ch3hwsol - Embedded Controller Design W00 Problem Solutions...

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Embedded Controller Design, W00 Problem Solutions 4 Solutions to Chapter 3 Problems For the following problems, refer to the loading example and Figure 3-15. 1. A) If a 10K pull-up resistor is used, how many additional LSTTL loads can be connected? B) How many CMOS loads could be added? When the LSTTL driver's output is 0: The output is assumed to be at 0.4 volts. This is max V(ol), and represents a noise margin of 0.4 volts from V(il) max of the CMOS and LSTTL loads. The CMOS gate input is sourcing a negligible amount of leakage current--less than 1 uA. The LSTTL gate input is sourcing as much as 360 uA (I(il) max) to the LSTTL output. The voltage across the 10k resistor is 4.6 volts (Vcc - V(ol) max). Therefore the current sunk by the LSTTL output is (4.6/10k) + 360 uA = 820 uA. The LSTTL output is specified to sink at least 3.2 mA and maintain V(ol) max, so it can sink an additional 3.2 - .820 = 2.38 mA. Since each additional LSTTL input would source 360 uA, the 2.38-mA excess sink capability of the LSTTL input represents 2.38/0.36 = 6 additional inputs. When the LSTTL driver's output is 1: The output is assumed to be at 3.4 volts; the min V(ih) for the CMOS device (3.0v), plus the same 0.4 volt noise margin that applies to the TTL signals. This represents a voltage drop of 1.6 volts across the 10k resistor. By Ohm's law, that's a current flow of 160 uA. The CMOS gate input is sinking a negligible amount of leakage current--less than 1 uA. The LSTTL gate input is sinking 60 uA of the 160 available. The excess source capability is 160 - 60 = 100 uA. Since each additional LSTTL input would require 60 uA, the circuit can support only one
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ch3hwsol - Embedded Controller Design W00 Problem Solutions...

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