Interesting 8/L problem (more than you want to know about switching transistors)

I was sending blocks of data to and from my DF32x4 emulator when I noticed that Bit 5 was always writing a zero to the disk - turned out to be a bad bus driver (M623 in slot D28) and I found a short After destroying the transistor removing it, I found out the clamping diode was the actual shorted part :^P so after replacing it with a 1N4148, I grabbed the nearest 2N3904 switching transistor and put it in. Guess what - Bit 5 still didn't work, although it was now changing states! To make a long story short, I found that the rise time of a 2N3904 is reasonably fast (around 50 nS) but the storage time in the circuit DEC used (a 7402 NOR gate driving the transistor base through 390 ohms) is on the order of 500 nS. (Once a bipolar transistor is turned on, there is a significant time required to extract the charges from the base so it will turn off, especially if a negative drive voltage is not available). Although the 7402 output waveform was square, this particular transistor base voltage was dropping much more slowly than any of the others; I could see on the scope that this made the BMB5 output change from 0 to +5 volts AFTER my disk circuit wrote the data into the NVRAM! Some research through the data books showed that a 2N5769, for example, would have been a much better replacement, with a storage time spec of 13 nS. Fortunately, after a little experimenting with "peaking" capacitors across the 390 ohm resistor, an 820 pF gave a sufficiently fast turn-on and turn-off with the 2N3904. So the object lesson here is that all switching transistors are not created equal, even in 1960's technology! Naturally since the DEC part numbers are no longer available, some thought is needed before just deciding on PNP or NPN... -Charles (Relearning long-dormant EE skills!)