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08/26/05 12:26
Modified:
  08/26/05 12:38

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Msg Score: +1
 +1 Informative
#100142 - But nevertheless totally unsuited!
Responding to: ???'s previous message
Peter said:
The schematic with direct driven bipolar transistors can be found on the original Intel, Philips, AMD, ... application notes.

So it must be proven as reliable.


Well, when the strong internal active pull-up turns-on an enormeous current will flow. Ok, it's only for two oscillator periods, so that there's no danger for the micro and possibly for the connected base emitter junction, but an enormeous kick is given to the ground routing resulting in annoying ground bounce.
You are right, the older application notes show this driving technique, but todays micros are much faster showing considerable smaller transition times at outputs. So, if you really want to run into avoidable problems refering to CE, then follow these old techniques. But I wouldn't do that.

Alexander,

whenever you want to switch-on something by the help of a standard 80C51-micro, then this should be done by using logical low state at port outputs. The reason is, that the micro emits logical high state whenever reset condition is invoked. So, what you like to switch-on would always be switched-on during reset condition, which not really makes sense in most cases.
So, if your micro runs with +5V, if all the anodes of your LEDs need to be connected to Vcc (+5V) and if you insist on using bipolar transistors for driving purpose, then I would recommend you the following circuit:




How to design this circuit?

A bipolar transistor used for switching loads shoud be driven by a current through the base emitter junction which is about 1/20...1/10 of load current. This means a base current of more than 1.3mA, when your LED current is 25mA.
To help to discharge the base emitter junction capacitance fast enough, when the transistor is turned-off, a 4k7 resistor is connected from base to Vcc.

Assuming that the voltage drop across LED is 1.8V, the resistors are calculated then as follows:

Collector resistor:
(5V - 1.8V) / 25mA = 128 Ohm, so 130 Ohm is adequate.

Base resistor:
(1.8V + 0.7V) / 4.7kOhm + 1.3mA = 1.8mA
(5V - 1.8V - 0.7V) / 1.8mA = 1.4kOhm, so 1.3kOhm is a good choice.

Using bipolar transistors to switch-on LEDs seems to be old fashioned, as a simple bus driver in 74LS technique would also do the job. But this bipolar transistor shown above is much slowlier than these digtal gates, resulting in turn-on and -off times in the microsecond range. This can considerably help to eliminate CE related problems!

Kai

List of 17 messages in thread
TopicAuthorDate
I need help with LED Driver            01/01/70 00:00      
   LED            01/01/70 00:00      
      LED            01/01/70 00:00      
   use a LED driver            01/01/70 00:00      
   74LS drivers            01/01/70 00:00      
   LED Driver - My Solution            01/01/70 00:00      
      quite risky            01/01/70 00:00      
         MCU-resistor-NPN            01/01/70 00:00      
            8051 is open drain !            01/01/70 00:00      
               not good idea            01/01/70 00:00      
                  totally wrong !            01/01/70 00:00      
                     This I would take with a grain of salt.            01/01/70 00:00      
                     But nevertheless totally unsuited!            01/01/70 00:00      
            Where in Atmel errata or other informati            01/01/70 00:00      
               That's standard performance of quasi bi.            01/01/70 00:00      
      Try This            01/01/70 00:00      
         updated            01/01/70 00:00      

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