Gain Stage
Gain Stage
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Since the Hall sensor and solenoid generate such a low level signal, a good gain stage must be required to boost the signal to a useful form. The signal can be as low as 3uV or equivalent to the LSB in 20 bit converter with 3V range. Unfortunately, the receiving end of the signal is not as precise as a 20bit converter. The precision of the receiving device is a tester or a DMM with a typical precision of 8bit or 12 bits with 3V range. Thus, a gain of roughly 8 bits or 50dB is required. To make matters worse, 3uV is a very low level signal and its signal can be corrupted by noise. The noise generated by the Hall sensor can be approximated by assuming the noise is coming from the internal resistance of the N-well used in the Hall sensor. From previous discussion of the N-well Hall sensor, a square Hall sensor would have a typical resistance of 1000 ohms. Assuming a noise power bandwidth of 10 kHz and the device at room temperature (300K), there is 400 nV rms (root mean square) generated by the Hall sensor. This is only 8.75dB away from our signal !! Signal Stage Gain Single stage gain involves one big amplifier that has a gain of 50dB. The single stage amplifier is not easy to design since most amplifiers use feedback to maintain fixed gain. Since the gain is high, the feedback network requires large resistors and low power or small resistors and high power. Since our signal is only 3uV, smaller resistors should be used to keep the amplifier's noise figure low. Hence, high gain, low power and low noise figure are difficult to achieve together. Suppose we wanted to design a single stage gain amplifier of 50dB and a NF (noise figure) of 6dB. NF is generally known as degradation of SNR. If your original signal had 12 dB of SNR, a amplifier with a NF of 6dB would reduce output SNR to 6dB. Suppose we use a non-inverting op-amp configuration as the basis of the design. To implement a gain of 50dB, the resistor ratio will be approximately 1:315. If the smaller resistor is 100 ohms, then the larger resistor would be 31.5 Kohms. The second requirement is a NF of 6dB. NF means that the amplifier's output noise will be 10dB higher than the amplified input noise source. In our situation, the amplified input noise source is 50dB X 400nVrms = 126uVrms. The NF requirement allows the total noise output of the amplifier, including the noise amplfied from the source, to be 6dB higher than 126uVrms or a total of 504uVrms at the output. The resistor network values are governed by the 504uVrms noise requirement. The resistor network is attached to the negative input of the op-amp; hence, the resistor network's thermal noise will be amplified by a factor of 50dB at the output. Doing some simple math, 504uVrms at the output and 400nVrms at the positive inputs, the negative input must have less than 1.2mVrms of noise. 1.2mVrms of noise is equivalent to a 9kOhm resistor (T=300K, F=10kHz). Hence, the total resistance of the feedback network with respect to the negative terminalmust be less than 9kOhm. The final feedback resistor values are 28 ohms and 8.82 kOhm.
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