# Wideband fully differential amplifier noise improved using active match

*Since its first appearance in 1999, the single to differential application of wideband* fully differential amplifiers (FDAs) have used a resistor to ground as part of the input match at the cost of higher input referred noise voltage. If that resistor could be removed, with an input impedance match set solely by the path into the summing junction, a much lower input referred noise should be possible. This would be a viable option when the input match can be maintained to very high frequencies via a >1GHz common mode loop bandwidth. The design equations for both approaches will be shown here with a comparison of input referred noise parametric on target gain developed.

**Single to differential conversions using a fully differential amplifier **

One of the more useful functions supported by the growing range of FDA amplifiers is to convert from a single ended source to the differential output required by all modern ADC inputs. These can be either DC or AC coupled designs. When DC coupled, a careful attention to input common mode range is required and bipolar supplies can be useful in that case for many FDAs. For higher speed requirements, single supply is more common and an input match to some source impedance is often desired to limit reflections and/or SFDR degradation. While single supply FDAs can provide a DC coupled path, an AC coupled approach will be shown here to remove input common mode range considerations from this development. These same results will apply to a DC coupled design as long as the inputs stay in range. A typical AC coupled implementation of a doubly terminated 50Ω input design is shown in figure 1. This is targeting an example design set to a gain of 5V/V starting with a 499Ω feedback element and using a free Spice simulator to generate the schematic (reference 1).

**Figure 1: Gain of 5V/V (14dB), 50Ω input impedance, AC coupled, Single to Differential Design**

Several considerations are common to this type of circuit -

- The feedback resistors are equal
- The input impedance is set as a combination of R
_{t }and the impedance looking towards R_{g1}. - The impedance looking towards R
_{g1}will be increased over just the R_{g1}value by the action of the common mode loop within the FDA (reference 2). That loop acts to hold the output common mode voltage fixed which will then cause the input common mode voltages to move with the input signal increasing the apparent input impedance looking into R_{g1}. - The R
_{g2}resistor is set to get differential balance as R_{g1}+ R_{t}||R_{s} - With R
_{g2}set, the noise gain (NG) for this circuit is 1+R_{f}/R_{g2}. - With the AC coupling on the input paths, the DC I/O operating voltages default to the internally developed V
_{cm}reference (1.2V for this 3.3V single supply device). That V_{cm}controls the output common mode voltage but, since there is no DC current path back to the input, it will also set the DC input common mode operating voltage.

The particular example above is using a very low noise, 4GHz gain bandwidth, FDA – the ISL55210 where, in this case, the design started by selecting an R_{f} value then proceeded to solve for the R_{t} and R_{g1} elements. There has been very little vendor guidance in splitting that input match contribution between the R_{t} and R_{g1} elements. The tradeoff is that driving the R_{g1} element down (R_{t }up) will reduce the input noise and extend the bandwidth (for voltage feedback based FDAs). Going in that direction then depends more on the common mode loop bandwidth to set the input match into the R_{g1} path (reference 2). While most reported approaches to deriving the resistor values for the circuit of figure 1 have either been iterative or approximate, picking an R_{f} for a target gain (A_{v}) and input impedance (R_{s}) can be manipulated into a quadratic solution for R_{t} (reference 3).

Solving the coefficient denominator for zero will give a minimum R_{f} to send R_{t} to infinity depending then only on the R_{g1} input path for the match. For the example here, this solves to 160.71Ω.

As the R_{f} is reduced towards this R_{fmin}, the R_{g} elements will decrease while R_{t} will →∞. Using equation 1 to get R_{t }as a decreasing R_{f }is selected, then the 2 other resistors are set by these expressions –

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