Data Sheet

ADA4930-1/ADA4930-2

Rev. D | Page 21 of 25

Terminating a Single-Ended Input in a Single-Supply

Applications

When the application circuit of Figure 50 is poweredby a single

supply, the common-modevoltage at the amplifier inputs, VP

and VN, may haveto beraisedtocomply with thespecifiedinput

common-moderange. Two methods areavailable: a dcbias on

the source, as shown in Figure 51, or by connecting resistors

RCM between each input and the supply, as shown on Figure 54.

Input Common-Mode Adjustment with DC Biased Source

To drive a 1.8 V ADC with VCM = 1 V, a 3.3 V single supply

minimizes the power dissipationoftheADA4930-1/ADA4930-2.

The application circuitofFigure50 on a3.3 V singlesupply with a

dc bias added to the source is shown in Figure 51.

ADA4930

RL VOUT, dm

1.990V p-p

3.3V

RS

50Ω

RG1

142Ω

VP

VN

RG2

142Ω

RF2

301Ω

RF1

301Ω

VOCM

VS

2V p-p

VDC

RT

64.2

Ω

64.2

Ω

50

Ω

Figure 51. Single-Supply, Terminated Single-Ended-to-Differential System

with G = 1

To determinethe minimumrequireddcbias,thefollowing steps

must be taken:

1. Convert theterminatedinputs totheir Theveninequivalents,

as shown in the Figure 52 circuit.

ADA4930

RL VOUT, dm

1.99V p-p

3.3V

VON

VOP

RTH

28.11Ω

RG1

142Ω

VP

VN

RG2

142Ω

RF2

301Ω

RF1

301Ω

VOCM

VTH

1.124V p-p

VDC-TH

RTH

28.11Ω

Figure 52. Thevenin Equivalent of Single-Supply Application Circuit

2. Write a nodal equation for VP or VN.

(

)

TH

DC

TH

ON

TH

DC

TH

P

V

V

V

V

V

V

−

−

−

−

+

+

+

+

+

=

28.11

142

301

28.11

142

)

(

28.11

142

301

28.11

142

TH

DC

OP

TH

DC

N

V

V

V

V

−

−

−

+

+

+

+

=

Recognize that while the ADA4930-1/ADA4930-2 is in its

linear operating region, VP and VN are equal. Therefore,

both equationsin Step 2 give equal results.

3. To comply withtheminimumspecifiedinputcommon-mode

voltage of 0.3 V at VS = 3.3 V, set the minimum value of VP

and VN to 0.3 V.

4. Recognize thatVP andVN areat theirminimumvalues when

VOP and VS are attheir minimum(and thereforeVON is at its

maximum).

Let

VP min = VN min = 0.3 V, VOCM = VCM = 1 V, VTH min = −VTH/2

VON max = VOCM + VOUT, dm/4 and VOP min =VOCM − VOUT, dm/4

Substitute conditionsintothenodal equation forVP andsolve

for VDC-TH.

0.3 = −1.124/2+ VDC-TH + 0.361 ×(1 + 1.99/4+ 1.124/2 –VDC-TH)

0.3 + 0.562 − 0.361 − 0.18 − 0.203 = 0.639 VDC-TH

VDC-TH = 0.186 V

Or

Substitute conditions into the nodal equation for VN and

solve for VDC-TH.

0.3 = VDC-TH + 0.361 × (1 − 1.99/4 − VDC-TH)

0.3 – 0.361 + 0.18 = 0.639 × VDC-TH

VDC-TH = 0.186 V

5. Converting VDC-TH from itsThevenin equivalentresultsin

V

0.33

0.186 =

×

+

=

TH

TH

S

DC

R

R

R

V

The final application circuit is shown in Figure 53. The

additional dcbias of 0.33 V at the inputs ensuresthat the

minimuminput common-mode requirementsaremet when

the source signal is bipolar with a 2 V p-p amplitude and

VOCM is at 1 V.

3.3V

ADA4930

RL VOUT, dm

1.990V p-p

RS

50Ω

RG1

142Ω

RG2

142Ω

RF2

301Ω

RF1

301Ω

VOCM

VS

2V p-p

RT

64.2Ω

64.2Ω

VP

VN

50

Ω

VDC

0.33V

Figure 53. Single-Supply Application Circuit with DC Source Bias