Datasheet

Inverting Stage (Channel 1) Operation
3-2
3.1 Inverting Stage (Channel 1) Operation
Inverting stage operation was shown in Chapter 2 of this manual. The default
configuration isa 50-W terminatedinput (R2), a 50-W matchingresistor output
(R5B), and a 100-W load (R6B). The terminating and matching values can be
changedif theuser requires differenttermination andmatching—for example,
75-W video applications. The load resistor, however, should not be changed
because many op amp specifications are measured at this load. Gain of the
stage can be modified by changing the values of R3 and R4.
Figure 3--1.Inverting Gain Stage
R2
49.9
W
R13
0
--VS
R4
750
W
+VS
J5
--IN1
--
+
SD
U1A
THS31x5PWP
2
3
14
4
1
9
6
J6
OUT1
R6B
100
W
R1
0
R3
750
W
J10
SHUTDOWN
R5B
49.9
W
The gain of this stage is –1 from J5 to J6 for a high impedance load, and –1/2
from J5 to J6 for a 50 W load.
Note:
The inverting configuration affects the level of voltage applied to the EVM.
Whilethefunctiongeneratorindicatesalevelof±0.5V(1V
PP
),only±0.484 V
(0.968 V
PP
) is applied to the board. The user is justified in wondering, what
happened—where did the rest of the voltage (±0.016 V) go?
There are two ways of looking at this problem:
- From thefunction generatorside. The function generator contains a 50-W
source resistor. The EVM contains a 49.9-W termination resistor, and
therefore the output of the function generator may see (approximately) a
2:1 voltage divider. The function generator anticipates this and scales its
output accordingly.
In the inverting configuration, however, the inverting input presents a
ground potential to the gain resistor R
g
. This is because the ideal op amp
model forces both inputs to the same voltage potential. The noninverting
input is connected to ground, and therefore the inverting input is also at
ground. The resulting impedance resistance for the stage is therefore
equaltothetermination resistorinparallel toR
g
,inthiscase750W.There-
fore, the instrument output is actually: