High current draw causes 7808CT Regulator to shutdown?

Hi guys, I'm currently working on a parallel parking system project for my final semester. To make things short, i need a constant 8v DC supply to my motor driver so that i'm able to monitor the delay when i have a constant speed.

Anyways, before trying the voltage regulator with it's capacitors and diode connected, my robot was able to move forward and brake(reverse DC motor for 50ms at max speed) and proceed with the code. However when i implement the 7808CT regulator, it would work up till forward and resets my Arduino every single time.

My suspicion is that, the current limiter in 7808CT causes the 8V regulator to shutdown since the current draw would be much higher to reverse the DC motor at full speed, therefore shutting down the 8V regulator and turning it back on.

I've spent hours on this and i'm just clueless. Not sure if my 7808CT is broken or do i need to add BD536 for higher current output preventing it from rebooting my arduino?

All help is appreciated! thanks!

-Matthew

Update:
- Hand-drawn 8V L7808CT connected to L298N Motor Driver leading to right/left motors in separated parallel connections. [/i][/b]

When reversing a brushed DC motor at full speed, it briefly draws twice the stall current. So, don't. Wait til it comes to a stop before reversing.

Isn't braking normally done by shorting the motor terminals?
Maybe with the NC contact of a relay.
Leo..

Hi,

Can you please post a copy of your circuit, in CAD or a picture of a hand drawn circuit in jpg, png?

What is the spec of your motor?
Does it have to use the regulated 8V supply?

If you remove the motor, does the problem still occur.
What I am trying to see is if it is a motor problem or a circuit configuration problem.

Tom.....

A circuit and program would make it easier to see what you are doing.

Weedpharma

@zinxer, please do not cross-post. Threads merged.

jremington:
When reversing a brushed DC motor at full speed, it briefly draws twice the stall current. So, don't. Wait til it comes to a stop before reversing.

well that i'm sure pretty much draws more current than the 8v reg can supply at only 1A.

Wawa:
Isn't braking normally done by shorting the motor terminals?
Maybe with the NC contact of a relay.
Leo..

I'm using a brushed DC motor if it makes any sense

TomGeorge:
Hi,

Can you please post a copy of your circuit, in CAD or a picture of a hand drawn circuit in jpg, png?

What is the spec of your motor?
Does it have to use the regulated 8V supply?

If you remove the motor, does the problem still occur.
What I am trying to see is if it is a motor problem or a circuit configuration problem.

Tom.....

Above is a sample Image of a hand-drawn rough sketch of my circuit. The 8V regulator and the connections to the Brushed DC motor can be seen at the top right. The motor is rated at 6-9V. I need it at 8V because it provides sufficient torque for my motors at this level and i also need it to be at constant 8v. When i remove the motor, the problem does not occur. Does this mean i need to include a transistor to increase current output?

weedpharma:
A circuit and program would make it easier to see what you are doing.

Weedpharma

got it up(;

Oh i'm terribly sorry, it won't happen again.

p.s. Thanks everyone for the quick response!

-MatthewPrag

Hi,
Yes it does sound like a powersupply problem.
Try connecting the input of the LM7805 to the battery, rather than the output of the LM7808.

Have you got a heatsink on the 7808 regulator, what sort of wire are you using to supply power to the motor drivers.
Are the wires going directly from the drivers to the LM7808 and not via connections on other components.
Can you post a picture of your project so we can see your layout please.

Also place 0.1uF capacitors across the terminals of each of the motors please, to reduce any commutator noise.

Tom...

Hi Tom,

I've actually tried connecting the input of a 7806 to the battery and it works perfectly fine except that the motors are struggling to move due to the low starting torque power provided by the 6V reg. The 7808CT is fitted with a heat-sink tho, and the wires i used are the regular ones :

Yes, the wires are directly connected to the motor driver without any components to bypass from a 9.6V(11V @ full charge) Nicd battery. I'm not sure if it means anything but both right wheels are in parallel connection as well as the left ones.

Thanks Tom!

-MatthewPrag

You don't use linear anything between the power source and a motor, all the control is
switch-mode. High power circuitry needs to have low power loss which means its
either very low resistance or very high resistance, not somewhere in the middle (which
would mean dissipating as much power as the load).

So your motor driver H-bridge should be powered directly from the power source.

And it has to be rated to take twice the stall current if you abruptly reverse (hint,
never abruptly change motor drive, always ramp it up and down).

[ Note with very large motors the motor will typically be destroyed if you don't limit
the current below the stall level ]

Hi,
Why when you connect the input of the LM7806 to the battery do the motors then run on 6V.
They should be powered by the 8V regulator all the time?

Please explain?

Tom...

MarkT:
You don't use linear anything between the power source and a motor, all the control is
switch-mode. High power circuitry needs to have low power loss which means its
either very low resistance or very high resistance, not somewhere in the middle (which
would mean dissipating as much power as the load).

So your motor driver H-bridge should be powered directly from the power source.

And it has to be rated to take twice the stall current if you abruptly reverse (hint,
never abruptly change motor drive, always ramp it up and down).

[ Note with very large motors the motor will typically be destroyed if you don't limit
the current below the stall level ]

Hi Mark, thanks for the feedback. Unfortunately if the motor driver were to take the direct power from the battery then the supplied voltage to the motors will also receive the direct voltage. Unless i could imagine having four L7808 regulators 2 for each parallel circuit motors where each will be used to drive when the polarity is switched(to turn back and forth). As far as things goes, i am pretty sure the resistance is too high for my l7808 to handle and thus switching it off and then back on. Could this actually be the cause?

