Designing an FSR 40*40 Matrix that resolves ghosting and other medical problems - Page 1
Forum section: electronics / beginners

author: andreahmed
author_rank: Newbie
timestamp: July 27, 2025, 06:43:29 pm
content: Hello,I'm not that good at electronics, coming from software background but I studied EE.I'm trying to replicate and use the same sensor matrix as shown in the product hereT-Scan Training Software 9.0 | Tekscanwww.tekscan.comwww.tekscan.comThe sensor matrix will be used for biting on it, and it's insulated. The resistance with small bite is around 100Kohm to 10Kohm. A Reset it its in megahoms.Due to CE and medical problems I don't want a DC current that can hurt the patient, and the nominal voltage should be less than 1V.So I designed the following circuit and not sure if there are problems with it, regarding ghosting, it's function,..etc Before I print it, I don't trust in myself.Shift registers will select Rows/Colums to be sensed by the Current to Voltage Converter. Then digital lock in will be sensed by ADC and written in Software to measure the ADC value.I attached rows section and columns procession sections.

author: xvr
author_rank: Frequent Contributor
timestamp: July 27, 2025, 08:56:06 pm
content: Opamp connection completely wrong. There is no negative feedback, only positive. They will operate (before burning out) as triggers, not buffers.

author: andreahmed
author_rank: Newbie
timestamp: July 28, 2025, 05:59:46 am
content: @xvrThanks so much for your input. Here is an updated schematic.My concern now is the AC Excitation correct ? also the lock in amplifier ?

author: andreahmed
author_rank: Newbie
timestamp: July 28, 2025, 06:20:45 am
content: My only issue is with ROWs side, so when a ROW is selected, the output from MUX will be voltaged followed, which is connected to selected rowand X terminal from MUX will be excited with PWM from teensyFor columns its easier, what ever selected will be current to voltage converted for column sidebut for columns how do I chain the output to just one OpAMP ?

author: xvr
author_rank: Frequent Contributor
timestamp: July 28, 2025, 10:26:15 am
content: QuoteYou have not follower, but amplifier with gain of 2.QuoteYes, selected amplifier will be feeded by PWM from teensy, but all other will have they input floating. So they output will be in unpredictable state.QuoteIn current configuration it is not current-to-voltage converter. Swap '+' and '-' inputs.QuoteMux on input will work (as in your current design).

author: MrAl
author_rank: Super Contributor
timestamp: July 28, 2025, 10:44:58 am
content: Quote from: andreahmed on July 28, 2025, 05:59:46 amHi,You should study op amp circuits a little more.What you have labeled as "current to voltage converter" has the inverting terminal connected to ground, and some positive feedback. That means that when a small positive voltage comes out of the 74HC4067 the output of the op amp goes positive, then makes the non-inverting terminal go more positive, which makes the output go even more positive, which then makes the non-inverting terminal go even more positive, etc., until the output saturates at the maximum capability of the op amp.The usual way to do something like that is to connect the non-inverting terminal to ground, then go from there, but that would mean you would need a negative power supply also and be able to deal with negative voltage outputs.You may not realize this but a resistor is a natural current to voltage converter. If you pump a current through a resistor, the result is a voltage across the resistor. The voltage obeys Ohm's Law:v=i*RWhat you might be able to get away with is to connect the inverting input to the output of the op amp creating a voltage follower. Then use a resistor to ground to sense the current, and use the op amp non-inverting terminal to buffer the voltage across the resistor. The output voltage will then be:v=i*R+VoffsetwhereVoffset is a small offset voltage like 2mv to 10mv approximately, depending on what op amp you use.Note I did not look up what you are using to create this current to be sensed however. If it has some peculiarities that may have to be addressed also.

author: andreahmed
author_rank: Newbie
timestamp: July 28, 2025, 12:45:12 pm
content: My Aim is to select ROW, ground it when the mux is high impedance to remove the ghosting effect (unintended current paths when touching the sensor), and excited the columns by a Clock and use Digital Lock In Amplifier.The same for Columns, ground it via the mosfet when its high impedance, otherwise convert the current to voltage and sense that by the ADC. While exciting the Op-AMP to have digital lock in loop.Please let me know if there is issues again with that design, I'm newbie in EE btw.

author: xvr
author_rank: Frequent Contributor
timestamp: July 28, 2025, 03:57:52 pm
content: QuoteYou can archive this by adding pull down resistors for each amplifier input (just after mux). They will bring not active rows to ground.D1 and D2 now connected towards each other - they will not conduct. Just remove them.QuoteAs I understand from schematic you have only one current-to-voltage converter. So there is nothing to ground - there is only one input.May be you want to ground not selected Columns output (left part of mux in schematic)? This is no so easy

author: andreahmed
author_rank: Newbie
timestamp: July 29, 2025, 04:41:52 pm
content: I have drawn the schematic.Basically I excite a clock with unity gain amplifier at 2.5VPP, and I have resistors network to pull down rows when the mux is in high impedance selectors.At columns side, I use a current to voltage converter with sense resistor 10k, diode is used to rectify the signal I get, and I used a charged pump for negative voltage source to op amps.All the muxes are at ground of -VPUMP, OpAMPs.Please if you have time to review it would be much apperciated.

author: MrAl
author_rank: Super Contributor
timestamp: July 29, 2025, 05:10:34 pm
content: Quote from: andreahmed on July 28, 2025, 12:45:12 pmIt looks like U6 is still not right. It is set up as an inverting op amp.The problem is, if you input a positive current, the output will try to go negative. Since you do not have a negative supply for the op amp, it will go to zero for any positive current.If you need to convert a positive current then use a resistor, then a voltage follower. You'll need to check the value of the resistor to make sure it will work with your available current maximum from the output of the previous stage.As a quick example, if your max current is 1ma positive and you use a 1k resistor, the voltage developed across the resistor will be 1 volt. If that is fed to a voltage follower op amp, the output of the op amp will be 1 volt positive. If you use a 2k resistor then the output will be 2 volts positive, etc.The op amp as a voltage follower is wired with the output connected directly to the inverting input. The input for the voltage developed is then the non-inverting input.