Reverse-engineering the 1977 Unisonic 21 calculator/game (part 4)

A while ago I stopped working on reverse engineering the Unisonic 21. I think I was toddling after a shiny balloon. Anyway, I was recently talking with one of the creators of the MOnSter 6502 [see also Evil Mad Scientist's blog post] who mentioned he was looking for another chip to dis-integrate. I thought the Unisonic 21 CPU would make a nice target! But it needs to be fully reverse-engineered first. So, here we go again.

When we last left our reluctant hero, we had just seen a complicated circuit whose purpose was to provide a lot of oomph to create a high-powered clock driver.

Here is the full circuit:


It's pretty huge, but we already know that all that capacitor stuff in the middle of each section is just there to drive the transistors hard. Let's remove all that clutter, and also replace the other bootstrap loads with plain old resistors and see what we're left with.


OK, it's still a little cluttered. First, notice at the top that I've renamed the incoming clock signals according to the phases they are active in. This might help later.

Notice that Q16 is immediately recognizable as an inverter. That means the signal coming out of it is just P0_P1_P2: the other three of the six clock phases.

Now let's split the circuit down the middle and analyze the right half:


If we take a look at the top half, we see that P3_P4_P5 feeds the gate of Q1. Q1 is on if its gate is low, which means during P0_P1_P2. This means the signal X1 is pulled high during P0_P1_P2.

What about Q2? It is pulled high if Q11 is on, which happens when the gate of Q11 is low, which is during P0_P1_P2. But there's that pesky Q13 in there which could modify the behavior at the gate of Q2. So let's take a look at the bottom half.

The P1_P4 signal is gated at Q22 by P0_P1_P2. It will only pass if P0_P1_P2 is low, or if P3_P4_P5 is high. So the other side of Q22 is essentially P3_P4_P5 AND P1_P4, which is just P4. This is the signal Y-, which gets inverted at Q6 to form P0_P1_P2_P3_P5 at output Y1.

Now, P4 gets inverted at the gate of Q24, and then push-pull buffered at Q23/Q19, and low-buffered again at Q5. So this is still P0_P1_P2_P3_P5 at output Y1. So far, so good.

So the gate of Q13 is what comes out of Q24, which is P0_P1_P2_P3_P5. Q13 will only be on if its gate is low, which means P4.

Now look at Q11 again. It is on only if its gate is low, which means P0_P1_P2. So the signal X+ is pulled high if Q11 is on (P0_P1_P2) and low if Q13 is on (P4). The rest of the time it appears to just be floating.

However! Because Q13 has intrinsic capacitance, it will not immediately stop conducting. It will probably keep its state until some other thing drains it. So we can probably say Q13 is on, pulling the signal X+ low, during P3_P4_P5.

From this, we can talk about Q2. Q2 is on if its gate is low, which only happens on P3_P4_P5. And from before, Q1 is on if its gate is low, which means during P0_P1_P2. So in the end, we definitely have Q1/Q2 being a push-pull buffer for P0_P1_P2.

So why Q13 and not just an active load? Maybe because that would require essentially duplicating the circuit below, which requires four transistors aside from the push/pull output. The top circuit only needs two transistors.

Okay, we have reasoned out what this circuit should do. The conclusion is that signal X1 is P0_P1_P2 and signal Y1 is P0_P1_P2_P3_P5 (or /P4).

Is it what it really does? I'm dealing with unfamiliar PMOS territory here, and that appeal to Q13 retaining its state through two clock periods fills me with unease. One step to take is to actually build this circuit and see what happens. But that will be another blog post.

In the meantime, there's the left half of the circuit to deal with. It is nearly the same thing:


We can proceed in the same way, except we've already done most of this work. This circuit is identical to the right half, except that the signal feeding Q25's gate is P3_P4_P5 and the signal at X- is P0_P1_P2. But we will go through the exercise again. We start with the bottom half.

Q25 is on only if P1_P4 is high and if P3_P4_P5 is low (i.e. P0_P1_P2 is high). This means Q25 is on during P0_P1_P2 AND P1_P4, which is P1. This is signal Y-.

This is inverted by Q27, so Q27's output is /P1. Q30 is on only if /P1 is low, or during P1. And Q37 is on when Y- is low, which is during /P1, so Q30/Q37 is a push-pull buffer for /P1. This is signal Y+.

So, Q41/Q42 thus form a push-pull buffer for /P1. This is signal Y2. Note that signal Y1 was /P4.

Signal X- is low during P3_P4_P5, which pulls signal X2 high via Q44.

Q43 is on when signal X+ is low. This happens when Q29 is on, which happens when Q27 is off. This is during P1.

Q31, however, is on, pulling signal X+ high, when signal X- is low. This is during P3_P4_P5. Again, from the same reasoning, we can consider Q29 on, pulling X+ low during all other phases (especially P1 of course).

