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Subject:		Re: Re: Re: MOTOR DRIVER CHIP
Name:		Rob
Date Posted:		May 8, 05 - 11:36 AM
Message:		The easiest way to calculate the resistor needed for an LED is using this formula: (Vsupply-Vled)/Iled=R Where: Vsupply=The supply voltage across which the LED is to be connected Vled=The voltage across the LED (called Vf on datasheets - if unknown 1.6 is a good guess) Iled=The current in AMPS to drive the LED (called If on datasheets, 10ma should give reasonable brightness ) I not exactly sure which bit of the circuit you are looking at but assuming you want an LED across the 12V line then the above formula is (12-2)/0.01=1K but an LED across the 5V line would be (5-2)/0.01=300R The resistor 'drops' the voltage across it (if you measured the voltage across one of the above resistors it would either be 10V for the 12V supply, or 3V for the 5V supply - the resistor has effectively reduced the supply voltage to a safe level for the LED). I know we are going a bit off topic but the excess volatge is actually disipated by the resistor as heat, this would be calculated by P=IV where P is the power in watts, I is the current drawn by the LED (in A) and V is the voltage dropped across the resistor eg in the the 12V cct this would be P=0.01*3=0.03W thats why you should check the power rating of the dropper resistor when doing calculations (In our case the power is so low, unless you have a really tiny resistor it will be fine) Your motor current sounds fine so I think it is safe to assume the IC is fried.
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Re: Re: Re: Re: MOTOR DRIVER CHIP by Rob · May 10, 05 - 9:21 PM

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