I have done an experiment that I cannot explain and would like some help.
I have a circuit that is used to generate a clock signal for multiple TLL-based D type flip flop registers (SN74LS173). There are 555 timers and also a push button to have an automatic clock or a manual clock, and a multiplexor (SN74LS157N) is used to select either the manual clock or the automatic clock that is fed to the rest of the circuit.
SN74LS157N recommended max output currents:
High (source) = 800uA
Low (sink) = 16mA
My issue is with the output of the SN74LS157N multiplexor. I wish to drive an 3mm 1.8V LED off the clock signal just for indication. With no resistor and LED connected to the clock signal the output voltage (marked Vx on the diagrams) is around 4.3V, a good high for TTL. But when I add the series resistor and LED, the voltage drops to around 3.5V. Here is the circuit below:
I first tried using a hex inverter (SN74HC04N) connected as in the diagram below and it worked perfect. Even with the LED illuminated, the clock signal voltage remains at 4.3V.
I also used a tri-state buffer (Toshiba 74HC245AP) connected like the diagram below and it behaved exactly like the inverter above. No problems here.
Finally I tried to use an NPN BJT (2N3904) to replace the ICs to drive the LED and it is connected like the diagram below. I worked out the 18K base resistor from taking beta as 100. Assuming a 5mA LED current, Ib = Ic / Beta, Ib = 5mA / 100 = 50uA. So Rb = (Vcc - Vbe) / Ib, Rb = (5 - 0.7) / 50uA = 86kohm. I have used 100Kohm in the experiment as shown.
The problem I have here is that the when the LED is illuminated, the voltage at Vx only gets up to 3.8V, and not the 4.3V like the inverter and buffer method. Why is this 2N3904 loading the 157 mux more so than the other methods? I have not got a current meter that can currently measure down to uA, but the voltage across Rb = 3.03V, so the base current must be around 30uA, which is under the max recommended current for the 157 mux.