I tried to make a simple NPN transistor inverter so that I can use the strobe line of the parallel port to store data in two separate flip-flop IC's. (74HC174 and 74HC574) The IC that data will be stored in is based on whether the strobe is on the falling edge or the rising edge.
I connected everything including this circuit and there is no smoke, no heat or anything. I could assume my resistor choices are terrible, but I'm not sure, especially with how many changes there are to PC port designs.
For reference, the parallel port I'm using is coming from a dell laptop and I specifically disabled ACPI.
When I tested the particular circuit shown, The voltages I get from the NPN base are 0.09 and 0.69V. However, the NPN collector voltages are acceptable (0.0 something, and 4.9 something).
What would cause the NPN base voltage to be so low in both cases and how do I fix it?
Best Answer
The transistor base is not an output of this circuit.
It is where the input signal connects to the transistor. There is not expected to be any gain between the input and this node.
If you want a non-inverted output with gain, you can add another CE stage.
Or if you want both inverting and non-inverting outputs with equal amplitudes from a single circuit, you could add an inverting buffer circuit (perhaps based on an op-amp).
Or if you want inverting and non-inverting outputs with equal amplitude and equal phase delay, you will need to go to a fully differential amplifier circuit, like the emitter-coupled pair. Or just buy a fully differential op-amp.
Or if you just want a non-inverted replica of the input signal to drive another high-input-impedance circuit, you should just connect to LPT_STROBE rather than to NORMAL. If you want to connect more than 2 or 3 of these same circuits to a single CMOS output, you'll probably want to increase the value of the input resistor to something more than 1 kohm, though.
Note: As Bimplerekkie points out in comments, you could have much higher fan-out if you used an n-channel MOSFET in place of the BJT in your circuit.
Or just buy a CMOS buffer and inverter chips. One-gate versions are typically less than $0.10 in reasonable volumes, so buying the one-chip solution is usually cheaper than what you pay to place all the extra resistors in a discrete inverter circuit.