THS4521 Datasheet, PDF (36/65 Pages) Texas Instruments – VERY LOW POWER, NEGATIVE RAIL INPUT, RAIL-TO-RAIL OUTPUT, FULLY DIFFERENTIAL AMPLIFIER

THS4521, THS4522, THS4524
SBOS458H – DECEMBER 2008 – REVISED JUNE 2015
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Device Functional Modes (continued)
8.4.1.3 Resistor Design Equations for the Single-Ended to Differential Configuration of the FDA
The design equations for setting the resistors around an FDA to convert from a single-ended input signal to
differential output can be approached from several directions. Here, several critical assumptions are made to
simplify the results:
• The feedback resistors are selected first and set equal on the two sides.
• The dc and ac impedances from the summing junctions back to the signal source and ground (or a bias
voltage on the non-signal input side) are set equal to retain feedback divider balance on each side of the
FDA.
Both of these assumptions are typical for delivering the best dynamic range through the FDA signal path.
After the feedback resistor values are chosen, the aim is to solve for the R_T (a termination resistor to ground on
the signal input side), R_G1 (the input gain resistor for the signal path), and R_G2 (the matching gain resistor on the
nonsignal input side); see Figure 74 and Figure 75. The same resistor solutions can be applied to either ac- or
dc-coupled paths. Adding blocking capacitors in the input-signal chain is a simple option. Adding these blocking
capacitors after the R_T element (as shown in Figure 74) has the advantage of removing any dc currents in the
feedback path from the output V_OCM to ground.
Earlier approaches to the solutions for R_T and R_G1 (when the input must be matched to a source impedance, R_S)
follow an iterative approach. This complexity arises from the active input impedance at the R_G1 input. When the
FDA is used to convert a single-ended signal to differential, the common-mode input voltage at the FDA inputs
must move with the input signal to generate the inverted output signal as a current in the R_G2 element. A more
recent solution is shown as Equation 1, where a quadratic in R_T can be solved for an exact value. This quadratic
emerges from the simultaneous solution for a matched input impedance and target gain. The only inputs required
are:
1. The selected R_F value.
2. The target voltage gain (A_v) from the input of R_T to the differential output voltage.
3. The desired input impedance at the junction of R_T and R_G1 to match R_S.
Solving this quadratic for R_T starts the solution sequence, as shown in Equation 1:
R
2
T
-
R
T
2R _S(2R _F
2R _F(2 + A_V ) -
+
R_S
2
A
2
V
)
R _SA_V(4 +
A_V )
-
2R _F(2 +
2R _F R_S2 A_V
A_V ) - R _SA_V(4 + A_V)
=0
(1)
Being a quadratic, there are limits to the range of solutions. Specifically, after R_F and R_S are chosen, there is
physically a maximum gain beyond which Equation 1 starts to solve for negative R_T values (if input matching is a
requirement). With R_F selected, use Equation 2 to verify that the maximum gain is greater than the desired gain.
é
ù
ê
A _V(MAX)
=
æ
ççè
R
R
F
S
ö
- 2÷÷ø ´
ê
êê_{1 +}
ê
êë
1
+
4R^F
R_S
æ
ççè
R
R
F
S
ö²
- 2÷÷ø
ú
ú
ú
ú
ú
úû
(2)
If the achievable A_V(MAX) is less than desired, increase the R_F value. After R_T is derived from Equation 1, the R_G1
element is given by Equation 3:
R _G1
=
2R^F
A_V
1+
- R_S
R_S
R_T
(3)
36
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