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???
10/05/10 01:11
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#178911 - Try a 'Search'.
Responding to: ???'s previous message
I did a search, found what was written before, as I stated in my edit. Nobody has really answered my question. I see how everyone recommends to do this, I just want to know if simply putting a resistor in series with the base will work and not damage my trainer. I also want to know WHY it seems to work when I put a resistor there. By definition the current goes down and then it allows my relay to see the needed current across CE? If an output port puts approx. 1mA out, wouldn't that sufficiently saturated the transistor and allow the needed current? Because adding a 4.7K makes that current now at, OH! I just did the math. Now its 2.5 mA. It seems to have gone up. Now I am really confused.

Thanks for the help, by the way!!!

List of 10 messages in thread
TopicAuthorDate
8051 driving 2n3904            01/01/70 00:00      
   NPN transistor...from port pin...            01/01/70 00:00      
   The better way..            01/01/70 00:00      
   Best is to reply inline in another posting.            01/01/70 00:00      
      sorry!            01/01/70 00:00      
   An efficient relay driver...            01/01/70 00:00      
      Zetex, RIP.            01/01/70 00:00      
   8051 driving relays            01/01/70 00:00      
      Try a 'Search'.            01/01/70 00:00      
         Bidirectional port driver topology            01/01/70 00:00      

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