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Coupling Capacitors for DRV8662 Piezo Haptics Driver

Other Parts Discussed in Thread: DRV8662
We have been working with DRV8662 Piezo Haptics Driver Evaluation Module to get familiar with how the DRV8662 works. Our final goal is to build our own circuit with the DRV8662 Piezo Haptics Driver. We will be using analog signals as the input to the chip; therefore, according to the datasheet in section 3.5.2, we wouldn't need the second-order filter. It is also mentioned in the datasheet that we still need the coupling capacitors, C4 and C5. So here are my questions:
1. what is the main functionality of the coupling capacitors, C4 and C5?
2. The coupling capacitors along with the gates input impedance create a high-pass filter rejecting all input frequencies below 10Hz. So if we use a larger capacitor value, say 1000 nF, in order to reduce the filter cut-off frequency, would it interfere with the DRV8662 functionality?
I would greatly appreciate your help in this regard.
  •  

    Hello Nima,

    Thank you for looking at the DRV8662. Before I answer the questions, I just wanted to clarify that the datasheet says that the filter may not be needed, as it still depends on the frequency components of the analog waveform being supplied. With that said:

    Q: What is the main functionality of the coupling capacitors, C4 and C5?

    A: The C4 and C5 caps are essentially DC blocking capacitors so that the amplifier is only amplifying the AC signals. These are used in the circuit to provide better amplifier performance (CMRR, PSRR, VOUTerror, etc...) as well as lower power consumed by the DRV8662.

    Q: The coupling capacitors along with the gates input impedance create a high-pass filter rejecting all input frequencies below 10Hz. So if we use a larger capacitor value, say 1000 nF, in order to reduce the filter cut-off frequency, would it interfere with the DRV8662 functionality?

    A: During operation, the DRV8662 brings up the IN pins to a common mode. This means that because your caps are 10x larger, it will take longer for those to charge. Additionally because the time constant will be larger, if there is a mismatch between those caps, you are more susceptible to see an offset on the output.

    Please let us know if you have any other questions.

  • Thank you very much for your response Rodney. Is there any other solution you would recommend in order to solve our problem stated in the second question?

    Thanks again,

  • Hello Nima,

    Since we have been in discussion through direct email, I wanted to follow up to the forum as well. To answer your question, yes the DRV8662 will function correctly at those lower frequencies. However, this will come at a cost of an offset at startup (in the mS range). This offset will occur if there is a mismatch between the two input capacitors. Of course, if you can assure well matched capacitors, then the offset can be mitigated.

    I hope this help explain, and let us know if you have any more questions.

  • Hello Rodney,

    Happy new year. I have a few other questions regarding the DRV8662 EVM that I would appreciate if you could help me with. Here is a short summary of my application: I want to isolate vibrations on top of an isolation table which consists of piezo actuator and a sensor. I use the sensor measurements as a feedback in my control system. The output of the controller inputs the EVM and the EVM drives a piezo actuator.

    First of all, is there any specification on the type of the anolog signal that I can use with EVM? In the datasheet, it says that input should be sinusoidal. What if it is not? In my case, the input to EVM is the output of the controller which is not necessarily a sinusoid and usually has a combination of different frequency contents.  Will the EVM work for this signal and amplify it?

    Secondly, according to the datasheet, the input range is 0-3 V. So, I can't send an AC signal and should add some offset to it, right? However, the coupling capacitors are cutting the DC offset and the chip is still getting an AC signal. Doesn't this contradict the range specification in the datasheet? Will something disastrous happen if I get rid of the coupling capacitors?

    Thanks again for your help,

  • Hello again Nima,

    Thank you and I hope your new year was good also. See the answers below.

    [Q] First of all, is there any specification on the type of the anolog signal that I can use with EVM? In the datasheet, it says that input should be sinusoidal. What if it is not? In my case, the input to EVM is the output of the controller which is not necessarily a sinusoid and usually has a combination of different frequency contents.  Will the EVM work for this signal and amplify it?

    [A] The input signal does not need to be sinusoidal. However, because the audio range we typically hear is from 20-20kHz with the majority being centered at 1kHz. Therefore, if you provide a tone signal at a very low frequency (200Hz), then it will stay away from the majority of what we can hear. Additonally because square waves have all frequencies due to the inpulse response, then the DRV8662 will be amplifying all frequencies as well. So, to answer your main question, Yes, it will amplify the signal.

    [Q] Secondly, according to the datasheet, the input range is 0-3 V. So, I can't send an AC signal and should add some offset to it, right? However, the coupling capacitors are cutting the DC offset and the chip is still getting an AC signal. Doesn't this contradict the range specification in the datasheet? Will something disastrous happen if I get rid of the coupling capacitors?

    [A] You are correct, but just to clairify the 0-3V is between the input terminals. Because the DC blocking capacitors are removing the DC components of the signal, then you can send it an AC signal centered around ground. Additionally, you can also remove the DC blocking capacitors and drive those same signals with a DC offset. However, this is a differential amplifier and is designed to amplify only the AC components. If you provide the common mode voltage, then the DRV8662 can have some slight offset even though the two inputs are equal (commonly known as common mode gain).

    Hope this helps.