I am very, very new to electronics. I do understand some, but I lack any real foundation and am hoping someone can answer this.

Here is what I want to achieve. I have a motor driver (LMD18200). The chip can handle 3 amps continuous, and 6 amps peak. I have the chip driving a DC motor. The DC motor may draw up to 7 or 8 amps when the motor is stalled. Running, it only draws about 2 to 2.5. The problem is that my mechanical system relies on the motor stalling when it has reached a hard limit. This is a normal condition for this mechanism. That 7 or 8 amps can burn up my motor driver....

After reading some stuff online, I though that I could connect a GIANT 2 ohm resistor between the power source (12 volt car battery) and the input to the motor driver circuit. Using ohms law, I = 12v / 2 ohm, I = 6 amps. So that means at MOST only 6 amps will reach my motor driver? Am I missing something - or is it really that simple? I understand cars really put out more than 12 volts, and perhaps I want to tweak the resistor value to NOT be so close to the limit of the chip. Any input would be helpful. THanks so much,

Does that effect current flow under regular conditions? In other words, am I goign to be starving my motor under normal conditions?

 

Your calculations are close, but you also need to take into consideration the resistance of the motor itself. If it's designed to draw 8 amps at stall at 12 volts, then the resistance of the motor is 1.5 ohms. Since you need total max current to be 6 amps you would only need an additional 0.5 ohms of resistance, but it would need to be a HUGE resistor. The resistor would drop 3 volts and since the max current through it would be 6 amps that gives you a power dissipation of 18 watts. You need to double that for safety so you would need to find a 36 watt 0.5 ohm resistor. That's a big darn resistor!

I'm sure somebody else will chime in, but I've heard a good way to regulate current to a motor is with Pulse Width Modulation. This way there is no need for huge current limiting resistors, but it does involve some fairly complicated electronics. I'm not at that point yet, but I'm sure someone else can point you in the right direction.

 

Rather than using external means like series resistor to limit the current, why not take a closer look at the LMD18200 datasheet from National Semiconductor and see what is already built in.

It has a section mentioning what happens if there is an overload due to shorted motor turns or locked rotor. Quoted directly from the datasheet:

Current limiting protection circuitry has been incorporated
into the design of the LMD18200. With any power device it is
important to consider the effects of the substantial surge
currents through the device that may occur as a result of
shorted loads. The protection circuitry monitors this increase
in current (the threshold is set to approximately 10 Amps)
and shuts off the power device as quickly as possible in the
event of an overload condition. In a typical motor driving
application the most common overload faults are caused by
shorted motor windings and locked rotors. Under these conditions
the inductance of the motor (as well as any series
inductance in the VCC supply line) serves to reduce the
magnitude of a current surge to a safe level for the
LMD18200. Once the device is shut down, the control circuitry
will periodically try to turn the power device back on.
This feature allows the immediate return to normal operation
in the event that the fault condition has been removed. While
the fault remains however, the device will cycle in and out of
thermal shutdown. This can create voltage transients on the
VCC supply line and therefore proper supply bypassing techniques
are required.