Created it, 06/10/19
Update it, 06/11/02
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3. - SECOND EXPERIMENT : EXAMINATION OF A DIGITAL CONVERTER / ANALOGICAL A 4 BITS
While using two operational amplifiers and a resistance network, you will produce a converter D / A with 4 bits.
The principle of operation is based on the possibility of varying the profit of an amplifying assembly by modifying the value of the resistance connected to the inverseuse entry of an operational amplifier.
It is a converter with balanced resistances.
3. 1. - REALIZATION OF THE CIRCUIT
The electric diagram of the assembly to be realized is given figure 5-a.
a) Remove matrix the whole of the connections and components relating to the first experiment.
b) Carry out the assembly located figure 5-b, without connecting for the moment the negative pole of the pile.
c) Put switches SW0, SW1, SW2 and SW3 on position 0.
3. 2. - OPERATIONAL TEST
a) Connect the negative pole of the pile to the circuit using the black crocodile clip, then put the digilab under tension.
b) Prepare the controller for the measurement of tension D.C. on the gauge 10 volts.
Measure the output voltage S of the circuit. For that, connect the negative test probe to the mass and the positive test probe on the terminal of R7 connected to pin 10 of the LM 747. You must find a tension zero or very low (tension of offset).
c) Put all the switches on position 1.
d) Regulate the trimmer P so that the output voltage S is equal to 3,75 volts. You calibrated the converter.
e) Put reversers SW1, SW2 and SW3 on 0 and leave SW0 on position 1.
f) Measure the output voltage S. It is equal to approximately + 0,25 volt. This value corresponds to the least significant bit (LSB) of the binary number present at the entry of the converter.
g) Put SW0 on 0 and SW1 out of 1.
h) Measure the output voltage.
You must find approximately + 0,5 volt.. The weight of this bit (bit 3) is worth the double of that of the LSB.
i) Give SW1 on 0 and put SW2 on position 1.
j) Measure the output voltage of the converter D / A.
You find + 1 approximately volt, this value corresponds to bit 2.
k) Put SW2 on 0 and SW3 out of 1 and take same measurement as previously.
You must approximately find + 2 volts, this value corresponds to the MSB (bit 1). This series of measurements enabled you to check the weight of each bit.
It is very probable that you did not completely find the values indicated and this, for two reasons.
First of all, the values of resistances R4, R3, R2 and R1 should follow a geometric progression of reason 2, i.e. to be worth 10 kW, 20 kW, 40 kW and 80 kW. However, the standardized values of the E12 series closest are 10 kW, 22 kW, 39 kW and 82 kW.
Then, it there with the tolerance of resistances and the precision of the controller which come into play the found actual value.
Nevertheless, this experiment has value only by its didactic side.
l) Now carry out all the possible combinations using the switches and for each one of them, measure the output voltage.
Defer the whole of the values in the third column of the table of figure 6.
The second column indicates to you the theoretical values of the output voltage.
m) Defer these theoretical values on the graphics of figure 7 which shows the ideal characteristic of transfer of this converter D / A to 4 bits.
As example, the graphics of figure 8 represent the real characteristic of a converter D / A with 4 bits.
You notice that this characteristic is not linear and that the converter has a nonnull tension of offset (0,34 volt for 0000 in entry).
n) Put the digilab not under tension and disconnect the pile.
The digital / analogical converter that you have just tried out is one of simplest than one can realize. It has a weak resolution and moreover, its precision is not very high.
On the market, there are converters D / A carried out in integrated circuits.
We will examine the operation of this converter.
Figure 9 makes it possible to see how the binary number is applied to the entry of the converter using the 4 switches.
When a bit is worth 0, corresponding resistance is connected to the mass thanks to the switch. When it is worth 1, resistance is connected to the + 5 volts thanks to this same switch.
Let us examine the circuit of figure 9.
R4 resistance corresponding to the MSB is connected to the + 5 volts and other resistances (R3, R2 and R1) are connected to the mass.
The inverseuse entry (-) of amplifier A is with potential 0 since the noninverseuse entry (+) is cabled with the mass. Therefore, in the calculation of the profit of A, three resistances R1, R2 and R3 do not intervene.
This G1 profit is worth :
G1 = - (R5 / R4) = - (2,7 kW / 10 kW) = - 0,27
The output signal on pin 12 is in opposition of phase compared to the input signal on pin 1.
The tension VA on pin 12 is worth :
VA = G1 x VI = - 0,27 x 5 = 1,35 volt
This first stage A operates conversion itself D / A. Nevertheless, one wishes to thus obtain a positive tension this first stage A is followed of a second stage B amplifying of tension and reverser.
The output voltage VB must be worth + 2 volts. The G2 profit of the amplifier B is thus worth :
G2 = VB/GOES = 2/- 1,35 = - 1,48148
This G2 profit is equal to - (R7 + P) / R6
P varies between 0 W and 10 kW.
For P = 0 W, G2min = - (R7 / R6) = - (5,6 kW / 4,7 kW) = - 1,2
For P = 10 kW, G2max = - (R7 + P) / R6 = - (5,6 kW + 10 kW) / 4,7 kW = - 3,32
To obtain the profit necessary (- 1,48148), it is enough to regulate the trimmer P.
The total profit G of the converter is thus worth :
G = G1 x G2 = (- 0,27) x (- 1,48148) @ 0,4
However, G is equal to VB / Ve, from where :
For Ve = 5 volts, G2 = 1,48148, R5 = 2,7 kW one obtains :
It is enough to apply this formula (3) to calculate tension VB according to the binary number applied to the entry.
For bit 1 (MSB), one will have VB = 20 / R4 = 20 / 10 = 2 volts
For bit 2, one will have VB = 20 / R3 = 20 / 22 = 0,9 volt
For bit 3, one will have VB = 20 / R2 = 20 / 39 = 0,5 volt
For bit 4 (LSB), one will have VB = 20 / R1 = 20 / 82 = 0,24 volt
For an unspecified binary number, one can use two methods:
Either one calculates the resistance of equivalent
entry, equal to the resistances put in parallel and connected to the
+ 5 volts. Then, it is enough to apply the formula (3).
Either one adds the output voltages corresponding to
each bit with 1.
For example, for 11002, bits 1 and 2 are to 1. One thus adds + 2 volts and + 0,9 volt and the result is worth : VB = + 2,9 volts.
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