[sdiy] Driving LEDs from CMOS

[Happy Harry]

Hi Dave....

That 47 ohm resistor is much too small. Assuming a gain of
100 for the 2N3904... and an LED current of 4 mA (set by your
680 ohm)... then the base needs only .04mA to drive the transistor
full on. A 100K resistor is probably a minimum.  I'd use a 10K
resistor, which would draw only .5mA from the Q output, well within
what CMOS can handle.

The 47 ohm and the base-emitter junction diode are sucking the
CMOS output down.

H^) harry


>From: DCMagnuson at aol.com
>To: synth-diy at node12b53.a2000.nl
>Subject: [sdiy] Driving LEDs from CMOS
>Date: Fri, 16 Feb 2001 09:01:01 EST
>
>Hi everyone,
>
>I've been messing around with some logic circuits to convert a momentary
>switch to a DPST switch with a schmitt trigger and D flip-flops.  The core 
>of
>the circuit is simple.  Switch feeds the inverting schmitt trigger, which
>then gets sent to the flip flop.  Using a logic probe, everything is fine.
>Pushing the switch will alternate the Q output between high and low.
>
>My problem comes from trying to drive an LED from this Q output.  I went 
>from
>the Q output through a 47 ohm resistor to the base of a 2N3904.  Emitter to
>ground and collector through a 680 ohm current limiting resistor, then to 
>the
>  LED.  The other LED leg goes to V+
>
>When this portion of the circuit is connected, the LED appears to work 
>fine,
>but measuring the Q output now shows the logic high state at only 1.6V,
>rather than 5V.  Apparently when the LED is lit it drags the voltage down 
>at
>the Q output of the flip flop.  Are there any work-arounds for this?  This
>1.6V logic high won't properly drive the next stage of the circuit.  When I
>disconnect the LED, everything works fine.
>
>Is there a better method of driving the LED than the 2N3904 method I used?
>
>I could upload a schemo if anyone needs to take a peek, but it's a pretty
>standard LED driver scheme.  Thanks in advance
>
>Dave Magnuson

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