LMH6555_14 Datasheet, PDF(20/37 Page) Texas Instruments – LMH6555 Low Distortion 1.2 GHz Differential Driver

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LMH6555_14 Datasheet, PDF (20/37 Pages) Texas Instruments – LMH6555 Low Distortion 1.2 GHz Differential Driver

◁

LMH6555

SNOSAJ1C â NOVEMBER 2006 â REVISED MARCH 2011

www.ti.com

VCC

IcQ + IPP

12.6 mA

39:

LMH6555

Å¸ Vx

25:

12.6 mA

25:

IN = RL + RS + RG

(VCC Â± IcQ RL) IPP RL

COMMON DIFFERENTIAL

MODE

RN = RL + RS + RG

Figure 35. Norton Equivalent of the Input Circuitry Tied to Q1 within the LMH6555 in Figure 34

IN = IN (common mode) + IN (differential)

IN (common mode) = (VCC â IcQ * RL) / (RL + RS + RG)

IN (differential) = IPP * RL / (RL + RS + RG)

(23)

The entirety of the Norton source differential component will flow through the feedback resistors within the

LMH6555 and generate an output. Therefore:

IN (differential) * RF = 800 mVPP

â RS = (RL* IPP * RF/ 0.8) â RG â RL

where

â¢ RF = 430â¦

â¢ RG = 39â¦ (RF and RG are internal LMH6555 resistances)

(24)

So, in this case:

RS = (169 * 15 mAPP * 430/ 0.8) â 39 â 169 = 1154â¦

Choose 1.15 kâ¦, 1% resistors for RS

(25)

c. With RL and RS defined, ensure that the U1 collector voltage(s) minimum is not violated due to the loading

effect of the LMH6555 through RS. Also, it is important to ensure that the LMH6555's CMVR is also not

violated.

The âVxâ node voltage within the LMH6555 (see Figure 35) would need to be calculated. Use the Common

Mode component of the Norton equivalent source from above, and write the KCL at the Vx node as follows:

Vx / RE + Vx / RN = 12.6 mA + IN (common mode); with RE = 25â¦

Vx / RE + Vx / RN = 12.6 mA + (VCC â IcQ RL )/ (RL + RS + RG)

âVx = 0.4595V

(26)

With Vx calculated, both the input voltage range (high and low) and the low end of the U1 collector voltage

(VC) can be derived to be within the acceptable range. If necessary, steps âaâ through âcâ would have to be

repeated to readjust these values.

VC = VX RL / RN + IN (RS + RG)

(27)

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IN_High = 7.05 mA, IN_Low = 5.19 mA (based on the values derived)

âVC_High = 0.4595 * 169 / 1358 + 7.05 mA (1150 + 39) = 8.44V

âVC_Low = 0.4595 * 169 / 1358 + 5.19 mA (1150 + 39) = 6.22V

(28)

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