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01-17-2016

Understanding and using charge amplifiers

Jennifer Calhoon

Design World

A charge amplifier is not the most common type of amplifier, but very useful in the right circumstances, it is really a current integrator which produces a voltage output proportional to the integrated value of the input current. This is useful when the sensor is capacitive such as a piezo device which could be a microphone, hydrophone or if the sensor is a photodetector. An opamp based charge amplifier looks like this:

Understanding and using charge amplifiers

The resistor R1 is to provide a DC operating point for the opamp – without it the output of the opamp would drift either up or down until it hit the supply rails, depending on the polarity of the opamp bias current. It is important that the resistor is low enough to provide a suitable operating point for the opamp without affecting the desired performance. The resistance must be higher than the impedance of C1 at the lowest frequency of interest. With a low bias current opamp such as a FET input or CMOS input one, the value can be very high as shown.

The “gain” of the charge amplifier is not obvious because it relies on the capacitance of the signal source. For example, a Brüel & Kjær 8103 hydrophone has 3850 pF of capacitance so the gain would be 20•log(3850 p/100 p) in this case or 31.7 dB. “Gain” is not necessarily the obvious way of describing the transfer function because the output voltage will be proportional to the transducer charge. However, it is often convenient to think of the signal as a voltage and that voltage comes from a specific capacitance and hence has a certain charge.For example, taking the B&K 8103 specifications, the sensitivity is quoted as 0.12 pC/Pa and also as 30 µV/Pa. As the capacitance is 3850 pF, the charge sensitivity should be based on Q = CV so 30 µV × 3850 pF which is 0.12 pC – the same figure quoted by B&K, it has simply been expressed in two different ways.

The AC performance of the example circuit is shown below.

Understanding and using charge amplifiers

The gain is 31.7dB, as expected. The 3 dB cutoff point is determined by the opamp gain-bandwidth limitations. To simulate this amplifier the transducer was modeled as a voltage source with series capacitor:

Understanding and using charge amplifiers

The transducer could also be modeled as a current source in parallel with a capacitor which would be appropriate if the transducer was a reverse biased photodiode.

The low frequency 3 dB point will be defined by the R1 and C1 which in this case, with 100 pF and 50 Mohms will be 31.8 Hz. If the simulation is extended to that frequency you will see that it is indeed the lower cut off frequency.

If the signal source is a photodiode you will only get an input current in one direction so the charge amplifier output will jump to a new value and stay there until the feedback resistor/capacitor combination discharges the feedback capacitor. So, you end up with a pulse such as this:

Understanding and using charge amplifiers

The injected charge determines the pulse height (in this case a negative pulse because current was injected into the opamp input). The decay of the pulse is determined by the feedback capacitor/resistor combination. So, in this case you may want a quicker decay and hence a lower feedback resistor value. This sort of circuit and pulse output is useful in devices such as a Radiation Isotope Identifier where both the pulse height and number of pulses are important. Some pulse shaping will often be carried out after the initial charge amplifier stage before measuring the peak height.

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