# Power Supply Issue, Voltage Drop Under Load

I have the Ardupilot Mega flight control board (APM), which outputs 4.8V. I want to power CC3200 MCU with it, which needs at least 3V for stable operation. I power APM using a power module connected to an adequate power source and can output up to 2Amps for APM and all the peripheral devices connected to it. That's way more than enough. When CC3200 is operative under full stress, it can draw up to 200mA.

I'm doing a level shifting on the 4.8V line, dropping it to 3.3V using two resistors 220Ohm and 100Ohm connecting them appropriately to form a voltage divider. 3.3V is the perfect voltage input for the MCU, which under full load should drop the voltage by maybe 0.1-0.2 Volts which is understandable.

However, when i make this voltage divider using the resistors i measure the voltage on the CC3200 and it's 0.3V, instead of ~3.1V! I've tried 2.2kOhm + 1kOhm with same results. Tried an other module (WiFi bluetooth module) the voltage there goes to 2V, kind of better, but still this is 1.3 voltage drop! I also tried powering led's and other stuff from the APM, but in all cases the voltage drops heavily when a load is connected to it when using a voltage divider for level translator. Nothing's wrong with CC3200, normally i power it using 2 AA batteries. But i would much rather power it from the APM.

But, if i power say the WiFi Bluetooth module without level shifting (it can handle up to 6V) the voltage remains almost perfectly stable to ~4.8Volts!

This issue has been troubling me. Is there not enough current to power the external device when using resistors as voltage divider (unfortunately i don't have another level translator device to test this theory)? Not much else comes to mind, but i don't know how to make sure. I need the help of more experienced and knowledgeable people. What could be the reason? What is wrong? What am i missing? Thanks in advance!

• Using a resistor divider as a supply doesn't work, that isn't connecting them properly. Calculate the voltage drop over your 100 Ohm with peak current use of 200mA: 20V. The 100 Ohm will want to drop 20V! Be sensible and use a normal Linear Regulator. – Asmyldof Apr 5 '16 at 22:38
• As VSV Mani suggested, Voltage dividers won't work as voltage regulators, as their output impedance is HIGH. If you want to take more current from it the voltage dividers should take 4-5 times more than the actual current, then only voltage is stable. If voltage divider resistors ~ 100 Ohms, Parallel impedance is 50 Ohms, So, current taken by output should be less than 3.3*/5*50 = 13.2 mA (these all approx assuming 5 times). Use LDO or Zenar Parallel regulator. – user19579 Apr 6 '16 at 6:34

simulate this circuit – Schematic created using CircuitLab

Figure 1. Voltage divider open-circuit. Figure 2. Micro attached in parallel.

• Figure 1 shows the voltage divider when tested unloaded. All appears well and 3.3 V is measured on the divider terminals.
• Figure 2 shows the voltage divider loaded by the micro. The micro is represented by a 33 Ω resistor which would pass 100 mA if connected to a stable 3.3 V PSU.
• The combination of R4 and R5 in parallel is about 30 Ω (28.7 Ω).
• Now our potential divider mid-point will give $4.8 \frac {30}{100 + 30} = 4.8 \cdot 0.23 = 1.1 V$

The moral of the story is that a potential divider will not give you a stable voltage under changing load conditions such as a micro whose current requirements will vary with time. You need to use a voltage regulator.

Take a look into buck converters. In your application, you need to save as much energy as possible. Another option is to use a linear regulator, however, that will consume energy dropping the voltage turning it into heat, which isn't favorable. This resistor chain voltage divider will require a lot of energy. You want to conserve as much as you can if you're flying a drone.

the resistor divider does not work because it has a too high source resistance = about 100 ohms. the concept of dynamic resistance is important here and try using a diode to drop the voltage by .7V. You would require two diodes as the drop required is 4.5 to 3.3 or about 1.2V. the diodes have a dynamic resistance of much lesser than 1 ohm when used at currents of 25 mA or more. If you are keen on knowing about dynamic resistance here is the data. It is change in voltage / change in current. Dynamic resistance of a diode is 26 ohms at 1ma [Se grey and meyer]. that is why a zener is much better than a voltage divider! Zeners have dynamic resistance of about 10 to 15 ohms. So if you have a 15V and want 5V, do not use a voltage divider but use a 10v zener in series that can drop the 10V. the diode has a drop of 600mV at 1ma, 660mV at 10mA and 720mV at 100mA. It has decreasing dynamic resistance of 26,2.6 and .26 at 1mA, 10mA and 100mA respectively. thus it is way better than a voltage divider when you have to drop voltages. use a 1N4007 {two in series} for your drop of 1.2 from 4.5 to 3.3

• Please use punctuation, proper capitalisation and break your answer into paragraphs. – Transistor Apr 6 '16 at 7:51
• I did exactly what you said. Thankfully i had 1N4007 diodes. I put 2 in series and the result was perfect! Thanks! – Nikos Apr 6 '16 at 17:22