each chip dissipated the same wattage.

    Just letting you know it can be done when needed for a wide voltage range adjustable regulator with plenty of excess voltage.
     
  12. KJ6EAD

    Senior Member

    Apr 30, 2011
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    Of course there's no need. I just prefer it to a resistor and know it will run cool even if the load is higher than expected. I thought overkill solutions and personal preference were allowed.
     
  13. Andreas

    Thread Starter Active Member

    Jan 26, 2009
    68
    4
    Yes, I concur with your logic. I chose a load current of 100mA to be on the safe side (it's more like 80 to 90mA in reality).

    I found the data sheet from Nat Semi for the LM2940T (TO-220 package) more useful than the LE50Cs (TO-92) data sheet from ST and so found that the Power Disipation could be anywhere from 600mW and higher depneding on what I_{G} could be.
    In any case feeding worst case and best case values into the equation led the Junction-to-Ambient Thermal Resistance for the LM2940T to be greater than the 53°C/W so no heat sink should be needed. The LE50C data sheet however shows that for a TO-92 package this to be 200°C/W which is higher and "might" shut down due to its internal thermal limit.
    I think I might err on the side of caution and go with the LM2940T as it simply has more robust specs and comes with its own inbuilt heatsink.
    I can't see a need for an input cap as it's running off a battery, not a mains driven PSU. Incidentally, a 1W resistor is only 11.2mm x 4.2mm so I can squeeze that in.

    Thanks for all the replies.
     
  14. #12

    Expert

    Nov 30, 2010
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    Most voltage regulators require a capacitor at their input to compensate for the inductance in the wire from the battery (or main capacitors) and another on the output to suppress oscillation. Be careful to evaluate whether you will need "local" capacitors in your particular circuit.
     
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