Voltage drop question

[robotnut]

22 Dec 2009 21:16 Voltage drop question
Hi All;

Total noob, first post:smile:

I'm working on a small robotics project. I have an embedded computer (5V) and 2 motors with a controller. It's all powered from a 4.8V NIMH battery.
The problem I'm having;

As the battery gets somewhat tired and I load the motors for starting or climbing a steep obstacle, the voltage drops a little from 5V to about 4.5 momentarily but it's just enough to restart the computer.

What I'm thinking of to solve this problem is adding a capacitor to the power cable of the computer. Would that help the problem? Like I said the drop only happens for a fraction of a second.

I know I could use a bigger battery but I have a room issue already as it is.

Thanx in advance.

[vk6kro]

You would need to give the currents of the motors and the computer.

If you just put a capacitor across the power supply, it would discharge into the motors when they turned on. You would get some benefit if the capacitor was large enough, though.

If you could operate the computer with 4.65 volts, this diagram shows a way that might help.

22672

Using a Schottky diode, you could charge a capacitor which would then discharge only into the computer and not into the motors. Schottky dides have a drop of 150 mV (0.15 volts), so the 4.8 volts from the battery would drop to about 4.65 volts, but it would hold this voltage for short periods if the battery voltage dropped.

[Adjuster]

Could you modify the software so that the motors are always soft-started (progressively, not suddenly). Alternatively, try putting low-value resistances in series with the motors. Clearly if these are too big the motors will no longer work, but it might be worth trying.

[Mike_In_Plano]

I don't know what processor your using, but most modern ones have a large operating range (ie 3V - 5.5V). I'd check the part number on you processor and look up it's data sheet. Most likely, the operating voltage range will be listed in the first two pages (under features).
.
Most likely, your drop out is caused by an overly protective reset chip. These little three pin chips measure the voltage and reset the processor when it dips. The beauty part is that for a given reset part number, the there are any number of voltages that are available - most from Digi Key.
.
. Best Luck, keep me updated,
.
. - Mike

[whome9]

I = C dV/dT, and dV in your case is 0.5v. Plug in your values to calc. your min. cap. size.

For 1 A, 1 second and a max 0.2v difference you'd need 5 F.

[robotnut]

Thanx for all the help, I did a cheap fix. Went with a 2 cell 7.4V 1000MAH lithium polymer battery with a 5V voltage regulator.....