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Hi everyone, I hope you are fine.
I have a question.
I have generated a 4Vpp signal with a 5MHz frequency using the following sequence: AD9851 (DDS) → AD8001 (Op Amp) → Ferrite bead (2KΩ) → Microfluidic sensor
This signal is connected to a microfluidic device whose impedance varies from 3KΩ to 10KΩ.
When I attach any through-hole TVS diode with a Reverse Stand-Off Voltage (V_R) rating of 6V to 30V across the signal, with one pin grounded and the other pin across the signal for circuit protection, it drops the signal to less than 1Vpp.
My questions are:
Why does my signal amplitude drop significantly, even though the Reverse Stand-Off Voltage (V_R) is much higher than the signal amplitude?
I want to match the impedance between my microfluidic device, which has an impedance ranging from 3KΩ to 10KΩ, and my signal, which currently has an impedance of 50Ω without the ferrite bead (2KΩ).
a) How can I increase the impedance of the signal?
b) When I add some series resistance across the signal, it greatly reduces the amplitude. Please suggest a solution to match the signal impedance with the microfluidic impedance without dropping the signal amplitude.
C) does connecting a ferrite bead of 2KΩ in series to the signal add to the overall impedance of the signal?
The AD8001 has high input impedance and low output impedance, making it perfect for this application.
Your response would be highly appreciated.
If your signal swings around zero (i.e. it does not have a DC-component), then you should use a TVS that is bi-directional (quite many of the semiconductor TVS’es come in two flavors, uni-directional and bi-directional).
At 5MHz you do not need to worry too much about the impedance matching unless your load is very far away from the driver… Have a low impedance driver (the Op-amp) and high impedance load, and you are good to go. Just make sure that the impedance variation of the load is compensated properly (e.g. if your driver impedance is in milliohms, then the sensitivity to load impedance variation (3KΩ to 10KΩ) more or less goes away automagically). You may consider removing the bead altogether too, unless there are noise issues etc.
Cheers, heke
I get this reply from someone. I don’t know about how to put in series, i will work on it now.
Hope so this will work
It is not the reverse standoff voltage that is causing the drop, it is the forward conduction voltage, which for most diodes is less than one volt. So if you put the two TVS diodes in series, opposing it will increase the protection threshold by a bit less than one volt. That may solve the problem.
I guess the problem is that you have a true AC signal (-2V to 2V) not pulsed DC (0v to 4V) like I thought. A regular TVS only works with DC, it causes the problem you describe with AC signals.
As @heke suggested, you can buy a special bidirectional TVS to clamp AC instead of DC.
In general agreement with the above comments. Using a unidirectional TVS would not work for an AC sine wave. As a unidirectional TVS would be forward biased for 1/2 the wave, that half would be limited to a forward diode drop of roughly 0.7V.
However, additionally, looking at the capacitance graphs for typical TVS diodes,
it appears that the capacitance would be significant. At 4V it would be well in excess of 1000pF. At 5MHz, a 1000pF load would be approximately 32Ω (Z = 1 / (2πfC) = 1 / (2 * π * 5MHz * 1000pF) = 31.83Ω
Putting that in parallel with a 3KΩ to 10KΩ load would completely squash it.
@David_1528 raised a good point. Here, it is important to use a TVS that is for signal lines. Supply line TVS’es have a large capacitance (there it is largely seen as a benefit). Say, Diodes’ D3V3L1B2WS-7 has capacitance of 11pF which should perform nicely(?)
For the cut-off frequency the driver’s output impedance is a major parameter.
Now, then there is the question that what “Microfluidic device” that OP refers to actually means? Is it an actuator or a sensor? Is the aim to measure the impedance change of the device (at the resonance frequency) due to ultrasonic energy absorbtion?
Cheers, heke
