arduinoarduino-shieldmotorohms-lawpower
I'm reading the specs for the SparkFun Monster Moto shield which specifies that
Given Ohm's law, does that mean that the maximum power is 480 watts? That seems like a lot!!!
Am I missing something? Please do excuse my ignorance, I'm only starting out with robots and electronics and still at the beginning of the All About Circuits book.
Two subjects are covered here. First, reality. Second, the questions posed.
First, Reality : About the most you can get from a U.S. 120 volt plug in is 1500 watts (generally speaking). So your heater element will draw more current (because of lower resistance) until the nichrome element heats up. During this transition time, one depends upon the circuit breaker characteristics to not shut off (trip off). Then, counting upon your forced air cooling (fan), the nichrome element should stabilize (temperature wise) at 9.6 ohms. That is as good as it gets for a constant applied voltage.
Second, your question : "How do I figure out the cold resistance". You need to get the manufacturer to supply you with the expected operating temperature where the resistance is 9.6 ohm ( or rather, at what temperature will the nichrome be at 1500 watts). From there, you can use the nichrome's temperature coefficient to calculate the resistance at room temperature.
A simple battery backup IC will do what you want.
Essentially, a comparator samples two voltage inputs, the main power, and a backup power supply (Batteries). Based on that, two cascading inverters paired with two p-channel fets power the circuit at Vo, while two n-channel fets are used as open-drain outputs. Pull them up with a resistor, and they are the perfect interrupt source for your arduino.
The specific one shown is the ICL7673, but it's maximum current is 38mA. It's designed for rtc and ram with a coincell, but an extra pair of P-channels can be used to increase the current.
There are other battery backup or power switches which will do higher current builtin.
Best Answer
The board is rated for a peak current of 30A, not continuous driver current of 30A. The continuous drive current spec'd is 14A.
In section 4 of the datasheet for the motor driver, it has the package and thermal dissipation information for the driver.
Figure 40 shows the junction-ambient thermal resistance of the board in natural convection (i.e. no fans). The board itself has roughly an area of \$3.5cm \cdot 6cm = 21cm^2\$, which is shared between the two motor drivers. For the sake of simplicity, let's assume that we only have 1 driver running, and we only get an effective ~\$15cm^2\$ of PCB heat sinking (close to the max values shown in the plot). At this level we have the following junction-ambient thermal resistances:
\$ R_{thHS} = 28 \frac{C}{W}\\ R_{thLS} = 26 \frac{C}{W}\\ R_{thHSLS} = R_{thLSLS} = 7.5 \frac{C}{W} \$
The temperature rises above ambient is then given in table 15. For this example, let's assume we're driving HSA and LSB, and we're analyzing \$T_{jHSAB}\$ (junction temperature rise of the high side gates).
\$ T_{jHSAB} = P_{HS} \cdot R_{thHS} + P_{LS} \cdot R_{thHSLS} + T_{amb} \$
Now let's refer to the electrical characteristics of device. The MOSFET gates can be modeled as resistors when on, with the following resistance values:
\$ R_{HS} = 28 m\Omega\\ R_{LS} = 10 m\Omega\\ \$
The power dissipation of a resistor given a current:
\$ P = I^2 \cdot R \$
So re-writing the junction temperature rise equation, we get:
\$ T_{jHSAB} = I^2 \cdot R_{HS} \cdot R_{thHS} + I^2 \cdot R_{LS} \cdot R_{thHSLS} + T_{amb} \$
Plotting this vs. current, we get:
At
~12Aof continuous drive we've exceeded the allowable thermal junction of the chip. At this current, A rough heat dissipation calculation for the driver chip is:\$ P_{d} = I^2 \cdot R_{HS} + I^2 \cdot R_{LS} = 5.47W \$
In addition to these calculation, DC motors don't have a constant current consumption. Rather, as the motor gets faster it generates a back EMF which will decrease the current flowing through the motor until the motor reaches the no load speed and consumes near 0 current. Maximum current is consumed at stall (reason why stalling DC motors is bad). Maximum mechanical power occurs at half the no-load speed.
So let's assume we're driving a motor with 16V and we want maximum mechanical power. Let's say for sake of argument this results in 12A flowing through the circuit. At half speed we get an 8V back EMF, resulting in a maximum mechanical power of (assuming 100% efficient motor):
\$ P_M = (16V - 8V) \cdot 12A = 96W \$
So as you can see the mechanical motor power is significantly higher than the heat losses of the motor driver.