Voltage drop

Thread Starter

alex116

Joined Apr 29, 2021
14
Hi Guys,

Cant get this right, I have 4 AA batteries total voltage 6v, 2500 mAh each. Then I have 2 devices one needs 6v another needs 3v power supply so I did basic voltage divider from 2 resistors.

10R resistors burned my hand.

Then I tried 1K resistors, I got voltage divided and multimeter shows 3v, However when I connect the load - in my case small microcontroller module to 3v output and servo to 6v output, the 3v output goes down to 1.5v as soon as I connect the module and that is not enough to power up the microcontroller.

Do you know why this might happen? am I choosing wrong resistors?

Thank you
 

Ya’akov

Joined Jan 27, 2019
9,071
You shouldn't use a voltage divider to get the 3V from 6V. It will dissipate a silly amount of current and use a lot of battery for nothing but waste heat. Ideally you want a switching regulator, a buck converter, which will do things much more efficiently.
 

BobaMosfet

Joined Jul 1, 2009
2,110
Hi Guys,

Cant get this right, I have 4 AA batteries total voltage 6v, 2500 mAh each. Then I have 2 devices one needs 6v another needs 3v power supply so I did basic voltage divider from 2 resistors.

10R resistors burned my hand.

Then I tried 1K resistors, I got voltage divided and multimeter shows 3v, However when I connect the load - in my case small microcontroller module to 3v output and servo to 6v output, the 3v output goes down to 1.5v as soon as I connect the module and that is not enough to power up the microcontroller.

Do you know why this might happen? am I choosing wrong resistors?

Thank you
Instead of tell you that you shouldn't, it's important for you to understand WHY-

10R or 10-Ohms is ridiculously low in terms of resistance. It burned your hand because of the amount of energy you were making the resistor dissipate. Here's the Math: Ohm's Law, and Watt's Law:

Ohm's Law for Current
I = E/R
I = 6/10
I = 600mA (just over 1/2 Amp). Your Resistor must be rated (and sized) for the amount of energy it's going to handle:

Watt's Low for Power (aka Watts):

P = IE
P = 0.600 * (6 - 3)
P = 0.600 * 3
P = 1.8W for each of the resistors. Which means the resistors you should have used would each be 4W resistors.

As you can see- A voltage divider is not useful for POWER distribution. It's real use is in controlling small (aka 'signal') levels of current.

Instead, as others have suggested, you want to use a Regulator. In your case, you could use an LM7806 powered by a 9V battery, or a couple of 9V batteries in parallel to increase amperage. The regulator uses about 2V to do its job, leaving 7 left for it to regulate- which is perfect for breadboard work. It will take the 7V and regulate that to 6V and provide up to about 500mA or so.
 

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wayneh

Joined Sep 9, 2010
17,496
thank you guys!
FWIW, I find it useful to distinguish between information signals and power signals. I consider information signals to be those where a voltage is being communicated with only as little current as required. Voltage dividers can work fine for these tiny information signals. They are quite poor for regulating power through a load.
 

MrChips

Joined Oct 2, 2009
30,712
As everyone has already said, voltage dividers are ok for reducing voltage when little current is required. They are not appropriate for reducing power. Use a buck converter.

Let's do some back-of-the-envelope calculations.

With 1kΩ resistor the voltage dropped 4.5V. So your MCU needs at least 5mA. Let's allow it 10mA.
Rule of thumb is the voltage divider should supply 10 times more than required. So we design for 100mA.

6V @ 100mA = 60Ω

If you really wanted to go this route or to understand the calculation and the power dissipated by the resistors, then do the following.

Choose two 33Ω resistors for your voltage divider.
Current drawn = 6V / 66Ω = 91mA
Power dissipated by one resistor = 3V x 91mA = 273mW
Hence you need 33Ω ½W resistors for your voltage divider.

Another option is to keep one 33Ω ½W resistor and replace the resistor to ground with a 3.3V 500mW zener diode.
 

Thread Starter

alex116

Joined Apr 29, 2021
14
Thank you for all the feedback,

I purchased a small buck converter as advised, connected, now getting 6v & 3v.
However now I'm encountering another problem which did not exist earlier - when I'm sending PWM signal to the Servo motor it does not rotate as expected (90 degrees) and does 5-10 degrees only.

To make sure it is not related to my microcontroller, I disconnected buck converter and connected devices via 2 different sources (2 AA batteries for Microcontroller and 4 AA batteries to the Servo). Then I connected Negative cable of two power sources as otherwise PWM signal doesn't go out and it worked as expected.

Any Ideas why this might happen?


buck.jpg

buck2.jpg
 
Last edited:

sghioto

Joined Dec 31, 2017
5,380
If the voltage is dropping low enough for the micro to drop out then maybe you need a higher current supply for the servo.
 

Thread Starter

alex116

Joined Apr 29, 2021
14
I think the setup with DC-DC converter should work, I just need help to understand what goes wrong with PWM signal in that setup, center tapping doesn't seem ideal.
 

Thread Starter

alex116

Joined Apr 29, 2021
14
Fresh batteries? Microcontroller has all the recommended decoupling/filtering caps in place?
Batteries are new, perhaps capacitors are the issue? please look at the 2nd image, I soldered + / - directly to VCC/VDD and GND of the microcontroller
I didn't use any capacitors
 

sghioto

Joined Dec 31, 2017
5,380
OK, it works using separate battery packs .
Power the servo using the 6 volt battery pack but power the micro using the buck converter connected to the 6 volt battery pack.
 

Thread Starter

alex116

Joined Apr 29, 2021
14
OK, it works using separate battery packs .
Power the servo using the 6 volt battery pack but power the micro using the buck converter connected to the 6 volt battery pack.
I did that, buck messes up the PWM signal and I don't know why. That is exactly the problem with Buck setup.
 

BobaMosfet

Joined Jul 1, 2009
2,110
I think the setup with DC-DC converter should work, I just need help to understand what goes wrong with PWM signal in that setup, center tapping doesn't seem ideal.
A servo is a motor- you have to overcome it's initial stall current to get it to turn, which is always higher than its operating current. What is its stall current?
 

BobaMosfet

Joined Jul 1, 2009
2,110
From my calculations the micro only draws appx 6ma. Why not just connect it to the 3 volt junction in the battery pack.
That is not necessarily valid. It's a digital circuit- it takes current in spikes and fits, which is what the bypass caps are also for. An easy way to tell is put a 1-ohm resistor in series with ground and measure the voltage drop across it- this will give you the true current being used.
 

BobaMosfet

Joined Jul 1, 2009
2,110
I did that, buck messes up the PWM signal and I don't know why. That is exactly the problem with Buck setup.
@alex116 You cannot get energy for free. Current is consumed at the expense of volts, and vice versa in a system where you're at a limit. A buck-convert is at a limit. It is designed specifically to provide X-Volts at up to Y-Amps. So, you need to make the buck converter output enough amps to cover the stall-current of the motor AND the MCU circuit that controls it. If your motor exceeds the current draw that the buck can provide, the whole circuit collapses.

Instead of all this guessing, please provide us with information about the servo. We need to pull a datasheet and look at its specs.
 

Thread Starter

alex116

Joined Apr 29, 2021
14
those are two setups I tried with the Buck. both didnt work, as I said, servo moves a bit when powered from single source like 10 degrees instead of 90.
Microcontroller doesn't shut down.

Servo is MG996r Stall Torque: 9.4 kg/cm @ 4.8V. Stall Torque: 12 kg/cm @ 6V. Running Current 500mA - 900mA @ 6V. Stall Current 2.5A @ 6V.
 

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