Simple DC circuit, variable current: how to control voltage without blowing up my receiver?

Thread Starter


Joined Aug 19, 2020

I'm almost embarrassed to post this very basic question, but my knowledge of electronics is so poor that I'm stuck. I'm sure that my reasoning is faulty, so if someone could just point out the error, I'd be grateful.

I have a 7.4V lipo that I want to use to power the receiver and 3 servos in a RC glider. The receiver and servos take between 4.8V and 6V. So, I figured I could just put a resistor in series between the battery and the receiver to drop the voltage from 7.4V to 6V (see pdf attached). I used another battery (5V) to power the receiver and servos to measure the current they draw. When the servos are not moving, the current is about 50mA. When all the servos are moving, the current averages around 100mA with the occasional peak of 140mA.

So, I figured I want to drop 1.4V over the resistor to give me 6V for the RC gear.

V=IR. When the current is maximum, the resistor should be 1.4/0.14 = 10 ohms. So far, so good.

Then I thought I'd better check for the minimum current of 50mA. So,

V=IR. V = 0.05*10 = 0.5V.

Here is the problem. Does this mean that only 0.5V will be dropped over the resistor at minimum current, leaving 7.4V-0.5V = 6.9V for the RC gear? If so, I'll blow them up as they only take 6V maximum.

I'd be very grateful for some clarity.



Joined Dec 24, 2019
Using a single resistor in series with the receiver is going to limit current to the receiver, and though it will also drop the voltage, the voltage dropped by the resistor will vary depending on the current being drawn by the reciever in turn varying the voltage to the receiver, which... will in turn probably vary the amount of current the receiver is drawing. It quickly becomes a circular dependancy nightmare. I think the better approach, still keeping it simple, would be to use 2 resistors and create a voltage divider. Then your voltage supply to the receiver should remain the same, for the most part, independent of the current being drawn by the receiver. Keep in mind though that using resistor based voltage divider wastes current, and though a small amount could be significant factor in battery life. Another direction is to use a switching regulator to reduce the current consumption of the voltage division section if it becomes an issue. Just something to keep in mind.


Joined Dec 24, 2019
By the way, don't ever be embarrassed for not knowing something. Acknowledging that we don't know something is the first step to learning and that is exactly what this forum is all about.


Joined Jan 30, 2016
7.4v won't 'blow up'a 6v servo or Rx. It's only 7.4v when charged & drops to 6v when discharged. That's why RC gear is spec'd as 6v nominal. There's 1000s of systems out there happily working on a 7.4v LiPo pack. If you're really worried you need to use a voltage regulator not just a passive resistor or any such combination - a passive solution cannot cope with a variable load. All the servos etc I have are 6 - 9v and my receiver spec is:

Operating Voltage Range: 3.5V-10V
- Operating Current: 100mA@5V

Some servos are rated at 4.8v and for those it might be advisable to use a 5v regulator. Either a LM7805 or a switching regulator board such this one from Hobby king (similar & cheaper on eBay)
Last edited:


Joined Mar 19, 2019
A simple voltage regulator is pretty cheap. Depending on the current either a TO-220 or TO-92 package. Far more efficient and bombproof than divider resistors. May require some heatsinking if the current draw is very much. Take a look at the LM317L which in TO-92 can go 200mA easily.

Audioguru again

Joined Oct 21, 2019
A modern RC receiver draws a low current, far from 50mA or 100mA.
A 2-cells Li-PO is 7.4V only when it is half-discharged. It is about 6.4V when it should be disconnected and is 8.4V when fully charged.
Go to for modern RC products.


Joined Nov 13, 2015
If you determine that the radio/servos indeed can not take more than 6.0 volts, then.... Rather than a resistor, use two series connected diodes. Each diode will drop about 0.6 volts and that drop will be fairly consistant across your current draw range. A pair of 1N4001's should be adaquate.