# What is this chip enable circuit doing? (old TTL, with diode, with pull-up?)

I'm trying to understand a small portion of a (vintage, TTL) circuit design and I'm a little unsure of what's going on in this section*, which determines whether the Chip Enable signal on a particular IC is high or low:

The "typical" logic that should determine the CE signal state is behind that 7408. My understanding is that signal A acts as an "override".

"A" is a TTL signal coming from three-state line, so it's either high, or low, or possibly high-z, though of course it's only interesting when it's high or low.

The output of a 7408 will always be either low or high.

The diode is confusing me here. It clearly is related to the function of signal A, but feels like that defeats a normal high signal from the 7408. So by my rough understanding, CE would only be high when both the output at the 7408 is high and signal A is high, but that doesn't map to actual behavior and it also doesn't make much sense, so I'm clearly misunderstanding the current flow in this situation.

What is the diode's function and how does current flow in the "truth table" for CE (7408 high or low; "A" high or low)

Possibly related question: Why set it up this way, instead of (say) using an additional appropriate logic gate

This feels like basic EE ideas that I'd appreciate clarity on, if someone would briefly walk me through the different cases of what's happening here and how. Thanks.

*(Section is redrawn detail from this schematic, which is a bit of a mess. The IC is "E8" at the upper left of page two. For simplicity I've omitted another signal configured like "A" coming into the same CE line because it's disconnected/unused).

It's a wire-or circuit, where the gate can drive a dominant '0' in place of the signal in 'A'.

So you get:

gate   "A"   "CE"
1       0      0
1       1      1
0       x      0  <--- gate dominates


Why do this? It's an inexpensive way to add a function - it only costs one resistor and one diode.

I've done similar for reset circuits.

• Thank you. Two questions: one, confirming that I guess my original vibe was right that chip enable can only go high when both the gate and A are high? Feels like I might call this a "wire-and" instead, no? Two, why the diode? What is the current flow we are needing to prevent in which case?
– BZo
Commented Apr 4 at 18:10
• Yes, it forms an AND. What seems to be happening is when that gate is high, it's allowing the light pen to manipulate the character gen CE - perhaps to make a tracking cursor. Commented Apr 4 at 18:34
• The diode is there to allow "A" to toggle low when the gate is high. It blocks the gate high-drive. Commented Apr 4 at 18:43
• I should add that the 2513 CE input is active low. Thus, the gate input being low forces the char gen value onto the output, otherwise the light pen is allowed to control it. So it's a bit more useful to think of the function as being a negative-logic OR. Commented Apr 4 at 18:56
• found a schematic here: github.com/bbenchoff/2513CharGenAdapter/blob/main/Schematic.pdf Commented Apr 4 at 19:49

Your are correct, it is a wire-AND.

The classic circuit of this type would be two diodes (one for each input, logic and A) and one pull-up resistor. Both diodes have to be high for the result to be high. Having one of the inputs "power" the circuit eliminates one diode.