Q43 is on, pulling signal X2 low, when signal X+ is low, which is during P0_P1_P2. Q44 is on, pulling signal X2 high, when signal X- is low, which is during P3_P4_P5.

So we see that Q43/Q44 forms a push-pull buffer for P3_P4_P5. This is signal X2. Note that signal X1 was P0_P1_P2.

X1 is the opposite of X2, Y1 is a negative pulse in the middle of /X1, and Y2 is a negative pulse in the middle of /X2.


At least, this is what we hope it is. We'll just have to build the circuit and see.

The game of Chicken Kickin' (where chickens must not be kicked)

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Chicken Kickin'

From Wikipedia, the free encyclopedia

Chicken Kickin' is a game enjoyed by men and boys throughout the United States. It is played mainly in rural areas, although there is an urban version called The Colonel's Chicken Lickin' Good Kickin'.[1]


The game starts immediately upon the completion of several games of Cornhole[2] and several packs of beer, whereupon the players first attempt to holler louder than ever. One self-selected man, called the Corkscrew, then releases the chickens, which may take several tries as coordination is one of the first things to go. Another man, called the Bottle, begins the game by calling out, "Woohoo, chickens!" The remaining players, called Drinks, or sometimes Drunks, must laugh uproariously and attempt to catch a chicken. The chickens, called Chickens, must dodge the Drinks and the Bottle, and make it past the Corkscrew back into their pen -- a task made more difficult due to the wobbling nature of the Corkscrew.

Although the game is called Chicken Kickin', the Chickens are rarely Kicked, and if they are, only by accident. In any case, score is kept by the Ground, which is the surface on which the Empty Beercans are thrown, and which Drinks fall upon to be considered "safe." Scoring is generally random. The game ends a few minutes after the last beer is drunk, whereupon the Drinks tally up the score by cursing at the Chickens while picking up the Beercans and attempting to place them in a plastic bag brought out by the Corkscrew. The final scoring is rarely determined, because the players normally lose interest and weave to their cars, where they are kept awake by the sounds of sad country music.[3] In urban settings, the music of choice is Rebecca Black's Friday.


The ancient Romans had a similar game, called Gallinae Calcitrandae, or The Chickens Must be Kicked. The only mention of the game is in Fellatius, where it is said to take place on the Bacchanalia on the night of March 15. Although the rules are not described, it is lamented as "a corrupting influence on small boys" and "requiring much washing of togas." Fellatius also comments that the game was a favorite of Emperor Hadrian (76 - 138 A.D.), who was said to have replaced chickens with the British.[4]

The Romano-British created a derivative game in 409 A.D., called Romanae Excalcitrandae, or The (Effeminate) Romans Must be Kicked Out.[5]


Second try at USSR integrated circuit analysis

In the previous post, I logicked my way into the geometry of a transistor. But in fact, the geometry turned out to be wrong since none of the circuitry made any sense. So I decided to start from this diagram, which is similar to the 74141 diagram. Which makes sense since the К155ИД1 is the same as the 74141.

From .
From the 74141 datasheet.

From the 74141 datasheet.

So if I look at the A input, I should see an inverter and a second inverter. The outputs from these go to output transistors.

Next, I drew one of the transistors in the above circuits as a generic three-terminal device: 0 for a terminal with no region around it, 1 for one region, and 2 for two regions.

Then, I drew the inverter circuits based on what I saw on the chip.

Then I logicked my way to a reasonable schematic. The two-terminal question-marked component seems to be maybe a resistor. 0 seems to be the base, 1 is the emitter, and 2 is the collector.

OK, this makes some sense. The two transistors on the left are an inverter. X taps off the output of the inverter, and the third transistor on the right is what you'd typically see in a 7400-series input. It's just two diodes back to back in a current-steering circuit. Since it taps off the emitter of the output, it is the inverse of X.

That means that we have this interpretation of a transistor:

Now, let's take a look at the outputs.

OK, that's not right according to the datasheet. But on the other hand, it could work. If A is high, and if DCB = 100, then we want 9 to sink current and 8 to be off. If A is high then X is low, which means that as long as the base is active, 9 will sink current.

I still haven't identified the protective zener diodes on the outputs.

So if the terminal designations are correct, then it's possible that this is the doping schema.


So what I've been calling the "hillock" in the middle of the rectangular region is perhaps the p-doped base. Again, this is weird compared to what you usually see, which is the collector completely surrounding the base.

The collector probably has an n+ region, which improves connectivity. It is embedded in an n or n- region forming the collector.

Note that I think the base contact seems to overlap into the rectangular region. If the rectangular region is n-doped and somehow connects to the collector, which is perhaps an n- region, then this may form a Schottky diode, meaning that these transistors are Schottky transistors.

So technically the collector does surround the base, it's just that it does it through a differently doped region.

Again, I don't have enough knowledge of IC fabrication to know whether this would work. But it seems... logicky?