1250 F 14 1. Homework Circuit System Design Cards 2 In this problem, you will design a level-shifter that translates voltages coming out of a comparator with power supplies of ±15V to levels appropriate for driving a logic gate input with power supplies of +5V and 0V, as in the (partial) circuit shown below. According to datasheets, we have the following voltage levels for a LM324 output when used as a comparator and for a 74HC00 NAND gate input: LM324* Characteristic Symbol Min VOH 3.3 Typ 3.5 V Output Voltage-High Limit (VCC = 30 V) VOH 26 28 V Output Voltage-Low Limit (VCC = 5 V) VOL 5.0 20 mV Output Voltage-High Limit (VCC = 5 V)** Max Unit *LM324 datasheet: http://www.digikey.com/product-search/en?vendor=0&keywords=lm324ngos **Power supplies are V+ = VCC and V- = 0V 74HC00* Characteristic High-level input voltage (VCC = 4.5 V) Symbol Min VIH 3.15 Absolute Min VIN VIN Low-level input voltage (VCC = 4.5 V) VIL Typ Max V –0.5 Absolute Max VIN VIN *74HC00 datasheet: http://www.digikey.com/productsearch/en?x=0&y=0&lang=en&site=us&keywords=tc74hc00 Unit V 1.35 V 5 V Use the above data to determine, as well as possible, values for output high (vTh<Hi) and output low (vTh<Low) from the comparator and input high (vThHi) and input low (vThLow) for the logic gate. Use these values in the equations on the "Op-amp LevelShifter" card (see below) to find resistor values to shift from vTh<Hi to vThHi and from vTh<Low to vThLow. That is, calculate ΔvTh<, ΔvTh, vTh< , and vTh , and then use these values to calculate G1 and G2, from which ratios of resistor values may be obtained. Assume RTh< = 0 ΩΩ, V+ = 15V and V– = –15V for level shifter, and V– is used on the bottom of Rs. Also, use resistors that are as large as possible without exceeding 1 MΩΩ. ANS: 1. vTh<Hi = 13 V (shift 28 V down by 15V since power supplies are ±15V instead of +30V and 0V). vTh<Low ≈ –15V (shift 20 mV, which is almost zero, down by 15V since power supplies are ±15V instead of +30V and 0V). 3.15 * 5V/4.5V = 3.5 V ≤ vThHi ≤ 5 V * 5V/4.5V = 5.56 V (since we use power supply of 5V instead of 4.5V). A practical choice might be to choose vThHi = 5 V = power supply voltage and subtract 0.5 V to give a margin of error in case resistor values are slightly off. So my choice would be vThHi = 4.5 V. Your choice might be different. –0.5 V ≤ vThLow ≤ 3.15 V * 5V/4.5V = 3.5 V (since we use power supply of 5V instead of 4.5V). A practical choice might be to choose vThLow = 0 V = power supply voltage and add 0.5 V to give a margin of error in case resistor values are slightly off. So my choice would be vThLow = 0.5 V. Your choice might be different. ΔvTh< = 28V , ΔvTh = 4 V , vTh< = −1V , vTh = 2.5V G2 = (4)(−1) − (2.5)(28) 4 = 0.176 , G1 = = 0.121 (−15)(28) 28(1+ 0.176) Rs = 1 MΩΩ, Rf = 176 kΩΩ or 180 kΩΩ, Ra = 1 MΩΩ, Rb = 137 kΩΩ or 130 kΩΩ Check: vTh = 0.121(1+ 0.176)13V − 0.176(−15V) = 4.49V or 4.5 V (correct for 18V in) vTh = 0.121(1+ 0.176)(−15V) − 0.176(−15V) = 0.505V or 0.5 V (correct for -15V in)