TLC876M Datasheet, PDF (14/23 Pages) Texas Instruments – 10-BIT 20 MSPS PARALLEL OUTPUT CMOS ANALOG-TO-DIGITAL CONVERTERS

TLC876M, TLC876I, TLC876C
10-BIT 20 MSPS PARALLEL OUTPUT CMOS
ANALOG-TO-DIGITAL CONVERTERS
SLAS140D – JULY 1997 – REVISED MAY 2000
PRINCIPLES OF OPERATION
driving the analog input (continued)
The expanded input circuit shown in Figure 15 aids in understanding the voltage offset generation when using
the external input circuit in Figure 14.
The ac coupling capacitors, C1 and C2, integrate the switching transients present at the input of the TLC876
causing a net dc bias current, I_B, to flow into the input. The magnitude of this bias current increases with
increasing the dc signal level, V_B, and also increases with sample frequency. When the sample clock frequency
is 20 MHz, the dc bias current is approximately 30 µA^† at V_BIAS equal to 3 V dc. This bias current causes an
offset error of (R1 + R2) x I_B at the AIN terminal. Making R2 negligibly small or modifying V_BIAS to account for
the resultant offset can compensate for this error. Note however that R2 loads the signal driving source, and
the value must be sufficient for the application.
For example, as shown in Figure 15, when V_BIAS is 3 V and the resistor values stated above, the bias current
causes a 31.5 mV^‡ offset from the 3 V bias, V_BIAS, at the AIN terminal. For the TLC876, V_BIAS can be as low
as 1 V for a 2 V peak-to-peak input signal swing.
VIN
C1
VB
TLC876
R1
AIN
RSW
R2
IB
CE
C2
+
VBIAS
–
Figure 15. Bias Current and Offset
For systems that require dc-coupling, an op-amp can level-shift a ground-referenced signal to comply with the
input requirements of the TLC876. Figure 16 shows an amplifier in an inverting mode with ac signal gain of –1.
The dc voltage at the noninverting input of the op-amp controls the amount of dc level shifting. A resistive voltage
divider attenuates the REFBF signal and the op-amp then multiplies the attenuated signal by 2. In the case
where REFBF = 1.6 V, the dc output level is 2.6 V which is approximately equal to (V(REFTF) – V(REFBF)/2.
† _{IB(AVG) = CE (VB) fCLK ≈ 30 µA, with RSW = 50 Ω, CE = 5 pF, R1 = 50 Ω, and R2 = 1 kΩ}
‡ _{VOFFSET = IB(AVG) (R1 + R2)}
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