Tektronix CURVE-TRACER 576 Instruction Manual page 72

Circuit Description-Type 576
Display Switching
Once the desired voltages and currents have been sensed
by the display sensitivity switching circuit, and once the
desired positioning currents have been obtained from the
display positioning circuit, the resulting voltage signals and
positioning currents must be applied to the display
amplifiers. Before being applied to the display amplifiers,
however, these signals pass through the display switching
circuit (see the Display Amplifiers and Display Positioning
Switches schematics).
Under normal operating conditions with neither the DIS-
PLAY INVERT, the ZERO nor the CAL buttons pressed,
these signals and currents pass directly to the display ampli-
fers. If the DISPLAY INVERT button ispressed,however,
the signal and current (CENTER LINE VALUE Switch
and POLARITY switch positioning current) input lines to
both amplifiers are reversed. This causes the display on the
CRT to be inverted, both vertically and horizontally.
The ZERO button, when pressed, disconnects the signal
input lines from both pairs of high impedance inputs and
shorts the input pairs together. This provides a zero refer-
ence for both display amplifiers. If the DISPLAY OFFSET
controls are being used when the ZERO button is pressed,
offset positioning current is caused to flow as if the
CENTERLINE VALUE switch were set to 0 (see Display
Positioning schematic and discussion of positioning).
The CAL button, when pressed, disconnects the signal
input lines from both pairs of high impedance inputs and
applies a substitute voltage across each input pair which
should cause full graticule deflection (10 divisions by 10
divisions). This provides a means of checking the accuracy
of calibration of the display amplifiers. The substitute volt-
age is determined by R501 through R513 and by D507.
Since each display amplifier has three gains to check, three
substitute voltages must be available. Relays K537C,
K541C, K637C and K641C determine which voltages are
applied to the high impedance input pairs for various
settings of the VERTICAL and HORIZONTAL switches. If
the DISPLAY OFFSET current controls are being used
when the CAL button is pressed, offset current is caused to
flow as if the CENTERLINE VALUE switch were set to 10.
Display Amplifiers
The vertical and horizontal display amplifiers are identi-
cal with a few minor exceptions. They are both differential
amplifiers, each with two sets of differential inputs and one
set of differential outputs. One set of differential inputs is
high impedance and receives its inputs from the display
sensitivity switching circuit. The other set of differential
inputs is low impedance and their inputs are the differential
positioning currents from the display positioning circuit.
The differential outputs are connected to the deflection
plates of the CRT and control the potential on the deflec-
tion plates.
The simplified schematic in Fig. 3-12 will help in under-
standing the operation of the display amplifiers. The dis-
3-18
play amplifiers control the voltage between the deflection
plates of the CRT by controlling the currents through load
resistors R L 1 and R L2· The currents I L 1 and I L2 con-
ducted by the load resistors are controlled by two means:
differential current Is and positioning currents lp1 and lp2.
The differential current flows through source coupling resis-
tor Rs whenever there is a differential voltage signal applied
to the high impedance gate inputs of FE TS 01 A and 01 B.
Positioning currents lp1 and lp2 are determined by the
resistance between the emitter of 02A and -75 volts and
between 02B and -75 volts, respectively.
The relationship between the load resistor currents and
the other currents in the amp I ifier is as follows:
IL
=
lp- (ID
+
Is) (Equation 3-1)
Equation 3-1 pertains to the currents which flow in one
side of the amplifier. Is is either positive or negative, de-
pending on whether it adds to or subtracts from I D· I D
represents the FET drain current. It originates from a con-
stant current source and is the same in each side of the
amplifier. This equation also shows that the load current is
dependent on the interaction between the differential
current (Is) and the positioning current (lp).
To understand the operation of this circuit, first assume
that the amplifier is operating in a balanced cor)dition
where the two positioning currents are equal (I P1
=
I P2l
and there is no voltage difference between the two high
impedance inputs (Is= 0). In this case, the load currents on
each side of the amp I ifier are equal to I LO· Equation 3-1,
then, becomes:
ILO
=
IL1
=
IL2
=
lp1- ID
=
lp2- ID (Equation 3-2)
To illustrate the effect the high impedance inputs have
on the load current, assume that a difference in voltage is
applied across the gates of 01A and 01 B, making the gate
of 01A more positive. This voltage differential causes dif-
ferential current Is to flow throuQh source coupling resist-
ance Rs. With this additional current (Is) flowing through
01A, less current is needed from 02A to keep drain current
I D constant. The current conducted by 02A is thus
reduced to ID - Is. Since the positioning current lp1,
which supplies the current conducted by 02A, is also con-
stant, there is a surplus of positioning current created equal
to Is which must be conducted by 05, and therefore R L 1·
The load current is increased to I L 1
=
I LO
+
Is. On the
other side of the amplifier, the current through 02B is
increased to I D
+
Is, which decreases the load current
through 06 and R L2 to I L2
=
I LO - Is. For this example,
it can be seen that whenever a differential voltage occurs
between the two high impedance inputs, the load currents
change, thus changing the voltage potential between the
deflection plates of the CRT.
To illustrate the effect the positioning currents have on
the load currents. assume that the voltages at the high
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