Bio impedance spectroscopy

To be more specific this is my requirement

Please suggest me which of your product will be suiting my requirement?

Hi Vikram123 ,

We have products that can support impedance spectroscopy but these also have limitations. Here are my suggestions:

1. AD5933 (1MSPS, 12-Bit Impedance Converter, Network Analyzer)

• This chip has a built in EVAL which is user friendly and get as accurate as ± 1% accuracy on impedance but it is limited to some frequency ranges.
• The main points of this board is that it is user-friendly, can measure accurate readings for impedance measurement but limitations in bio-impedance application might need extra stuff e.g. (Analog Front End Circuit)

Based on a test we made, using a 470Ω as a calibration resistors, we were able to get ± 1% on the frequency 5-88kHz range.

2. ADuCM350 ( Configurable Impedance Network Analyzer & Potentiostat with Integrated Cortex M3 Core)

• This board has a specific application for bio-impedance spectroscopy. Although accuracy based on your metrics might be a bit out of range. From the 1-5% that you were able to set as a threshold for error is somehow cannot be achieve unless your test probes has not contribution to the impedance measurement and they are specially made for bio-impedance application. This application about a 4-wire bio-impedance measurement might also help you as well in giving you more information.

These are just few products on our portfolio that has bio-impedance application. Feel free to check each on out and check which one meets with your specified standards.

All the best,

Jules

Nothing seems to be wrong with your LTSpice circuit - should work as simulated. A couple of things to consider:
1. The AD5933 output as a voltage source is not that perfect. With your output voltage settings it has internal resistance of about 200Ω (and worse at all other settings), please see table 17 on p. 30 in the datasheet. Perhaps increasing the values of R3 and R4 (and R1 and R2 for that matter) to 10k or even up to 47k could be useful
2. Not sure if you need this Current_sense resistor in there

You can use the circuit with the instrumentation amplifier you outlined in the picture. As your "Body" is connected to the ground, perhaps you can use even simpler circuit as you only need to shift the DC baseline in the voltage from zero to  Vdd/2 so that the AD5933 can process it.

Hi Snorlax,

I really appreciate your knowledge sharing and helping attitude.

I am new to analog circuit design. I am stuck with few more doubts(I am not able to understand) while reading the AD5933 datasheet. I hope you help me to understand these concepts better.

Q1. What is DC bias of ac excitation signal which they are talking here in below table?

Q2. Seriously I didn't understand this table also?

what is output dc bias level

Our supply is 5V, what 3.3V has to do here

Q3. What does this graph signifies?

I mean what does that no. of devices on y axis is talking about

What does mean and sigma signifies

Vikram123 said:

Q1. What is DC bias of ac excitation signal

The AD5933 synthesizes voltage sine wave working from unipolar power supply while the sine wave is bipolar by definition. When sine wave is coming from a unipolar device - the baseline of it by necessity has to be shifted to accommodate the entire waveform somewhere between 0V and Vdd. So instead of pure AC sine wave: Vexc * Sin(ω t) it is the same sine wave plus some DC offset:  Vexc * Sin(ω t) + Voffset. With the AD5933 this Voffset depends on the Vexec that you program into the chip and that dependence is tabulated in the tables you are referring to. Moreover, both this Voffset and the amplitude of the sine wave Vexc depend on the Vdd voltage. Both voltages are tabulated for Vdd = 3.3V, but if you feed your AD5933 with Vdd = 5V you can calculate the resulting Vexc and Voffset by multiplying the values from the table by 5/3.3. The data sheet specifies p-p values for excitation voltage, which is double the amplitude: 2*Vexc.
In addition, the internal resistance of the output voltage source also depends on the selected excitation voltage, see table 17 on p. 30 in the datasheet.

Vikram123 said:

Q3. What does this graph signifies?

It is something like a probability of your particular AD5933 to have a certain Voffset when you select 2V p-p range for the excitation. In other words, if you take a lot of chips, select 2V excitation range on each of them and measure the Voffset and then plot the number of chips with a certain Voffset as a function of Voffset - you will get a histogram like this graph. Basically it is telling how much scatter in Voffset from the table value one should expect .

Hi Snorlax,

I understood this concept.

I was trying to simulate the below circuit in LT spice.

www.instructables.com/.../

According to his explanation, I am supposed to get AC voltage. But when I try to simulate the same in LT spice I am not getting. Here is the results.

RC filter 

AD8041 datasheet

www.analog.com/.../ad8041.pdf

Can you please tell me where I went wrong.

Simple enough problem. You have the + input of the AD8041 connected to the output rather than the - input, thus positive feedback and the amplifier will saturate to the rail. Swap the + and - input so you have negative feedback.

Heyy, 
Sorry I was referring this circuit which I found on engineering zone.

But after changing to what you told also not fetching the result.

Yes, the diagram in that post was incorrect. The author said it was correct in the physical wiring of the circuit. Also, in case you are going to use this diagram, the inputs are flipped on U6-1 as well.

Vikram123 said:

But after changing to what you told also not fetching the result

What is the result you are looking for that this circuit does not  produce? If you are trying to shift the baseline for the sinewave up by 1.5V without changing its amplitude and phase, your R1 and R2 are too low. At 10 KHz the C1 has equivalent impedance of about 1.6k and it produces a voltage divider with 50Ω (R1 in parallel with R2), so the voltage is attenuated approximately 33 times (a factor of 50Ω/(50Ω+1.6k)). If this is not  the desired behavior, consider increasing R1 and R2 to, say, 100k. 

Hi,
I tried multiple iteration with different set of resistors and capacitors.

I used this formula to calculate cutoff frequency

But I am not getting my required output. Could you please tell me the method(or values) of resistors and capacitor value which will suit for my requirements.

here is the my requirements

We will be using Arduino supply to power these Ics. 

VDD =5V

frequency = 100khz

we will be operating in range 1

output excitation voltage =3V(Vp-p)

Dc bias voltage = 2.24V

output we wanted is  3V(p-p) with DC bias voltage =0 V

Basically we want AC signal.

Regarding your model: remove V1 and R1 and increase C1 back to 10nF (where it was previously). 
When you calculate the cutoff frequency for a low-pass filter you typically want to keep it considerably lower than your operating frequency or the filter will unnecessarily suppress and phase-shift your signal. The AC voltage source in your model has frequency of 10KHz, which is a decade lower than the cutoff, so the signal is suppressed.
The AD8041 might not be the best choice for your design as it has low input resistance of 160k, which will have to be taken into account when considering the accuracy of your amplitude voltage and DC offset.

Regarding your model: remove V1 and R1 and increase C1 back to 10nF (where it was previously). 
When you calculate the cutoff frequency for a low-pass filter you typically want to keep it considerably lower than your operating frequency or the filter will unnecessarily suppress and phase-shift your signal. The AC voltage source in your model has frequency of 10KHz, which is a decade lower than the cutoff, so the signal is suppressed.
The AD8041 might not be the best choice for your design as it has low input resistance of 160k, which will have to be taken into account when considering the accuracy of your amplitude voltage and DC offset.