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Regarding an excerpt from this text:

Some accelerometers feature an integrated electronic circuit which converts the high impedance charge output into a low impedance voltage signal.

What is meant by low/high impedance here? Output impedance?

Why is charge output a high impedance and what does it mean that it is converted to a low impedance voltage signal? Can you give an analogy or basic circuit example to understand these? And why does it matter?

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  • \$\begingroup\$ whirlwindusa.com/support/tech-articles/… \$\endgroup\$ – jsotola Sep 21 '18 at 1:40
  • \$\begingroup\$ @jsotola The reference you linked says "A high impedance microphone or guitar will usually output a greater signal (voltage) than a low impedance microphone. " I don't get this at all. If some device has high output impedance it would have more opposition to the current, so how come high impedance microphone output more voltage? Again a text without any diagrams very confusing, \$\endgroup\$ – cm64 Sep 21 '18 at 1:48
  • \$\begingroup\$ read the section about the water hose \$\endgroup\$ – jsotola Sep 21 '18 at 2:11
  • \$\begingroup\$ are you familiar with voltage dividers? \$\endgroup\$ – jsotola Sep 21 '18 at 2:12
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The conversion from high impedance to low impedance is known as signal conditioning. It's necessary for most sensors in some way to match their output to the desired load.

For theses purposes, high and low impedance are a way of expressing how much load a source can handle while maintaining voltage (and therefore signal integrity.)

A low impedance load draws lots of current, so it needs to be paired with a low impedance source that can sustain it.

Some sensors generate a high signal voltage with very small current. This makes them a high impedance source. If connected to a low impedance load, there would be a significant drop in the source voltage. This is a bad thing because it's the voltage we usually want to measure.

A buffer (and/or other conditioning method) is placed at the sensor output to measure the voltage and reproduce it from a high current (lower impedance) power source. This allows the sensor signal to drive longer runs and larger loads without distortion.

National Instruments has a good tutorial and guide with theory and example circuits for the various methods of signal conditioning.

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  • \$\begingroup\$ Thanks and could you add couple of simplistic equivalent circuit examples for a high and low impedance sensor? Are you saying a high impedance sensor can be modeled as a voltage source with a very high output impedance? And low impedance sensor is a voltage source with a very low output impedance like opamp? \$\endgroup\$ – cm64 Sep 21 '18 at 1:52
  • \$\begingroup\$ @cm64 An opamp is a perfect example. The inputs are high impedance and draw practically no current. They are ideal for a high impedance source, such as a sensor. The output can source a reasonable current, making them a relatively good lower impedance source. The overall goal is to turn what you're measuring into something useful without distortion. \$\endgroup\$ – Phil C Sep 21 '18 at 2:02
  • \$\begingroup\$ @cm64 I edited in some links with more detailed explanations and example circuits. \$\endgroup\$ – Phil C Sep 21 '18 at 2:20
  • \$\begingroup\$ Sensors tend to generate a signal voltage with very small current. That's not about impedance. You had to put it some sensors and high signal voltage. All capacitive sensors, for example. And vice versa for magnetic sensors. \$\endgroup\$ – Janka Sep 21 '18 at 2:50
  • \$\begingroup\$ If you "match " the impedances, you get 6dB drop in amplitude, and that 6dB is not exactly 6dB, thus all the money spent on that sensor to ensure many bits of precision have been wasted. Thus high impedance load means the precision will be fully utilized. \$\endgroup\$ – analogsystemsrf Sep 21 '18 at 4:25
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A permanent coil and moving magnet are used as a seismic geophone. The coil has low impedance like an 8 Ohm speaker.

The typical (high dielectric constant) ceramic capacitor is well-known to generate piezoelectric currents, also known as microphonic in relatively high impedance circuits. A 1nF capacitor is about 150kohm at 1kHz which is "relatively" high impedance.

PZ accelerometers are even more sensitive to vibration with the crystalline ceramic materials used.

A cheap electret microphone has a FET buffer rather than a charge amplifier so it basically detects sound as air pressure velocity, not acceleration. With backside partial cancellation. But the derivative of that signal might be an indicator of acceleration, but rather crude and non-flat frequency response.

Comparing accelerometers is like comparing mics.

An electret mic needs an internal FET to buffer the audio with a 10k to 33k load resistor to provide bias and gain from the modulated charge current from the PZ element.

A magnetic mic is a relatively low impedance using a moving coil and permanent magnet. The PZ electric mic is similar with a stored charge that is modulated by vibration but higher impedance due to the fact the ceramic is an insulator and the moving coil is a conductor.

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  • \$\begingroup\$ I really want to understand your analogy between accelerometers and mics. But can you provide two equivalent circuit examples and encircle what part is actually high or low impedance. I need to see the part where it gets the name from aka low or high impedance. There is high or low impedance accelerometers like mics. Please provide some simplistic basic circuits for me understand what part gives the name low or high impedance. Thanks in advance. Treat me someone only knows the meaning of output impedance and input impedance of a circuit. \$\endgroup\$ – cm64 Sep 21 '18 at 23:18
  • \$\begingroup\$ No easily, but consider a speaker is 8Ohms mainly due to wire resistance of 4 Ohms and the rest is acoustic pressure. Then a crystal is relatively high impedance due to the series capacitor . Look at my answer here and the LRC vs f values are shown in graph electronics.stackexchange.com/questions/369378/… \$\endgroup\$ – Sunnyskyguy EE75 Sep 21 '18 at 23:34
  • \$\begingroup\$ If you can calculate a resistance divider, you can see the attenuation comparing R vs C chosen at different f and see visually the impedance ratio =1 at the breakpoint. \$\endgroup\$ – Sunnyskyguy EE75 Sep 21 '18 at 23:37

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