According to the data sheet linked by knossos it says 2.8A max each output (I assume he just added them together to get to the 5.6A).
I was referring to this line here:
5.6A OUTPUT PEAK CURRENT (2.8A DC)
which is for both outputs. I should have been more clear about that.
i doubt that small IC can handle 18 + amps
what can i do so i can test the motor without blowing up IC.
The section Soeren pointed out gives the information you need. It can protect itself and the motor from overcurrents. You need to connect an external resistor and capacitor. The information for resistor and cap selection is also in the documents.
(This is all pretty much paraphrased from the appropriate sections:)
The Delay Time (tDELAY
) before overcurrent protection kicks in is determined by this capacitor (CEN
The cap also provides noise protection for the enable (EN) signal. The cap should be as big as possible without making the Delay Time too long.
The resistor (REN
) value determines how long the circuit should stay disabled (tDISABLE
The datasheet recommends a 100KΩ resistor and 5.6nF cap. This would give you a 200μs Disable Time.
If you need a longer disable time, or a shorter delay before disabling, you can look at figures 19 and 20 to select more appropriate values. For example (values chosen just because they line up neatly on the charts) a 3 nF cap and a 220KΩ resistor would give about a 300μs disable time and about a 0.9μs delay before disabling the circuit.