A problem faced using L7805CV IC

Discussion in 'The Projects Forum' started by MerlinM1, Dec 5, 2010.

  1. MerlinM1

    Thread Starter New Member

    Dec 5, 2010
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    Hey all, I am a beginner to the world of robotics, and have decided to make my first wired robot using L293D and L7805CV I have made all connections but there is this problem of heating up of L7805CV as soon as i ground the connections to get logic 0V, also there is this problem that my L7805CV doesnt give me a proper 5V when load is applied using a motor, it gives 5V perfectly under no load condition...
    Could anybody please guide me..
    also that if i use a resistor in between the ground and the circuit will it be beneficial?
    awaiting your replies
     
  2. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    Do you have a schematic of your circuit?

    Also, a 7805 can only supply about 1/2 amp without a good heatsink. What is your input voltage, what voltage are the motors being driven at, and how much current does the control/logic need?
     
  3. MerlinM1

    Thread Starter New Member

    Dec 5, 2010
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    the diagram i am posting is the one i'm using, instead of the microcontroller outputs, i am using switches to run the thing...
    My input voltage to the 7805 is 12V and the motors are 12V DC motors and i have connected the pin 8 of L293D to 12V. the logic needs to be at 5V when active high, and 0V when active low, but now due to sme changes i've made, iam getting 5.09V as active high and 1.43V as active low.. now maybe due to this 1.43V as active low, the motors are not working.. and also during my cross checking i gave direct GND to the logic pins, and at that time the motors worked, but without direct ground it doesnt work.. i think the problem is 1.43V instead of 0... please help me..
    thanx...
     
  4. retched

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  5. spinnaker

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    Oct 29, 2009
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    And what is your battery? That 7805 is not very efficient. With proper battery selection and / or getting creative with dropping voltage with diodes, there may not be a need for a regulator if you are using a battery.
     
  6. retched

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    Regulators like the 7805 work best when dropping a voltage JUST above the dropout voltage. This voltage in noted on the datasheet as Vd.

    Vd is average 2v MAX 2.5v, so trying to supply the 7805 with as close to 7.5 the better.

    Why? Every volt a linear regulator like the 7805 drops, it gets rid of as heat.

    So you need 5v from 12v.
    12v-5v=7v
    So you have to get rid if 7v at whatever current, say 1v.
    Thats 7W as heat. Thats hot.

    For comparison, if you used 5 AAA, AA, C, or D 1.5v batteries, you would be suppling 7.5v to the regulator.

    The dropout is 2v to 2.5v and that results in the voltage you want.

    Another option, using a beefy resistive divider to split the 12v to 6v and use a LDO Regulator.

    LDO stands for Low Drop Out. They need even less voltage to "do their job" (the Vd)

    So the resulting 6v with a 5v LDO will get you your 5v with minimal waste in the regulator.

    Here is a 5v LDO regulator with a Vd (dropout voltage) of .45v at 1Amp for $1.63us for single quantity.

    http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=497-6695-5-ND

    The good thing about using an LDO like this, is as your batteries drain, you keep a regulated 5v even when the battery pack falls to 5.45v. With the 7805, once the battery got below 7v, there is no more regulation.

    So in a design aspect, you spend a little more for an LDO, but you can use fewer, or less expensive battery sizes, and therefor save space and excess regulation heat, resulting in longer runtime between battery changes/charges. Obviously, CURRENT still plays a lead role in this production. If you need less than 1 amp, there are even cheaper, smaller LDOs with an even smaller Vd.
     
    Last edited: Dec 5, 2010
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  7. MerlinM1

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    Dec 5, 2010
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  8. ablemicky

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    Dec 6, 2010
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    heating up of your 7805 when powered simply means that excessive current is drawn by your robot circuit,i suggest you use a current amplifier for your regulator.
     
  9. MerlinM1

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    Dec 5, 2010
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    Alright, your idea of using lesser battery input seems fantastic, i'll try this out first, maybe the heating produced will decreases... I hope it works....
    but yet i would prefer i get to know about those capacitors that can be used to decouple the power supply...thanx again..
     
  10. MerlinM1

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    Dec 5, 2010
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    now simply how would a current amplifier help me out???... u mean to say i should reduce the input current drawn by the L293D IC somehow????
     
  11. MerlinM1

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    Dec 5, 2010
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    hmm...
    now simply how would a current amplifier help me out???... u mean to say i should reduce the input current drawn by the L293D IC somehow????
     
  12. retched

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    Dec 5, 2009
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    He doesnt know whats going on.

    Your not drawing too much current.

    The 7805 is oscillating.

    The 7805 datasheet tells you what you need.

    Top of page 4 in the datasheet shows that you need a .33uf and .1uf caps. It also shows you where they go.

    A .33uf from input to ground and a .1uf from output to ground.
     
    Last edited: Dec 6, 2010
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  13. MerlinM1

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    Dec 5, 2010
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    well, thanx for your advice, i read the thing, and now, i used them too in my circuit, but then again the same problem of heating up persists... i donno what is going on now, but still i feel there is some big problem concerned with grounding of the circuit, as soon as proper ground is applied, i see that L7805CV gets heated up to the extreme..., i tried to cross check the circuit, and i got to know that if i apply appropriate ground to the active low inputs of the controller, the motor runs, but the L293D itself is unable to do it itself, and it does so when i apply direct ground to L7805CV, but here L7805CV starts to heat up and i gotta shut the whole thing down... so u see the whole thing is connected, as i solve the problem of grounding, My L293D works, whereas, L7805 Heats up, and when I dont give GND directly, i suppose L293D doesnt get the right input, and so doesnt work... i just gotta make both the things run simultaneously if ever i wanna complete my circuit.... AaahhhHHHH I am frustrated... could u please help?... I am uploading the improved version of the ckt diag the one am currently using..
    Thanx again..
     
