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Extra info for Lessons in Electronic Circuits - Digital

Example text

However, the right steering diode is conducting current through the base of Q2 , through resistor R1 : 38 CHAPTER 3. ” More specifically, it will be saturated by virtue of the more-than-adequate current allowed by R1 through the base. 2. THE NOT GATE 39 level. Due to the presence of diode D2 , there will not be enough voltage between the base of Q3 and its emitter to turn it on, so it remains in cutoff. Let’s see now what happens if we reverse the input’s logic level to a binary ”0” by actuating the input switch: Vcc = 5 volts Vcc + Vcc R4 R1 5V R2 - Q3 Input D1 Q2 D2 "Q1" 0V Output Q4 R3 ≈5V Now there will be current through the left steering diode of Q1 and no current through the right steering diode.

Conversely, they output a ”low” (0) logic level if the inputs are at the same logic levels.

5. OVERFLOW 25 The answer (1001002 ), interpreted with the sixth bit as the -3210 place, is actually equal to -2810 , not +3610 as we should get with +1710 and +1910 added together! Obviously, this is not correct. What went wrong? The answer lies in the restrictions of the six-bit number field within which we’re working Since the magnitude of the true and proper sum (3610 ) exceeds the allowable limit for our designated bit field, we have an overflow error. Simply put, six places doesn’t give enough bits to represent the correct sum, so whatever figure we obtain using the strategy of discarding the left-most ”carry” bit will be incorrect.

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