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)