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I have connected 6v to the power jack of my arduino, and when I tested the output voltage from the Vin, it turns to be 5.3v with no connection and 4.8 with some connections to the I/O pins.

IS that normal? Because I was trying to supply a 6v motor through H-bridge, but the voltage at the motor decreased down to 3v!

Is it possibly because of a protection diode in both the arduino and the H-bridge? Is it possible that these diodes can pull down the voltage from 6v to around 3.1 v ?

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2 Answers 2

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The most likely thing that is wrong is you don't have enough current.

I would add another battery in series. (You said you had four in series now.)

If the current supplied by the power source is used up, and the motor is pulling more, the voltage will drop for wall adapters. I don't know for sure about batteries, but my guess is that will happen too.

Your batteries may not have enough current to be supplied at once. Batteries have internal resistance (that varies). What that means is there is limits to the instantaneous current it can supply. EX: You cannot draw 500A from one battery.

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    \$\begingroup\$ VIN is before the regulator. A supply to the DC Barrel goes through a diode, to the VIN pin. None of the 5v/Regulated parts apply. \$\endgroup\$
    – Passerby
    Commented Aug 11, 2013 at 0:41
  • \$\begingroup\$ Iam using 2 (6v DC motors}) with stall current 1amms. I'm using 4 AA Alkaline batteries with 1800-2600 mAh charge stored. Nothing connected with the motor except a blutetooth module. The connection is as follows: 4 controls pins of the H-bridge (L298n) are connected to 4 output pins on the Arduino. 2 Enable pins connected directly to another 2 output pins. The battery pack with 4 AA Alkaline is connected to the power jack, the (Vin) oin is connected directly to the (Vcc) (power supply) of the H-bridge. The GND of the H-birdge am=nd 5v are connected as well. \$\endgroup\$
    – Adel Bibi
    Commented Aug 11, 2013 at 7:56
  • \$\begingroup\$ What is the battery brand you're using? You may not have enough current supplied. Batteries have internal resistance (that varies). What that means is there is limits to the instantaneous current it can supply (1 AA can't full run a 500A motor at full speed). Try hooking it up to a power supply that is 2.5A+. (1A (Motor 1) + 1A (Motor 2) + 1/2A (Arduino, Bluetooth, and extras)). In fact, I doubt that four AAs battery can provide that much current at once. I don't know how many you would need, but you may have to use lots of 9Vs or many AAs. \$\endgroup\$ Commented Aug 11, 2013 at 15:31
  • \$\begingroup\$ I'm using ALKALINE batteries (AA-AM3-MN1500) I have just tried out using 5 batteries in series, and it worked pretty much fine! I guess the problem was with the (Inrush current) that is first drawn by the two motors, the extra battery helped in solving the issue. Thanks alot for your help! :) \$\endgroup\$
    – Adel Bibi
    Commented Aug 11, 2013 at 21:47
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    \$\begingroup\$ @AdelBibi you have 24 points now, now you can vote. \$\endgroup\$
    – Passerby
    Commented Aug 12, 2013 at 19:08
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The protection diode on the DC Barrel socket will not provide a 3v drop. The M7 Diode has a maximum Vf of 1.1V at 1A power draw. At less, the Vf is lower. What will provide that much of a drop is an unregulated power supply. Or a power supply that isn't strong enough for your motor. Or a faulty power supply.

Update: OP has mentioned that they are using 4x AA batteries, and 2 motors with 1 Amp stall current. Considering in-rush current, passive h-bridge components, and figuring 75% of stall current as no-load current (this is just a random high figure), that is still 1.5 Amps. First, some batteries, especially your average AA batteries, cannot provide 1A or more current. Those that can, will quickly deplete themselves. Duracell Procell line of AA batteries (Datasheet), a 1 Amp draw will drop a fresh battery to 1.1V from 1.5v in under 30 minutes. Down to 1V in an hour. Completely dead. 1V * 4 = 4v. Factor in the diode voltage drop of 1V at 1A, it's down to 3V, what you measured. Add to that the Arduino's current draw itself through the 5v regulator(Figure 100~200mA) .

Further more the M7 Diode (Datasheet) has a Maximum average forward rectified current at TL =55° of 1 Amp. If it wasn't for the voltage drop on the batteries, which also means a current drop, that diode would be dead.

In short, most batteries are not very well suited for driving large Amperage draws, like the motors plus everything else, nor is it good to drive more than 1 Amp through the protection diode on the barrel jack. This is your problem.

To fix: 1, Consider a wired power supply, or lithium or other high current battery packs. 2, Do not power the motors through the protection diode, instead connect them directly to the power source. Use the DC jack just for the Arduino and other small parts (like the bluetooth module) 3, Consider smaller motors. 4, Separate power supplies for the Arduino and the Motors. Ideally, you want 5 or 6 1.5v AA batteries at the DC Jack or VIN pin for proper regulation. 4 1.5v AA batteries is cutting it right at the bare minimum for the 5v regulator on the Arduino Uno. If you can only use the 4 1.5v batteries, then connect it to VIN pin, NOT the DC jack, because the protection diode will drop 0.7v making the problem worse.

Just as a side note, the 5v Regulator on the Arduino has a 2v drop on it. 7v is the minimum for proper regulation on it. Powering a 6v Motor with a 7v supply is okay though in most cases, just results in a slightly larger current draw.

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  • \$\begingroup\$ So Mr Passerby, Do I understand that powering my arduino from 6v battery pack will cause this much a voltage drop! If I used larger values will this simply be regulated and has a voltage drrp of 0.7v due to the protection diode? The arduino data sheet says that the recommended power supply is 7-12v ... Does this mean any values below 7v will cause a problem. i.e a great voltage drop \$\endgroup\$
    – Adel Bibi
    Commented Aug 11, 2013 at 8:01
  • \$\begingroup\$ @AdelBibi you didn't mention battery. I thought you had a power supply/ac adapter/wallwart. Yes, depending on the batteries, a huge power draw can cause the voltage to drop a lot. Cheaper batteries will be worse than high end batteries. ~0.7V is the average diode drop yes. And any thing below 7v at the barrel socket or the vin pin will not be properly regulated by the 5v regulator, but in your case, this is not the problem, since you are taking power from the vin pin, which is before the regulator. \$\endgroup\$
    – Passerby
    Commented Aug 11, 2013 at 8:10
  • \$\begingroup\$ @AdelBibi please see the update. \$\endgroup\$
    – Passerby
    Commented Aug 11, 2013 at 10:18
  • \$\begingroup\$ But why does the voltage at the (Vin) of the arduino loses some of it's voltage! I measured the output pin of the (Vin) and it was around 1.3 v less than the supplied voltage! \$\endgroup\$
    – Adel Bibi
    Commented Aug 11, 2013 at 21:48
  • \$\begingroup\$ @AdelBibi The VIN is after the Protection Diode so you have that voltage drop (~0.7V to 1V depending on current). Measuring after the Diode will cause it to measure (Voltage - Diode voltage drop). Add to this that the VIN pin is directly connected to the 5v regulator, so that it starts regulating, drawing current (100mA ~ 250mA to power all of the Arduino when it is not in a low power/sleep mode). This causes the voltage on the batteries to lower in response. That covers the rest of 1.3v difference you see. \$\endgroup\$
    – Passerby
    Commented Aug 12, 2013 at 0:03

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