Micro-Servo don't work when using 7.4V Battery + diodes

Hi, i have a 7.4V LiPO battery (5000mah), some Towerpro SG90 micro-servos and eight 6A10 diodes. If i attach the servo directly in the 5V out of arduino, it works, but i need to poweer it up with the battery, so ihad to decrease the voltage to 5V, i did it with seven 6A10 diodes, now the voltage is 5.2V, just like arduino, but when i connect the servo it dont work! It just kinda vibrates a little, but dont work at all... What can be happening?

Perhaps the supply isn't stiff enough? 7 diodes in series is going to be a pretty soggy voltage source. Try just 1, most servos will run from 7.2V.

[ just looked up the datasheet, those diodes drop 0.8V at an amp or so, so 7 of them will drop over 5V. Use just one or two ]

MarkT: Perhaps the supply isn't stiff enough? 7 diodes in series is going to be a pretty soggy voltage source. Try just 1, most servos will run from 7.2V.

[ just looked up the datasheet, those diodes drop 0.8V at an amp or so, so 7 of them will drop over 5V. Use just one or two ]

Really? Why then when i test with my multimeter, it only decreases 0.3V~0,4V for each diode? And by the way, i have tried with one, two, three... until seven diodes, and it did not work in any way, the servo just move a little then stops.

what are these diodes made of ?

michinyon:
what are these diodes made of ?

Just to decrease the voltage from the battery (fully loaded = 8.4V, servos run at 6V max.)

Do you have the arduino and servo grounds still connected together like below?

Really? Why then when i test with my multimeter, it only decreases 0.3V~0,4V for each diode?

The voltage drop across a diode depends on the current through it (logarithmically) so the relevant measurement would be with the servo(s) straining against a load. MarkT is right, just use one or two of those diodes.

Why don't you just get a 7806 regulator.....

zoomkat: Do you have the arduino and servo grounds still connected together like below?

This is my schematics:

jremington:

Really? Why then when i test with my multimeter, it only decreases 0.3V~0,4V for each diode?

The voltage drop across a diode depends on the current through it (logarithmically) so the relevant measurement would be with the servo(s) straining against a load. MarkT is right, just use one or two of those diodes.

Im gonna use 12 servos, so the current will increase (for testing i used only one servo)

JimboZA: Why don't you just get a 7806 regulator.....

I will use 12 servos, so the current will be higher than 1A (it's going to be about 4A)

This is my schematics:

Bad! the servos won't work (and possibly damaged) unless you have a common ground between the battery ground and the arduino ground. Connect the grounds together and only use two diodes.

mateusviccari:

MarkT:
Perhaps the supply isn’t stiff enough? 7 diodes in series is going to be a pretty
soggy voltage source. Try just 1, most servos will run from 7.2V.

[ just looked up the datasheet, those diodes drop 0.8V at an amp or so, so
7 of them will drop over 5V. Use just one or two ]

Really? Why then when i test with my multimeter, it only decreases 0.3V~0,4V for each diode?
And by the way, i have tried with one, two, three… until seven diodes, and it did not work in any way, the servo just move a little then stops.

The voltage drop depends on the current drawn - your multimeter draws a few fractions
of a microamp, at that current level a diode will drop much less than at its nominal rating.

Have you measured the actual voltage across the servo as you try to operate it?

If it drops a lot the batteries are dead, if not then the servo may be dead.

Voltage drop across a diode is not current dependent, the drop will remain constant until you physically destroy it via to much current. Its just silicon with opposing charges, this creates a voltage barrier, non conducting below its rating but once you have enough voltage to get past it, it will never change, and it has no measurable resistance inside, thus has no current limiting.

If you only measure 0.3-0.4V drop then its probably germanium diode.

Connecting the grounds will solve the issue almost certainly.

I will use 12 servos, so the current will be higher than 1A (it's going to be about 4A)

If your planning to draw 4A, are those diodes rated for 4A?

Is there a reason why you can't use a regulator rated for more then 4A? LM1084 I believe is a 5A regulator, that comes in lots of different voltage ranges or adjustable.

A switching regulator might be another option.

You can use multiple 7806, 2-3 servo per 7806, based on whatever its current rating is (1.5A?).

Voltage divider on the source is not recommended but crudely workable if you can calculate the proper resistances and wattage ratings.

Zener diode (6.2V) might be feasible.

randomname: Voltage drop across a diode is not current dependent, the drop will remain constant until you physically destroy it via to much current. Its just silicon with opposing charges, this creates a voltage barrier, non conducting below its rating but once you have enough voltage to get past it, it will never change, and it has no measurable resistance inside, thus has no current limiting.

If you only measure 0.3-0.4V drop then its probably germanium diode.

Connecting the grounds will solve the issue almost certainly.

I will use 12 servos, so the current will be higher than 1A (it's going to be about 4A)

If your planning to draw 4A, are those diodes rated for 4A?

Is there a reason why you can't use a regulator rated for more then 4A? LM1084 I believe is a 5A regulator, that comes in lots of different voltage ranges or adjustable.

A switching regulator might be another option.

You can use multiple 7806, 2-3 servo per 7806, based on whatever its current rating is (1.5A?).

Voltage divider on the source is not recommended but crudely workable if you can calculate the proper resistances and wattage ratings.

Zener diode (6.2V) might be feasible.

Every servo eat about 0.3~0.4 mah. So if i use many 7806, it should look like this? Sorry for these dumb questions and bad drawings, my knowledge in eletronics is next to zero.

I think that diagram's about right, except the middle pin from the 780x must also be at ground.

Every servo eat about 0.3~0.4 mah mA

mAh (note the capital "A", millli*Amp **hours) is *energy not power (mA, milli*A*mps).

As long as you ground the 7805 it should be fine.

Be very careful with that battery, its unforgiving to shorts or weak connections due to its ability to source a massive amount of current. Always store it in a fire bag, never leave unattended while charging/discharging!!!!!

If possible test with a wall plug or other power supply to verify. The servo's probably wouldn't even be drawing 50mA when free spinning, the 300-400mA rating is most likely its stall current.

randomname: As long as you ground the 7805 it should be fine.

Be very careful with that battery, its unforgiving to shorts or weak connections due to its ability to source a massive amount of current. Always store it in a fire bag, never leave unattended while charging/discharging!!!!!

If possible test with a wall plug or other power supply to verify. The servo's probably wouldn't even be drawing 50mA when free spinning, the 300-400mA rating is most likely its stall current.

What do you mean by grounding the 7805? Another question i need to ask, i am planning to build a "spider" robot (hexapod), so when the robot is stopped, the servos will be stopped too. What i mean is that sometimes the motors will not operate with 100% of the power. AShould i bother with this, or is it ok?

@randomname

Voltage drop across a diode is not current dependent, the drop will remain constant until you physically destroy it via to much current. Its just silicon with opposing charges, this creates a voltage barrier, non conducting below its rating but once you have enough voltage to get past it, it will never change, and it has no measurable resistance inside, thus has no current limiting.

Where did you hear this nonsense? Or do you just make these things up?

To avoid spreading this misinformation, here are a couple of links that describe diode current/voltage relationships:

http://www.st-andrews.ac.uk/~www_pa/Scots_Guide/info/comp/passive/diode/chars/chars.htm

What do you mean by grounding the 7805?

The middle pin is ground. The input voltage is across the left (+V) and middle (0V); output is across the middle (0V) and right (5V).