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I'm reading the specs for the SparkFun Monster Moto shield which specifies that

  • Max Voltage is 16V
  • Maximum Current 30A

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.

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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:

TjHSAB

At ~12A of 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.

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  • \$\begingroup\$ Hm, I looked at the datasheet, and I can't find where it lists Peak or Continuous current ratings. It lists Imax=30A as continuous under Absolute Max, and the very first page lists 30A as it's featured rating as well. The datasheet's specifications are all done at IOUT = 15A, not 14. But I'm only seeing 30A as the continuous, and nothing listed for a peak or pulsed. \$\endgroup\$ – Passerby Feb 13 '13 at 3:31
  • \$\begingroup\$ On the sparkfun page they list 14A max continuous drive for the board. It's more of a recommendation, though and from my experience I would treat that value with some suspicion. Probably best to stick under that at least. \$\endgroup\$ – helloworld922 Feb 13 '13 at 3:33
  • \$\begingroup\$ Ah, the "practical current". Yea, someone made a comment about the eagle files being iffy and someone else about how the copper weight needs to be a higher 6oz for proper heat dispersal to handle a high current rate. \$\endgroup\$ – Passerby Feb 13 '13 at 4:09
  • \$\begingroup\$ @helloworld922 I will need to do some further reading on the information you've provided because as mentioned in my OP I am still very much a beginner :) However, thank you for the elaborate answer it deserves my vote and tick no doubt. Cheers \$\endgroup\$ – Marko Feb 13 '13 at 7:25
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Yes it's a lot (but motors an take that or more). No, the board can't handle it without adding heat sinks and cooling. The average use case for these is more like 12V/6A or 72 watts (Stall current). Design goals should always be to (reasonably) over-design, for protection. You don't want to run parts at their maximum capacity, for safety, longevity, and ease of expandability.

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  • \$\begingroup\$ Thanks @Passerby, so how did you come up with the 12V/6A figures? The power you suggest seems to only be 15% of the maximum capacity. \$\endgroup\$ – Marko Feb 13 '13 at 3:11
  • \$\begingroup\$ Oh sorry. Just based on Sparkfun's target audience, the comments and everything. 12v 2A motors are standard hobbyist parts. (Plus stall current is higher). Biggest thing is that without heatsinks, more than a few continuous amps will quickly cause heat issues. The output current is dependent on the junction temperature, if you can't keep it cool, it won't work well. \$\endgroup\$ – Passerby Feb 13 '13 at 3:27
  • \$\begingroup\$ Not sure who downvoted you but you have my +1. Thank you for your answer. \$\endgroup\$ – Marko Feb 13 '13 at 23:13
  • \$\begingroup\$ @Marko, appreciate it. \$\endgroup\$ – Passerby Feb 13 '13 at 23:26
  • \$\begingroup\$ Twasn't me, but the motor driver almost definitely cannot handle 72W of heat dissipation. As I demonstrated in my post the amount of power dissipated by the driver should be small compared to the amount of mechanical power output from the motor. If it wasn't, that would be one lousy driver. This is probably the biggest reason motors are driven with PWM instead of using a linear driver. \$\endgroup\$ – helloworld922 Feb 15 '13 at 3:18

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