Q2 Implementation Details
All logic in the Q2 is implemented using 2N7002 n-channel MOSFETs with resistor pull-ups. Compared to a bipolar logic family like TTL, NMOS is nice because it requires fewer components: an NMOS inverter requires only a single transistor and a single pull-up resistor. Unfortunately, the resistor pull-up presents a trade-off between speed and power, where a lower value resistor pull-up will make the gate faster at the expense of more power draw when the output is low. The situation is made worse by the fact that most modern discrete MOSFETs have a rather high gate capacitance (likely because they are made for switching large loads rather than implementing logic), making fast logic tricky to implement without using excessive power.
Although NMOS has a nearly unlimited fanout due to the high impedance between the input and output, the gate capacitance means that as we increase the fanout, the propagation delay increases. For the Q2, we try to use mostly 10k pull-up resistors as a compromise between speed and power. If we assume a load of 50pF and a Vth of 2.5 (pessimistic values for the gate capacitance and threshold voltage of a 2N7002), this gives us a propagation delay of around 347ns and a current draw of 0.5mA at 5V for a fanout of one. As the fanout increases to, for example, 24 to clock the A register, the propagation delay increases to 8.32us just for the last level of logic. If we wanted to clock the Q2 at 100kHz, this would leave us only 1.68us of slack for all other logic. To avoid such issues, we track the fanout and insert 1k resistors instead of 10k resistors in strategic locations, which in this case brings the delay down to 832ns with a draw of 5mA.
Since USB adapters are prevalent and provide an easy 5v power source, the Q2 was designed to use USB for power. In general, this limits the power consumption to be 500mA. It is relatively easy to verify that the Q2 will not draw too much power.
Each LED is driven through a 4.7k resistor. We assume a 2v voltage drop through each LED, giving us a power draw of 0.64mA per LED. Each gate uses a 10k, a 1k, or a 100k resistor pull-up (depending on the fanout and required speed), so we assume 0.5mA for each 10k resistor and 5mA for each 1k resistor, and 0.05mA for each 100k resistor. This provides an absolute worst-case estimate since we don't expect all gates and LEDs to be drawing power at all times (gates only draw power when the output is low). We assume the RAM chips and LCD each use 20mA. This gives us:
- 71 LEDs = 45mA
- 376 10k resistors = 188mA
- 19 100k resistors = 1mA
- 17 1k resistors = 85mA
- 2 RAM chips and 1 LCD = 60mA
So we get a total worst-case draw of 379mA or 1.9 Watts.
The following sections detail the various sections of the Q2.