TomGeorge:
Hi,
Why when you connect the input of the LM7806 to the battery do the motors then run on 6V.
They should be powered by the 8V regulator all the time?

Please explain?

Tom...

Hi Tom, oh i'm sorry. The intention of swapping the l7808 with a L7806 was to test if it was the 8V reg problem and it seems that a L7806 reg seemed to work well but have issues starting the torque. With a little push i was able to let it run for abit and it was able to proceed with the program even with the sudden reverse(brake). I assume that the load at that time was not as high compared to the faster rotation of the reverse @ 8v. Therefore, the reversing current was not high as compared to the L7808 to shut it down?

-MatthewPrag

Wait-you have a 9.6 volt battery? The dropout for a LM7808 regulator at 1A is ~1.75 volts. Use a switch-mode regulator. I will post a link.

Hey Isaac, thanks for pointing that out, i forgot about that! Anyways i have tried to put in series with a fully charged 7.2V + 9.6V Nicd direct to the L7808 with a heatsink. For the first 3 runs, the robot moved as intended but after the 4th try, it returns to the same issue again where it would reboot my Arduino(L7808 shutdown and back on). The L7808 and heatsink seems to be quite hot if it help to explain any?

Anyways, i will be heading to a local electronic store tomorrow and construct this circuit:

-MatthewPrag

zinxer:

Wawa:
Isn't braking normally done by shorting the motor terminals?

I'm using a brushed DC motor if it makes any sense

See the L298 datasheet (click !), Figure 6 on page 6.
At the right,a truth table is shown, telling you what happens on which levels at the controlling pins.
See the function of C = D ?

Try that.
It will save some power, as well as other problems.

MAS3:
I'm using a brushed DC motor if it makes any sense

See the L298 datasheet (click !), Figure 6 on page 6.
At the right,a truth table is shown, telling you what happens on which levels at the controlling pins.
See the function of C = D ?

Try that.
It will save some power, as well as other problems.

void halt()

{
  digitalWrite(in1, LOW);  //right front and rear wheels
  digitalWrite(in2, LOW);

  digitalWrite(in3, LOW);  //left front and rear wheels
  digitalWrite(in4, LOW);

}

This is where "C=D", but because this only releases the dc motor, the motors still coast. i had to reverse the motors so that it would stop/brake exactly at a point of delay. I understand however that this may draw up to twice the amount of stall current. My only problem now is to find a way to bypass the 1A limiting "Io" from the L7808 as it gets really hot even with the heatsink. I also notice that my 1602LCD dims everytime it brakes, this probably shows that the Voltage regulator is struggling to supply enough current to overcome the load.

Did you try to do this:

void halt()

{
  digitalWrite(in1, HIGH);  //right front and rear wheels
  digitalWrite(in2, HIGH);

  digitalWrite(in3, HIGH);  //left front and rear wheels
  digitalWrite(in4, HIGH);

}

It shouldn't matter whether you do this by setting them both HIGH or both LOW, but it won't hurt to try this, right ?
You need to have the ENable pin HIGH to have this to work.
Best is to set EN low, change in1 through in4 to whatever you need them to be, and then set EN to HIGH.
Disabling EN will set the free running of your motor.

I'd say create 2 power supplies, 1 for the motor controller, and another one for the other electronics.

MAS3:
Did you try to do this:

void halt()

{
 digitalWrite(in1, HIGH);  //right front and rear wheels
 digitalWrite(in2, HIGH);

digitalWrite(in3, HIGH);  //left front and rear wheels
 digitalWrite(in4, HIGH);

}

It shouldn't matter whether you do this by setting them both HIGH or both LOW, but it won't hurt to try this, right ?
You need to have the ENable pin HIGH to have this to work.
Best is to set EN low, change in1 through in4 to whatever you need them to be, and then set EN to HIGH.
Disabling EN will set the free running of your motor.

I'd say create 2 power supplies, 1 for the motor controller, and another one for the other electronics.

This is what i have tried to do to perform the brake, but it seems to still coast the motors.

void forwardbrake()     

{
  //right motors:
  digitalWrite(in1, HIGH);        
  digitalWrite(in2, HIGH);
    //left motors:
  digitalWrite(in3, HIGH);
  digitalWrite(in4, HIGH);
  digitalWrite(enA, HIGH);
  digitalWrite(enB, HIGH);
}

Anyways i have also carried out some readings on the overall circuit and here are the data i've retrieved:

No load current (booting): 0.21A
Stall current(all motors forced stop) : 0.89A
Operating current( motors running): 0.62A

The current reading was retrieved without the use of the voltage regulator. As for now, my only resort is to add a power transistor to allow more current output and remove some load off my Voltage regulator as it gets really hot very quickly. Could anyone provide any suggestions or calculation for the implementation of a power transistor onto the circuit?

Thanks everyone for the help!
-MatthewPrag

Try parallel voltage regulators.

Isaac96:
Try parallel voltage regulators.

If two identical voltage regulators were to be paralled, the regulated voltage will still differ slightly and it will cause one regulator to withstand all the load and it would shutdown followed by the other. It doesnt seem like a solution for this Will update here later if it works by including a power transistor to take off some load off the 7808reg.

-MatthewPrag