Voltage regulator

Discussion in 'The Projects Forum' started by chimera, Nov 11, 2010.

  1. chimera

    Thread Starter Member

    Oct 21, 2010
    122
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    Hello

    I have been working on a self project--a car-robot that uses a proximity sensor to stop when faced with a solid obstacle and no more than that--The current that the entire system draws(as ive designed it) is 0.54 Amps. I am currently using a battery used for flying Rc planes. It can provide 9.6V at 1000mAh (as rated). I figured it can do that job with a voltage divider.

    1- Ive set up a voltage divider with 10Ω, 10 watt resistor. Well that is because i wasn't to sure of a voltage divider; so i went the other option. The heat generated by this ceramic resistors is insane and i don't like the design of it. I use the output of the voltage divider to power the car-robot

    2- The battery when charged by its charger charges up to 11.7 volts. When connected across the voltage divider, the car moves (after an initial push to over come the static friction).

    3- When i measured the battery at time full charge, it read 11.62 volts. That is alright since the voltage divider still did the trick. But the battery drains out too soon

    4- I only have two questions: 1- Why the does the battery drain so quickly as it says that it can provide up to 1000mAh. I only need 0.54 amps at any given time. 2- If i use the voltage regulator IC LM317, how do i calculate the current through the adjust pin and the output current. I ask this so that i can calculate the power dissipated through the resistors, if i go with the regulator,

    Thanks for the time and as always, any reply is a good reply..(as long its relevant..:p)
     
  2. eblc1388

    Senior Member

    Nov 28, 2008
    1,542
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    0.54 Amps at what voltage? 3.3V, 5V or 0.2V or you don't know? Is it going into a motor?

    It will never works using this setup. You are more likely to end up with a free space heater as well.

    Depending on the load voltage, your best option at 0.54A is a switching power supply to give you the required output voltage.

    What do you mean by too soon? 10 minutes, 1 hour or what?
     
  3. chimera

    Thread Starter Member

    Oct 21, 2010
    122
    2
    hmm..the voltage the circuit works on is 5 volts and the battery drains in less 7 mins. I forgot to mention these in my earliar posts. However, im more interested in knowing how to calculate currents through the LM317 IC. I want to limit the output voltage of the LM317 to 5 volts and have a current a of 0.55 Amps drawn through it (of course with the load attached).

    But since im dealing with big current values, i need to power resistors. But at what rating? To figure this out, i need to know how to determine the currents drawn through the adjust pin of the lm317 and the output. I hope this explains it.
     
  4. nomurphy

    AAC Fanatic!

    Aug 8, 2005
    567
    12
    9.6V - 5V = 4.6V
    4.6V * 0.55A = ~2.5W that the LM317 needs to dissipate.

    You can reduce the burden on the regulator by placing a series resistor at the input to help drop the voltage and dissipate some heat:

    0.55A * 4.7 ohms = 2.6V
    0.55A * 2.6V = ~1.5W : will need a 4.7 ohm >3W resistor
    9.6V - 2.6V = 7V : input to regulator at max current
    7V - 5V = 2V : regulator dropping voltage
    2V * 0.55A = 1.1W : LM317 dissipation (will still need heatsinking)

    Designs such as these are very inefficient. A switching regulator would be much better, but may be beyond your scope.

    Also, the above does not take into account all the aspects of what you are trying to do and is somewhat simplified.
     
    Last edited: Nov 11, 2010
  5. chimera

    Thread Starter Member

    Oct 21, 2010
    122
    2
    I like the disclaimer.:D. However, there isant much to it. The battery is used to power a motor and a proximity sensor. The sensor uses two IR leds usinf 70mA and a IR phototransistor using up almost 13mA. There is a power BJT used to act as a switch in between the motor and phototransistor. When the there is no obstacle, the motor will continue to run. When there is an obstacle, the motor will stop till the object is removed or direction is turned. Dosent seem too complecated. So would the regulator work instead of a power voltage divider.?
     
  6. SgtWookie

    Expert

    Jul 17, 2007
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    Since the load is a motor, you would likely be much better off to use PWM to control the current flow; the motor itself acts as the inductor. This would be a very efficient use of battery power.
     
  7. chimera

    Thread Starter Member

    Oct 21, 2010
    122
    2
    okay i never understood this thing. To PWM something, a 555 timer comes to mind. But how will the output from the timer chip power up the motor. As in what voltage and current will the 555 timer deliver at its output.

    But on a seperate note, designing the regulator would mean that i need power resistors and size is an issue. SO i was thinking of getting a rechargable battery that can deliver 6 volts with a 1000mAh rating. How does that sound?
     
  8. retched

    AAC Fanatic!

    Dec 5, 2009
    5,201
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    The output of the timer will connect to a transistor or MOSFET to switch a higher power buss to the motor.
     
  9. SgtWookie

    Expert

    Jul 17, 2007
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    Here's one way to do it using a 555 timer, a couple of transistors, and a few resistors/capacitors.

    See the attached schematic and simulation waveforms.

    You can even get all of the parts from your local Radio Shack store.

    Radio Shack only carries one 1 Ohm resistor, a rather large concrete power resistor. It'll work, though. As shown, the circuit will allow just over 1/2A current flow through the motor. It's not the most efficient design due to the power dissipation in R4 (about 270mW), but it's far, far better than you could hope to achieve with any linear regulator.
     
  10. chimera

    Thread Starter Member

    Oct 21, 2010
    122
    2
    Great reply Sgt! Thanks for the attachment :D. But I was going to try a battery before I go all fancy with PWM motor. my circuit needs 0.54 amps at 5V to operate. How do i calculate what battery i will need --- as in what rating of the battery should I be looking at?
     
  11. SgtWookie

    Expert

    Jul 17, 2007
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    Well, the battery you have is rated 9.6v @ 1000mAh, right?

    If you used the circuit I posted with it, that should give about an hour or so of run-time. You probably won't have to push-start it, either.

    What kind of run-time were you expecting, or do you need?

    The larger and heavier your battery is, the more power it's going to take to get your robot moving.
     
  12. chimera

    Thread Starter Member

    Oct 21, 2010
    122
    2
    yeah, thats the rating I have on the current battery. But i am using a voltage divider to drop it down to 4.2 volts. This is because the only power resistors that I have are 10 ohms at watts. So with this, the only voltage divider i could set up is of a 4.2 V. Since My circuit needs 5 volts to operate, I was thinking of just getting another battery at 5V @ 1000mAh (still need to find a place to buy it). In this way, I wont need the 555 timer circuit you suggested (to reduce complexity of the over all design)

    To answer your question, the run time i want is no more than 10 mins at the max. This is my first robot and I wanted to see how things go. So the performance stats such as run time and efficiency arent important. I hope ive portrayed some detail. If you want, I can attach a schematic of my circuit in the next reply. That way, u could see what ive set up.
     
  13. SgtWookie

    Expert

    Jul 17, 2007
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    Schematics of the circuits as they are originally, help to eliminate ambiguity that's all too common with text explanations.

    Are you using a microcontroller?

    Post up your circuit, and we'll go from there.
     
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