MC1495 ON Semiconductor, MC1495 Datasheet - Page 7

• Manufacturer Part Number: MC1495
• Description: LINEAR FOUR-QUADRANT MULTIPLIER
• Manufacturer: ON Semiconductor

Theory of Operation
which
transconductance. A detailed theory of operation is covered
in Application Note AN489, Analysis and Basic Operation
of the MC1595. The result of this analysis is that the
differential output current of the multiplier is given by:
where, I
respectively, and V
at the multiplier input terminals.
DESIGN CONSIDERATIONS
General
to a specific application by proper selection of external
components. External components may be selected to
optimize a given parameter (e.g. bandwidth) which may in
turn restrict another parameter (e.g. maximum output
voltage swing). Each important parameter is discussed in
detail in the following paragraphs.
Linearity, Output Error, E
output voltage from a straight line transfer function. It is
expressed as error in percent of full scale (see figure below).
percentage error is:
following methods:
nonlinearity in the X and Y input differential amplifiers. To
avoid introducing error from this source, the emitter
degeneration resistors R
enough so that nonlinear base-emitter voltage variation can
E
100 mV and the full scale output is 10 V, then the
R
The MC1495 is a monolithic, four-quadrant multiplier
The MC1495 permits the designer to tailor the multiplier
Linearity error is defined as the maximum deviation of
For example, if the maximum deviation, V
Linearity error may be measured by either of the
1. Using an X-Y plotter with the circuit shown in
2. Use the circuit of Figure 4. This method nulls the level
One source of linearity error can arise from large signal
=
Figure 5, obtain plots for X and Y similar to the one
shown above.
shifted output of the multiplier with the original
input. The peak output of the null operational amplifier
will be equal to the error voltage, V
V
V
O
E
A
(max)
(max
operates
and I
)
x 100 =
B
I
A
X
are the currents into Pins 14 and 2,
– I
on
and V
B
100 x 10
= I =
X
RX
the
Y
and R
10
are the X and Y input voltages
or E
2V
OPERATION AND APPLICATIONS INFORMATION
R
–3
principle
Y
X
RY
X
R
V
x 100 = 1.0%.
must be chosen large
Y
Y
I
3
E (max)
of
.
E(max)
variable
http://onsemi.com
, is
MC1495
7
be ignored. Figures 17 and 18 show the error expected from
this source as a function of the values of R
operating current of 1.0 mA in each side of the differential
amplifiers (i.e., I
3 dB Bandwidth and Phase Shift
and the stray multiplier output capacitance and/or the
operational amplifier used to level shift the output. If
wideband operation is desired, low value load resistors
and/or a wideband operational amplifier should be used.
Stray output capacitance will depend to a large extent on
circuit layout.
sources: phase shift common to both X and Y channels (due
to the load resistor-output capacitance pole mentioned
above) and relative phase shift between X and Y channels
(due to differences in transadmittance in the X and Y
channels). If the input to output phase shift is only 0.6 , the
output product of two sine waves will exhibit a vector error
of 1%. A 3 relative phase shift between V
in a vector error of 5%.
Maximum Input Voltage
Exceeding this value will drive one side of the input
amplifier to “cutoff” and cause nonlinear operation.
(observing power dissipation limitation) between 0.5 mA
and 2.0 mA, approximately 1.0 mA. Then R
determined by considering the input signal handling
requirements.
is valid. Reference to Figure 19 will indicate limitations of
V
will cause saturation or “cutoff” of the input transistors. See
Step 4 of General Design Procedure for further details.
is derived from I
with the assumption R
At T
X(max)
Bandwidth is primarily determined by the load resistors
Phase shift in the multiplier circuit results from two
V
Current I
Therefore, with R
The equation I
X(max)
A
= +25 C and I
or V
For V
, V
3
Y(max)
Y(max)
and I
X(max)
3
A
A
= I
– I
– I
due to V
R
2kT
qI
13
X
input voltages must be such that:
13
B
13
X
B
V
V
13
= V
= R
X
= R
=
=
Y
are chosen at a convenient value
= I
X(max)
= 1.0 mA).
(max) <I
=
>> 2kT
Y(max)
3
Y
R
Y
2V
(R
2kT
= 1.0 mA,
qI
1
X
= 10 k the above assumption
>
X
and V
qI
X
3
R
<I
1.0 mA
+ 2kT
13
V
Y
13
= 52 .
10 V
= 10 V;
3
Y
I
qI
3
and R
R
R
7
13
. Exceeding these limits
Y
Y
2V
) (R
= 10 k .
Y
X
>> 2kT
X
V
Y
X
X
Y
+ 2kT
and R
and V
and R
qI
qI
3
3
Y
Y
.
Y
) I
with an
results
3
can be