Voltage Regulator Death

Discussion in 'The Projects Forum' started by vh_man, May 10, 2010.

  1. vh_man

    Thread Starter New Member

    May 10, 2010
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    I am Constructing an Electric Bike for a school project, and I am running into the most peculiar issue.

    the PWM for the speed control is generated by a counter, Magnitude comparator, and A/D converter and a few other components, all requiring 5V for operation. the motor runs on a 24 Volt SLA battery, which I am switching with a 60A MOSFET. I have all protection on the motor, like freewheeling diodes and filtering capacitors. however, the issue does not remain here.

    scince the past monday I have destroyed about 10 5V Voltage regulators. I hook them up right, I attach them to a HUGE heat sink, and they constantly self-destruct after about a minute of use. For the record, If I run the 5V part off of my benchtop supply, I draw no more than 100 milliamps, so I'm not shorting the regulator. the regulator is supposed to have a Vin of 8 to 36 volts, which is fine for being powered from a 24 volt SLA battery.

    Getting tired of blowing fuses (thank god they were there... I would be replacing quite a bit of IC right now if they weren't) I simply hooked another regulator up without anything on its output except for a 1K resistor. I ran my circuit for a while. And the regulator blew up. With no load.

    I am at a complete loss. I can understand the regulator dying from over-current, or over-voltage, but there is nothing of the sort occuring. What could possibly be destroying my regulators? my best guess is back-emf, but I have freewheeling diodes to protect against that. And if that was the case my MOSFET's would fry, not my regulator.

    Also, if I JUST hook a regulator to the output of the battery, it works fine. It has to be something in the power area that is causing this.

    schematic:
    [​IMG]

    I dont have the PWM controller's schematic, but I do not believe that it is the cause of the problem.
     
    Last edited: May 10, 2010
  2. SgtWookie

    Expert

    Jul 17, 2007
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    Well, you haven't posted your circuit so it's going to be tough to help.

    Post your schematic. Make it easier on those who would like to help you. You'll get better answers a lot faster.
     
  3. SgtWookie

    Expert

    Jul 17, 2007
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    Just for grins, have a look at the attached simulation.

    I'm figuring that you have rather large inductive spikes on your supply, despite your use of caps and flywheel diodes. "Batt" represents your 24v battery. "Spikes" is a very rough wild guess at what kind of transient spikes you might be having from the motor's operation.

    The combination of Rbatt, C1 and C3 act as a low-pass filter to eliminate most of the transients. Rbatt being 1k Ohms, it won't allow much current flow - if your load is over around 12mA average, it will be too much resistance.

    The LM317, R1,R2 are a rough substitute for a 7805 regulator - close enough.
     
  4. vh_man

    Thread Starter New Member

    May 10, 2010
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    So your saying I need a low-pass filter on my power supply?

    what about some kind of cap on the input side of my regulator? either that or some kind of zener protection before the actual voltage regulator.
     
  5. SgtWookie

    Expert

    Jul 17, 2007
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    Your C2 is FAR too large for being across the motor, and also it's a polarized capacitor.
    It is very likely blown.

    For a cap across a motor, use <1nF (0.001uF) and non-polarized. The idea of such a cap is just to "buy time" while the flywheel diode switches. 220pF to 1nF ceramic would be OK. Start towards the smaller end.

    You have no cap on the 7805 input. Not good.

    Figure out what your average 5v load current will be.

    Then referring back to the schematic I posted, figure Rbatt needs to be about 15/(avg_load_current+6mA).

    For example, if your average load current is 30mA, then Rbatt=15/(30mA+6mA) = 15/.036 = 417 Ohms. 390 Ohms would be OK.
    Since you're decreasing Rbatt, increase C3 to keep about the same filtering ability.
    330uF*1k/390 = 846uF. 1,000uF would work.

    Don't forget to calculate the power requirement for Rbatt. In the above example, it's 15v*36mA = 540mW. Double that for reliability, and you get 1.08 Watts. In this case, you might "squeak by" with a 1W resistor, but it would be better to use a 2W resistor.

    C1 and C2 are mandatory, and must be as close to the regulator as possible. They should be ceramic or metal poly film. If you omit them, the regulator may oscillate at high frequencies.
     
  6. ronv

    AAC Fanatic!

    Nov 12, 2008
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    It looks to me like c2 was ment to be from +24v to ground not + 24 to the top of the FET. If it is really wired this way that's your problem.
     
  7. retched

    AAC Fanatic!

    Dec 5, 2009
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    ronv, Im missing what you are speaking about.

    The cap is across the motor leads, and is from 24v to ground. The PNP is switching on the motor to ground via the pwm signal.

    But there are two "C2" in the diagram, but only the right-most is on 24v. Is this what you were referring to?
     
  8. SgtWookie

    Expert

    Jul 17, 2007
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    I've already covered this. It's far too large, and it's a polarized cap.

    It should be a non-polarized cap less than 1nF.

    It's across the motor leads. It would have been fine if it were across +24v and GND; but as shown it's not good at all.
     
    retched likes this.
  9. eblc1388

    Senior Member

    Nov 28, 2008
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    @vh_man,

    The voltage regulator output capacitor value is exceptionally large and you have forgotten to put a protection diode from voltage regulator output to its input.

    This diode provides a current path and prevents the 7805 output capacitor from discharging back to its input through the 7805 internal. Without it the current spike will destroys the 7805 as soon as you removes the +24V.
     
  10. vh_man

    Thread Starter New Member

    May 10, 2010
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    Thanks so much for your help.

    I replaced the motor cap with a 1nf capacitor (and freed up some serious board room), and added a .01uf cap to the input of the regulator. I also decreased the output cap to 1uf, which was the manufacturers recommendations. This semi-worked, but I still killed the regulator. Thinking back to "spikes" I added a ferrite-core Inductor that I pulled from an old computer power supply into the positive supply going to my regulator. this effectively stopped my spikes from reaching levels that were dangerous to the regulator. (checked with my scope)
     
  11. russ_hensel

    Well-Known Member

    Jan 11, 2009
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    eblc1388 said:

    @vh_man,

    The voltage regulator output capacitor value is exceptionally large and you have forgotten to put a protection diode from voltage regulator output to its input.


    He is right. Also any time the motor acts as a generator it could produce a output voltage larger than the input.
     
  12. Norfindel

    Active Member

    Mar 6, 2008
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    9
    What about using a diode to make sure no current can go from the motor to the battery section, if it happends to work as a generator?

    What were you seeing in the oscilloscope? What voltage values are reaching the point before the inductor you added?
     
  13. vh_man

    Thread Starter New Member

    May 10, 2010
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    This is what the freewheeling diode does.

    Also, I was seeing spikes of upwards of 80 Volts. With the Inductor it reduced them down to about 38, which seems to work fine with the max voltage of 36.
     
  14. Norfindel

    Active Member

    Mar 6, 2008
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    9
    But with the diodes as drawn in your schematic, if you have 80v at the battery + terminal, they won't be solving that problem. I was thinking about a diode inserted into the wire that goes to the upper motor terminal, so that no voltage higher than 24v can reach the battery.
     
  15. vh_man

    Thread Starter New Member

    May 10, 2010
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    those diodes are the freewheeling dioes to protect for back EMF. they do nothing while the motor is running.
     
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