Show how you would modify the nanocode of the add microinstruction (figure 12.2) so that it can

Show how you would modify the nanocode of the add...

Key Concepts

Nanocode

Nanocode is the low-level, detailed set of instructions that directly generate the control signals for the hardware components of a CPU. It breaks down higher-level microinstructions into specific bit patterns that control gates, registers, and ALU operations. In modifying the add microinstruction, understanding how to alter nanocode is essential for leveraging new circuitry.

Microinstruction Modification

Microinstruction modification involves altering the predefined instructions that control the micro-operations within the processor. In this context, the 'add' microinstruction needs adjustments in its nanocode representation to activate the new circuits, ensuring that the processor's control unit directs the new hardware correctly during the add operation.

Control Signal Integration

Control signal integration refers to the process of mapping new hardware features onto the existing control signals within the processor’s microarchitecture. The modification of the add microinstruction involves generating additional or altered control signals that correspond to the newly introduced circuitry, thereby integrating it into the micro-operation cycle.

ALU Operation Enhancement

ALU operation enhancement implies updating the instructions that drive the arithmetic logic unit (ALU) to take advantage of new hardware capabilities or improvements. The modified add microinstruction must configure the ALU to use any new pathways or optimized circuits provided by the enhancements, thereby potentially improving performance or efficiency.

Microprogrammed Control Unit

The microprogrammed control unit is the part of the CPU responsible for interpreting microinstructions and generating appropriate control signals. When new circuitry is introduced, as in problems 12.17 and 12.18, the microprogrammed control unit must be reprogrammed to correctly decode and dispatch the revised control signals embedded in the modified nanocode for the add operation.

Transcript

00:02 Adding two batteries to the circuit in problem 19 decimal 5 can be done in two different ways.

00:12 That is either in parallel or in series.

00:18 So using the same circuit components, both series and parallel resistors, as in problem 19 decimal 5, we connect two batteries in parallel, parallel circuits maintain constant voltage whereas current fluctuates.

00:41 So as we can see here, starting from the negative terminal of the first battery, a wire runs to the negative terminal of the second battery as well as the positive terminals connected to each other and both batteries through the red wire here the positive and we've added diodes in here so we can keep the flow going one way then the positive of the second battery goes to a power distribution board and then starts on the circuit series circuit pathway running to the point zero three om resistor and in this case because we have a 200 amp current because two batteries are connected in parallel there will be a voltage drop of 6 volts leaving 6 volts to run down the parallel resistor right here rated at 0 .18 oms which will cause a 33 .3 amp current absorption so so that leaves us from 200 amps, 166 .7 amps going to the light bulb pushed by 6 volts.

02:30 So we're assuming the bulb can handle this.

02:33 However, the electrical dynamics considerably change.

02:41 So we no longer have 9 volts.

02:44 We have a voltage drop that is a 3 volt difference, of course, from problem 19.

02:54 Decimal 5, so we're left with 6 volts and of course we have a fluctuation of 116 .7 amps increase compared to problem 19 .5 so we're left with 166 .7 amps to the light bulb.

03:14 So if we were going to relate this to the piston driven water pump running a continuous water flow hydro mechanical circuit, we first have to consider that the four -stroke gasoline piston engine used obeys more or less a quadratic torque curve, which is somewhat actually considerably different from, let's say, diesel, or two -stroke gasoline piston, because our torque curves could be logarithmic or power functions, but with the four -stroke gasoline, piston, engine, single cylinder, we have.

04:01 A governor set at 3 ,600 rpm, which in this particular case is where the torque reaches its maximum of 2 .77 foot pounds.

04:17 So any rpm greater than or less than 3 ,600 will result in a drop of torque.

04:24 So because we have constant voltage, we can use this translational mechanical translation example to match the doubling of the battery in parallel equivalent to doubling putting two motors in parallel together so it is correct to rate the wattage of 2800 watts or 3 .8 brake horsepower instead of 4 ,200 watts and 1 .9 brake horsepower so therefore we're keeping a constant velocity or rpm or in this case voltage.

05:07 So the result will be a doubling of the torque right here to 5 .54 foot pounds at 3 ,600 rpm constant speed governed, which results in twice as much source system pressure, which is equivalent to these two batteries in parallel.

05:32 Thus, 5 .54 foot pounds of torque instead of 2 .77 at 3 ,600 rpm.

05:41 So we're using the same step -down pressure regulator, which is equivalent to the parallel resistor, and also the same step -down flow regulator, which is equivalent to the series resistor.

05:59 So the result is we'll have a drop in velocity.

06:04 However, a substantial increase in water pressure, extra water here, sorry, substantial increase in water pressure and a moderate decrease in water velocity.

06:20 That will be the translation to connecting two batteries in parallel from the same problem, problem 19 decimal 5.

06:34 So that's the first way to connect these two batteries.

06:39 The second way is to connect them in series, which will be demonstrated in the following.

06:56 So here now we have the batteries connected in series where the current remains constant.

07:06 However, the voltage has doubled itself to 24 volts.

07:16 So again, we start from the negative terminal of the first battery.

07:23 We have a connection to the positive terminal of the second battery, and of course the positive terminal of the first battery to the negative terminal of the second battery and again we have our diodes here to keep the flow moving from the second positive of the second battery flowing from positive to negative to the power distribution board then it takes off down the series pathway to the 0 .03 series resistor resulting in a 3 -volt drop just like in problem 19 decimal 5...

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