Charge amplifier DC problem - Page 1
Forum section: electronics / projects

author: EnzoF
author_rank: Contributor
timestamp: November 11, 2021, 03:01:28 pm
content: Hello! I designed a charge amplifier circuit for piezo conditioning. It is an AC circuit with a virtual ground to get the midpoint somewhere between 0-3.3V so the signal fits into the ADC range.I made this circuit:I simulated it to get the right frequency response and amplitude and tested it on a bread board. I had to change some values on the breadboard but I had to use the components I had on hand so I can test the amplification level I need.The problem I'm facing which did not show up in spice is that if I change the value of resistors R3-R5 I change the center point voltage reference. In spice the midpoint voltage is 1.65V no matter what compnents I use.Why does it not show up in spice and can I change something in spice to show me what the resulting midpoint voltage will be?As you can see I have 1.65V as a reference and the only difference to my real circuit is that I have an op amp supply the 1.65V virtual ground.The final circuit I will order as a PCB will look like this:I can't test it because I don't have the exact components at home.Can I somehow calculate what the resulting mid point voltage will be on this circuit so I can change the voltage reference point to give me the desired midpoint voltage? Can I change something in spice maybe?Thanks.

author: SuzyC
author_rank: Frequent Contributor
timestamp: November 11, 2021, 03:47:55 pm
content: R1 and R7 in both attached diagrams shunt/load the piezo with <1500 ohms, and <400 ohms to ground. 22-Meg to 45k voltage divider prevents setting 1.65 V operation midpoint level desired. Having an op-amp gain ~22-meg/45k (~489) may not be realizable with a single op-amp.Attach 2 is not a PCB layout.you might find some practical designs here:https://www.bing.com/images/search?q=Piezo+Amplifier+Circuit&mmreqh=v1g9Fug00dc4hzbVPglm3%2bc%2bSX2BSETqEVHdqWWKzE0%3d&form=INLIRS&first=1&tsc=ImageBasicHover

author: EnzoF
author_rank: Contributor
timestamp: November 11, 2021, 04:13:23 pm
content: Sorry for stupid questions but I don't really understand electronics.The R2 resistor is not a resistor, it's the piezo it self. Not sure what the resistance is, probably in the gigaohm range. It didn't seem to do anything in the simulation I just chose an arbitrary value.The piezo has a lot of amplitude and this is more of an atennuator circuit than an amplifier circuit.I tried compensating with a lower voltage reference value, that works well but as soon as I changed the resistors R3/R5 the midpoint voltage was completely different.So are you saying that the 45k and 22meg resistors form a voltage divider that change the midpoint voltage?If I say I want a 68k resistor in stead of the 45k. How can I calculate what voltage I should put on the voltage reference to get somewhere around 1.65V at the output of the op amp?

author: EnzoF
author_rank: Contributor
timestamp: November 11, 2021, 04:43:23 pm
content: Quote from: SuzyC on November 11, 2021, 03:47:55 pmNot a pcb design but the final values of the components will be those on the second circuit. 68k for the R3 nd R5 resistors.

author: SuzyC
author_rank: Frequent Contributor
timestamp: November 11, 2021, 04:48:39 pm
content: Correction: R1 and R7You may be overcomplicating the design. A single JFET , BJT or else a FET op-amp voltage follower may best interface with your sensor.I donno what type of transducer you are using or what is the sensor output you are expecting to detect.Your design of a circuit with a gain ~490 is not attenuating the piezo output.

author: EnzoF
author_rank: Contributor
timestamp: November 11, 2021, 06:10:04 pm
content: Quote from: SuzyC on November 11, 2021, 04:48:39 pm27mm piezo that outputs 0-10V in this case.The r1 and r7 resistors attenuate the signal and also act as a low pass filter.The resulting circuit takes a signal 0-10V and convert is to 0-3.3V. I just measured it on the oscilloscope. My microcontroller takes 0-3.3 V and the signal fits.The circuit also rejects common mode noise which is very important because I need to detect very small signals while not making the strong signals to max out.I have the same circuit on my other sensors which are about 0-90V but those I have already designed and they are working very well.If you have a suggestion for a better circuit that takes a 0-10V signal and converts it to 0-3.3V while rejecting common mode noise I'm all ears. Also it needs to low pass below 10kHz with the combination of the exact piezo I'm using (in this case the piezo capacity is 20nF which affects the low pass frequency).Here is the desired frequency response:In the meantime, if you know how I can calculate the change in midpoint voltage when changing the R3 and R5 values while keeping the other things constant that will be helpful.

