What's the best way to drop a voltage, a bit?

What's the best way to drop a voltage, a bit?

Or do I need to?


I'm playing with a small robot, made from FischerTechnik components,
that I'm driving with a Basic Stamp, through a Pololu Micro Dual Serial
Motor Controller.

Both Stamp and Motor controller are powered by an 8-pack of AA NiMHD
batteries, providing c. 10.5 V.

This is well within the capability of the Basic Stamp's onboard voltage
regulator, and the Stamp's regulated 5V output is within the 3.0-5.5V
capacity of the Motor Controller's logic supply.

The question is regarding the Motor Controller's motor supply. This is
spec'ed at 1.8-9.0V, with currents of up to 1 A per motor.

I'm using FischerTechnik's S-motors, which run at 9V, and have a .7A
stall current, so we're fine, there.

The question is whether I should reduce the 10.5V for input to the Motor
Controller, and if so, how?

I see four alternatives:

1. Let it run at 10.5V.

2. Stick in a couple of resistors to create a voltage-divider.

3. Use a voltage regulator.

4. How about a pair of diodes?


The problem is I don't know why which would be preferred over the others.

Any suggestions?

--
Only justice, and not safety, is consistent with liberty, because safety
can be secured only by prior restraint and punishment of the innocent,
while justice begins with liberty and the concomitant presumption of
innocence and imposes punishment only after the fact.
- Jeffrey Snyder

 

Subject: What's the best way to drop a voltage, a bit?

1) Only the manufacturer knows for sure. You might want to contact them. But
it does sound like 9V is the upper limit.

2) A voltage divider will only work if the load current is small in relation
to the voltage divider -- in other words, just pour current down the drain for
regulation. Probably not a good idea. Using a series resistor is a somewhat
better idea, and will not waste current. Let's say your peak current is 0.7A,
and you want to drop 1.4V. Using Ohm's Law, that would mean you would pick a 2
ohm resistor, at least 2 watts -- I^2*R = (0.7A)^2 * 2 ohms = 1 Watt -- pick 2
Watts for margin. The thing is, the supply voltage will vary linearly as
current varies, between 9V (0.7A) and 10.5V (0A). This is a better idea, but
probably not where you want to go if you want to stay below 9V for any load.

3) All voltage regulators have a specification called "dropout voltage", which
is the required minimum difference between the input and output voltage
necessary for the regulator to work as advertised. Standard regulators have a
dropout voltage spec between 1.5V and 3.5V., which means you'd probably need
what's called a LDO/low dropout regulator if you wanted 9.0V. If you were
willing to go down to 6V or so, you might want to use an LM317 adjustable
voltage regulator. If you had very tight specs for your load voltage, this
would be the right way to go if you had to use the 10.5VDC source. However, it
sounds like a little much, considering the wide specification for the
controller. Also, all regulators waste some current to run themselves. That
will reduce battery life.

4) How about 3 series diodes instead of 2? Forward voltage drop on 1N400X
diodes will be between 0.55V (for a couple of mils) to 0.9V or so (for 1A),
depending on manufacturer. A good rule of thumb average is about 0.7V for
reasonable currents. Since they can handle up to 1A forward current, you
should be O.K. there. So, you can expect a voltage drop of between 1.55V
(minimum current) to 2.7V (high current, maximum voltage drop across diodes).
You should only get about 0.4V change in load voltage between minimum current
and maximum current, which sure beats a series resistor. That will put you in
the safe range for all eventualities. The KISS principle (Keep It Safe &
Simple) applies here, too. As an added bonus, you're not wasting current for
the regulator. All current going through the diodes will be used by the load.

By the way, you can use any 1N400X diode -- whatever you can scrounge. The
difference between the diodes is mostly how much reverse voltage it can hold
before breaking down -- not a consideration here.

Have fun, and good luck
Chris

 

The way I would do it is either;

Create a dummy battery with a piece of metal rod to make the power pack a 7
cell, or Tap the regulator from some battery besides the end one.

 

If the specs are 1.8-9V then 10.5V is overkill. May be fatal.

A voltage divider is not an option. A series resistor may help but it
requires a minimum load to keep the the voltage <=9V and lowers the voltage
below 9V when the load increases. It has its use with known, fixed voltages
and a known, fixed load. Otherwise it's bad practice from a technical point
of view .

A voltage regulator should be an option if the available and the required
voltages differ for more then 1.5V. Some regulators will do the job but
require at least 1V between in- and output. This will work fine with a fresh
charged battery but will give up pretty fast when the battery voltage
decreases. I see two other options.
- As you spec says 1.8-9V you may go down to 8V or less. FAIK low drop
regulators for 8V/1A are available (from Linear?) Disadvantage: Your motor
get less then full power.
- Use a switched mode (buck) regulator. The one that (almost) shorts input
to output when the input voltage becomes too low.

Looks like the most simple method. You'll have to look at the voltage drop
at the minimum and at the maximum current. If two normal Si rectifiers let's
say 1N4007 have not enough voltage drop at he minimum current you may add a
third diode. It may be usefull to take a Schottky type for that third one as
its voltage drop is (less then) half the drop of a normal SI rectifier.

petrus

 

Why dont you limit it with a zener?