Just another design problem

A voltage amplifier must raise the level of its input signal to the peak peak input voltage range of an ADC. The following requirements apply when the amplifier is driven from a signal with a source resistance of 50 \(\Omega\) :

  1. The noise figure should be better than 1.5 dB

  2. The -3dB frequency range should be from 9 kHz to 30 MHz

  3. The dynamic range should be at least 94 dB

  4. Signals with crest factors up to 3 should be processed without clipping

  5. The ADC input voltage range is 5 V

  6. The ADC input impedance is 100 \(\Omega\)

  7. The mean power consumption must be less than 15 mW

Give your remarks on the feasibility of this amplifier.

Interpretation

We will study if a passive-feedback amplifier offers a solution. To this end, we need to investigate whether there is a conflict between the requirements for the power dissipation and the noise figure. A low noise figure requires small values of the feedback resistors, while a low power budget requires high values. Hence, the topic of investigation is a possible show-stopper due to the power dissipation in the feedback resistors. If this is the case, non-energic feedback, possibly combined with transformers at the input and/or the output of the amplifier, need to be considered. In this case, the total power disipation is at best equal to that of the load.

Power in the load

RMS output voltage \(v_o\) equals the peak-peak ADC voltage divided by two times the crest factor:

\[v_o = \frac{5}{2 \cdot 3} = 0.833 \, V\]

Hence, the power disdipation in the load equals 7 mW.

Signal modeling

In order to determine the required gain of the amplifier, we need to know the peak-peak value of the input signal to be processed. This value can be found from the dynamic range, the crest factor and the noise figure.

With a noise figure \(F=1.5\) dB, a signal source resistance of \(50\, \Omega\) and the frequency range from 9 kHz to 30 MHz, the input-referred RMS noise \(v_n\) equals:

\[v_n=\sqrt{4kTR_sB 10^{\frac{F}{10}}} = 5.86\, \mu V\]

With a dynamic range of 94 dB ( \(D=5\cdot 10^4\) ), and a crest factor of 3, the maximum peak-peak input voltage ( \(v_{pp}\) ) to be processed linearly is:

\[v_{pp} = 2 \times 3 \times 5 \cdot 10^4 \times v_n = 1.76 \, V\]

Required voltage gain

The voltage gain \(A_v\) of the amplifier is the ratio of the ADC input voltage range and the maximum value of the peak-peak input voltage. Hence, according to the above, we need:

\[A_v = \frac{5}{1.76}= 2.84\]

Values of the feedback resistors

In the best case (no noise cortribution of the controller), the feedback resistors determine the noise figure of the amplifier. With noise figure of 1.5 dB we write:

\[10^{\frac{F}{10}}=1+\frac{R_p}{R_s},\]

where \(R_p\) equals the parallel connection of the feedback resistors ans \(R_s\) the source resistance ( 50 \(\Omega\) ). From this we obtain a maximum value for the parallel connection:

\[R_p=20.6 \, \Omega\]

With a voltage gain of 2.84, we obtain the values of the feedback resistors as 58.5 \(\Omega\) and 31.8 \(\Omega\). The total resistance of the series connection is = 90.3 Ohm.

Power in feedback resistors

The mean power in feedback resistors is 7.7 mW.

Total power dissipation

With passive feedback the total power dissipation will be at least 14.7 mW. Although the total power dissipation is just within the requirements, it would require a power efficiency of the controller of 98%. This seems quite a challenge, and with the current state of technology rather impossible.

Discussion

In fact, the design of this amplifier becomes more challenging after we include a butget for the noise contribution of the controller. Hence, with passive feedback we need a much larger power budget for processing signals with the specified dynamic range and bandwidth, and with the required noise figure.