author: EnzoF
author_rank: Contributor
timestamp: November 11, 2021, 06:11:27 pm
content: Quote from: SuzyC on November 11, 2021, 04:48:39 pmThe suggested circuit is going to be very noisy when it comes to 60Hz hum.

author: SuzyC
author_rank: Frequent Contributor
timestamp: November 11, 2021, 06:13:09 pm
content: Now you tell me whadya doin'.R1 and R7 a low-pass filter?

author: EnzoF
author_rank: Contributor
timestamp: November 11, 2021, 06:21:57 pm
content: I'm making electronic drums. The requirement is to be able to detect very small signals and also very large ones. So the only way to do that is to have a circuit that rejects common mode noise.This circuit took me a lot of trial and error. Believe me I've tried a lot of stuff.Compared to the circuit you suggested (some kind of voltage amplifier) this is infinitely better.My problem is I'm not very well versed in electronics so I need an explanation how I can calculate how the midpoint voltage changes depending on the R3 and R5 resistors because LTspice doesn'doesn't seem to show it (my simulation is probably not 100% accurate).

author: EnzoF
author_rank: Contributor
timestamp: November 11, 2021, 06:23:04 pm
content: Quote from: SuzyC on November 11, 2021, 06:13:09 pmI provided you with the frequency response of the circuit if you don't believe me.Do you want the ltspice file?I believe that the piezo capacitance and the R1/R7 resistors form a low pass filter.So R1/R7 work as attenuation and filter.

author: SuzyC
author_rank: Frequent Contributor
timestamp: November 11, 2021, 07:30:19 pm
content: Consider this: A signal generator driving a capacitor with the output side of the capacitor connected to ground through a resistor is simply a high-pass filter. The higher the resistor, the lower the low-freq response. The actual load on the piezo transducer is R1+R7.You have 1.65V pushing current into a voltage divider of 1200-ohm +22Meg +68000 + 680 ohm total resistance. This creates a voltage at the positive input of ,0051 V. With a gain of 490 the no-signal ouitput of the op-amp would be ~490 * .0051 = ~2.5V (if op-amp bias and offset errors are ignored, they must be very small). N n e case, This ain't exactly 3.3V/2

author: EnzoF
author_rank: Contributor
timestamp: November 11, 2021, 07:44:09 pm
content: Well in practice it doesn't work like that. Or maybe it does but the result is desired.Here is what happens when you change the resistence:Note the frequency response change and also the gain.

author: EnzoF
author_rank: Contributor
timestamp: November 11, 2021, 07:48:55 pm
content: Quote from: SuzyC on November 11, 2021, 07:30:19 pmThank you, this is the answer to my question. This is consistent with what I'm seeing in my microcontroller. I need to lower the output to about 1.4V or so. I need a reference that is less than 1.65V.

author: SuzyC
author_rank: Frequent Contributor
timestamp: November 11, 2021, 08:34:55 pm
content: The op-amp has a -3db high freq. cutoff around only 7.8Hz, while the transducer has a cutoff around 4.9KHz (with R1=R7 680 ohm). Thisresults in a freq. response roll-off fight between the op-amp and the transducer. Reducing R1 and R7 doesn't create a low-pass filter, it just moves the low-freq cutoff frequency of the piezo transducer to a higher freq. which decreases the output of the transducer at lower frequencies and this results in a ~20 dB decrease in output voltage within the freq. response of interest(when R1=R7=100).As the transducer is excited with an increasing freq, it puts out more voltage, but the lower the value of R1=R7, the less is fed to the op-amp, but because the op-amp gain roll-off is rapidly increasing in gain with lower and lower frequency, it swamps out the decrease in response of the transducer capacitance-resistor input divider at the same time.

author: EnzoF
author_rank: Contributor
timestamp: November 12, 2021, 06:54:19 am
content: Quote from: SuzyC on November 11, 2021, 08:34:55 pmThanks for the explanation, I had no clue what is actually happening, it was pure trial and error in ltspice. The result is some kind of band pass response and the signal is much, much cleaner than any other circuit I tried.Anyway, thanks for the help. I didn't realise the resistors to ground made a voltage divider. I should be able to compensate for this.