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How to calculate δVout from Charge injection data?

Thread Summary

The user is experiencing large voltage glitches (about 2V) when using the ADG608 multiplexer in a sensor circuit. The final answer suggests considering the input capacitance of the amplifier and isolating the switch for accurate charge injection (Qinj) evaluation. The user's negative voltage drop at Vsource=2.5V during switching on is attributed to the test conditions differing from the actual setup. The ADG1208 is recommended for better Qinj performance, and the user should refer to the 'Ask the Applications Engineer-26' article for more details on charge injection.
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Category: Datasheet/Specs
Product Number: ADG608,ADG1208

Hello,


I hope you can help me. 

We need to use multiplexers in a sensor circuit. The circuit is like the drawing below.

I got a big voltage glitch at every edge of switching on when using ADG608.

Here are 3 questions.

  • Q1 Glitch Voltage Calculation 

δVoutQINJ/Cload

Because there is no Cload after MUX in my circuit,

Is it correct to only use Cd for calculation?

For example,QINJ2pC @Vsource2.5V Cd=40~54pF

So δVoutQINJ/CD=2/54*1000=37mV

After amplifier δVout will be 37*11=407mV

But what I see is that the voltage glitch is much bigger (about 2V),can you tell me the reason?

  • Q2 @Vsource2.5V why I got a negative voltage drop?

From the plot of QINJ vs Vsource, I can see the QINJ is always positive during VDD=+5 VSS=0V

I guess that's because the QINJ result got  in the test circuit is under the condition of switching off,

and my negative voltage is happening at the switching on. So it is the opposite.

Is my understanding correct?

  • Q3

If my understanding is correct, let's get back to Q1

Because the voltage glitch takes place at switching on,

does that mean I need to take the capacitance of input into consideration for δVout calculation?

or the glitch is happening at the just turning of switching on&off ,

the MOS channel is not thoroughly on ,so the input capacitance will not have an effect on the calculation?

 

After all, I'm now using ADG1208, the voltage glitch issue has improved quite a bit.

  • Hi Shibi, 

    Q1: In your application, it might be good to check the input capacitance of the amplifier. The Qinj occurs during switching, in this case you should use CD (off) = 40 pF plus the input cap of the amplifier. The ADG608 Qinj TPC plot was measured with a 1 nF capacitor load, which is not the same in your setup. If you want to evaluate the switch Qinj properly, I suggest isolating the switch to prevent other capacitive components from influencing it.

    Q2: The output figure in Test Circuit 8 is just a guide on how to test the charge injection. As you can see, the magnitude of Vout was not specified, so  ΔVout is not guaranteed to be positive all the time.

    Q3: Charge injection in the figure occurs at the drain and during switching, that's why your switch figure in Q1 shows only CD. You might want to check Q&A: What is "Charge injection"? You can learn more about charge injection and ways to mitigate it.

    The ADG1208 would be a better option if Qinj is critical to your application. This has better Qinj performance and is also fully specified. I would also recommend using ADG1208 in your application.

    Regards,

    Yoworex

  • Hi  Yoworex,

    ​I really appreciated your quick reply. That's very helpful.

    Q1:I understood there was a different capacitor load in the test plot and my setup.

           But the Qinj should not be changed because it only depends on the MUX inner parasitic capacitance.

           If Qinj in the formula Qinj=V*C  is constant , the V calculation should be easy. What I need to know is Cload.

           As you guided ,I looked for the input capacitance of amplifier, there's no data found,

           but normally it is considered as 10pF around.

           Then the calculation will be

           δVout=Qinj/(CDoff+CAMP)*GainAMP=2/(40+10)*11*1000=440mV

           It's still far from the actual measured value.

    Q2:I agree with you that ΔVout is not guaranteed to be positive all the time.

          But my understanding is Qinj should have reproducibility.

          So I expected the Qinj to be positive @Vsource=2.5V,but I got ​a ​negative result.

          That's why I asked the question.

    Q3:I'm really confused.

           Since I've read the article below. What I learned was the Qinj occurred during both Switch ON and OFF.

           https://www.analog.com/en/analog-dialogue/articles/ask-the-applications-engineer-26.html

           

          

     Best Regards,

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