  14. Kermit2

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    Feb 5, 2010
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    Check for shorts with your ohm meter.
     
  15. mtripoli

    New Member

    Feb 9, 2010
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    @Merlin: A quick look at your "schematic" raises a bunch of questions. I just downloaded the data sheet for the L293D and will have a look but I don't know that you have this configured properly. The very first thing I would do is go back to the control signals; are these signals in the correct state before you fire up the motor driver? If you have two control signals active at the same time you are going to short the motors and things are going to get hot fast. Usually a microcontroller will "come up" with it's outputs set as inputs (high impedance or "Hi-Z") and you need some kind of "passive" way to set the control signals (i.e. pull-up and pull-down resistors on the control lines). This prevents the control lines from floating into a bad state for the motor driver. Once your program is up and running make sure that the control signals are in a known "safe" state. This doesn't always mean "off". At a quick glance you have the "ENABLE" pins being passively controlled; I would control these with the micro as well. I'll pencil sketch this and see...

    EDIT:
    It looks like your schematic is ok. However, I would do the following:

    1.) Put pull-down resistors on all control lines. This is a resistor connected from the signal line to ground. Something like a 47K on pins 1, 2, 7, 9, 10, 15.

    1a.) Make sure you have a common ground between this circuit and your microcontroller circuit only.

    2.) Connect pins 1 and 9 (the enable pins) to your microcontroller.

    3.) In your code make sure all outputs are set to logic 0. If you have (for instance) pins 2 and 7 (or 10 and 15) "high" at the same time your motor is in a "brake" condition (will not turn) and the driver will get very hot. Do not do this.

    4.) To make the motor run from a "start" state (all pins are off or "cleared"):
    a.) "Set" (make high, turn on) the ENABLE pin for the side you want (pin 1 in this example).
    b.) Set pin 2, clear pin 7 (7 should already have been clear). The motor will run one direction. "Clear" pin 2, the motor will turn off. "Set" pin 7 (2 is off), the motor will run in the other direction. Clear pin 7, the motor will turn off. If you PWM pin 2 (or 7) you can control the speed of the motor (get it running before you start on this).

    5.) Put a decoupling cap on pin 16 (can't hurt). Bigger input and output caps on the voltage regulator are probably a good idea as well. Make sure you have the Vreg pins connected correctly.

    I'm willing to bet that if you have it wired the way you show with no shorts its a problem of proper microcontroller configuration and not the wiring. With the pull-downs as I indicated you can disconnect the circuit from the micro and run the circuit by just pulling the ENABLE and DIRECTION pins high.



    Mike T.
     
    Last edited: Dec 6, 2010
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  16. SgtWookie

    Expert

    Jul 17, 2007
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    When the grounding on your 7805 regulator was not proper, you subjected L293 inputs to excessive voltage, which probably zapped the L293. Absolute maximum voltage for the Enable input is 7v. A 7805 with 12v in and a floating ground would output around 10v.
     
  17. MerlinM1

    Thread Starter New Member

    Dec 5, 2010
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    Hey Mike...
    I did as u said, and believe me, and am happy to tell that it worked perfectly... I am glad that i finally have completed the thing, i removed the micro and provided direct GND seperately to seperate pins using a self made controller board consisting of 4 DPST switches, i used them to mimic the action of a microcontroller, and it really worked... Now no heating up or any discrepancies are occuring except 1, and that one's a problem that while switching, when i run the motors in reverse direction, their speed decreases considerably compared to its speed during its normal; forward direction, I tried to sort out the thing, but din find a clue... i even tried changing the L293D if there might've been any problem with it, but no, the problem still persists, well, for my bot, its not a big prob as it will go in the reverse dir slowly, but still to gain knowledge i would like to know the cause of this problem... this drawing of lesser current on reversing the direction... i have 2 12V DC CCW motors, and they altogether draw a current of abt 300mA while running in the forward direction... plz help me... thanx again..

    And also thanx to all who helped me out, and made me complete my project... thanx a lot..!!!! :D:cool:
     
  18. retched

    AAC Fanatic!

    Dec 5, 2009
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    Congrats on the success, and good call Mike (mtripoli).

    Try wiring the motor backwards to see if you still get full current in the "previously reverse direction" which is now the "forward direction"

    Also, you have placed protection(back-emf) diodes on the motor lines, correct?

    If you switch from forward to reverse quickly without them, ...bad things would happen.. ;)
     
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  19. mtripoli

    New Member

    Feb 9, 2010
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    I'm glad to here you're making progress. I agree with Retched; I guess I assumed (my fault) that you have the diodes in place (see the data sheet for proper installation). There should be a "dead-time" in between going from forward to reverse as well. It doesn't have to be too long; if you have a scope you can see when all is quiet before switching. I'd say a apuse of 10-100mS would do it. If you didn't have the diodes in place you could have started to "break down" the switching transistors.

    Does the IC run hot when it's running? If so, you'll need a heatsink. The PCB (when you get there) should have large copper areas attached to the center ground pins. I haven't seen the lead frame for this device but I suspect it's ground pins are connected together (under the plastic).

    Mike T.
     
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  20. retched

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    Dec 5, 2009
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    You agree with me about:
    ?

    ;)

    tee-hee-hee
     
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