Just another design problem

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.

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