| Electronics > Beginners |
| 74F08N current issue for ARDUINO |
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| vis5254:
I've a 74F08N AND GATE IC . I checked the datasheet and realised that the output voltage on HIGH condition goes to 5v and when output goes in LOW , current predominates and reach a value of 40mA . ARDUINO UNO digital pin has maximum withstanding capacity of 40mA . So what could be a better way in reducing 40mA to some 20 or 30mA ..?? Will a resistor in series will do the job ? |
| Moriambar:
Hi, I actually do not understand how you have connected this to the arduino ie: is it the inputs or the output of the and gate? How is the arduino pin mapped, and how is it physically connected. The datasheet I pulled out for a 74F08N specifies an absolute maximum of 40mA when in low state, so my first guess is that something is shorted out or connected wrong... |
| Ian.M:
For logic, MCUs etc. as long as the supply voltage is the same and you connect a single output to N inputs*, it typically just works# - the inputs don't draw excessive current from the output. However, you must never directly connect normal outputs% as if they are at different levels, its called contention and excessive current is likely to flow, which will probably damage them if it carries on for long (at a us timescale). If there is a significant risk of contention, a series resistor may help - 330R is a good value for a 5V logic system as it will keep the worst case current under 15mA. * Where N is less than the fanout limit for the logic family you are using. # Provided you make any accommodations required to handle the differing logic levels when interconnecting different logic families e.g a bipolar TTL output will probably need a pullup resistor to drive a 5V standard CMOS input. % The rules for wiring open collector or open drain outputs together are different - its safe, you need a pullup resistor, but the result is only high if *ALL* the outputs are high. If any go low the low level wins. |
| vis5254:
I've a 7805 ic for producing 5v . I was trying to check if my AND would work .. I provided 5v from 7805 as vcc as well as same voltage as 2 input . I'm concerned about the current flow now . |
| Ian.M:
When dealing with bipolar TTL chips (basically anything 74nn(n) or 74xxnn(n) where xx doesn't contain the letter C), you need a pullup resistor to take an input to logic '1' as direct connection to +5V risks damaging the chip. 1K is a suitable value. OTOH to take an input to logic '0' you must wire it directly to Gnd, not through a resistor. If you put a separate 1K pullup on each AND input, it should output logic '1'. Add a jumper wire to Gnd to put one of the inputs low, and the output should go to logic '0'. |
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