MC1495 Просмотр технического описания (PDF) - ON Semiconductor
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MC1495 Datasheet PDF : 20 Pages
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MC1495
To set currents I3 and I13 to the desired value, it is only
necessary to connect a resistor between Pin 13 and ground,
and between Pin 3 and ground. From the schematic shown
in Figure 3, it can be seen that the resistor values necessary
are given by:
|V−| −0.7 V
R13 + 500 Ω =
I13
|V−| −0.7 V
R3 + 500 Ω =
I3
Let V− = −15 V, then R13 + 500 = 14.3 V or R13 = 13.8 kΩ
1.0 mA
Let R13 = 12 kΩ. Similarly, R3 = 13.8 kΩ, let R3 = 15 kΩ
However, for applications which require an accurate scale
factor, the adjustment of R3 and consequently, I3, offers a
convenient method of making a final trim of the scale factor.
For this reason, as shown in Figure 21, resistor R3 is shown
as a fixed resistor in series with a potentiometer.
For applications not requiring an exact scale factor
(balanced modulator, frequency doubler, AGC amplifier,
etc.) Pins 3 and 13 can be connected together and a single
resistor from Pin 3 to ground can be used. In this case, the
single resistor would have a value of 1/2 the above
calculated value for R13.
Step 2. The next step is to select RX and RY. To insure that
the input transistors will always be active, the following
conditions should be met:
VX < I13,
RX
VY < I3
RY
A good rule of thumb is to make I3RY ≥ 1.5 VY(max) and
I13 RX ≥ 1.5 VX(max). The larger the I3RY and I13RX product
in relation to VY and VX respectively, the more accurate the
multiplier will be (see Figures 17 and 18).
Let RX = RY
then I3RY
I13RX
= 10 kΩ,
= 10 V
= 10 V
since VX(max) = VY(max) = 5.0 V, the value of
RX= RY = 10 kΩ is sufficient.
Step 3. Now that RX, RY and I3 have been chosen, RL can
be determined:
K=
2RL
RX RY I3
=4
10
, or
(2) (RL)
=4
(10 k) (10 k) (1.0 mA) 10
Thus RL = 20 kΩ.
Step 4. To determine what power supply voltage is
necessary for this application, attention must be given to the
circuit schematic shown in Figure 3. From the circuit
schematic it can be seen that in order to maintain transistors
Q1, Q2, Q3 and Q4 in an active region when the maximum
input voltages are applied (VX′ = VY′ = 10 V or VX = 5.0 V,
VY = 5.0 V), their respective collector voltage should be at
least a few tenths of a volt higher than the maximum input
voltage. It should also be noticed that the collector voltage
of transistors Q3 and Q4 is at a potential which is two
diode-drops below the voltage at Pin 1. Thus, the voltage at
Pin 1 should be about 2.0 V higher than the maximum input
voltage. Therefore, to handle +5.0 V at the inputs, the
voltage at Pin 1 must be at least +7.0 V. Let V1 = 9.0 Vdc.
Since the current flowing into Pin 1 is always equal to 2I3,
the voltage at Pin 1 can be set by placing a resistor (R1) from
Pin 1 to the positive supply:
R1 =
V+ −V1
2I3
Let V+ = 15 V, then
R1
=
15
(2)
V −9.0 V
(1.0 mA)
R1 = 3.0 kΩ.
Note that the voltage at the base of transistors Q5, Q6, Q7 and
Q8 is one diode-drop below the voltage at Pin 1. Thus, in
order that these transistors stay active, the voltage at Pins 2
and 14 should be approximately halfway between the
voltage at Pin 1 and the positive supply voltage. For this
example, the voltage at Pins 2 and 14 should be
approximately 11 V.
Step 5. For dc applications, such as the multiply, divide
and square-root functions, it is usually desirable to convert
the differential output to a single-ended output voltage
referenced to ground. The circuit shown in Figure 22
performs this function. It can be shown that the output
voltage of this circuit is given by:
VO = (I2 −I14) RL
And since IA −IB = I2 −I14 =
2IX IY
I3
=
2VXVY
I3RXRY
then
VO =
2RL VX′ VY′
4RX RX I3
where, VX′ VY′ is the voltage at
the input to the voltage dividers.
V+
RO
I2
V2
RO
+
V14
I14
−
VO
RL
RL
Figure 22. Level Shift Circuit
The choice of an operational amplifier for this application
should have low bias currents, low offset current, and a high
common mode input voltage range as well as a high common
mode rejection ratio. The MC1456, and MC1741C
operational amplifiers meet these requirements.
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