TransWikia.com

How to choose TVS/ ESD for sensitive application?

Electrical Engineering Asked by kobi89 on October 29, 2021

I am using ad8232 ECG with 3 electrode configuration,

Edited
ad8232 used schematic

I need to protect circuit from any voltage more than 5V, since ecg measure only around 300mV with common mode signals. the circuit should protect from even high voltage such as 5V to 150V surge current max 50mA at 50Hz-200Hz range,

I have a doubt,

I hope to use bidirectional TVS/ESD due to all leads are on body. and same GND is on body for another application

How to choose correct TVS that will not affect ECG measuring signal range (0-300mV) at all. or with minimum affect, what parameters of TVS/ESD should I consider?

One Answer

R3 and R4 serve for this purpose, they limit current through the in-built ESD diode which is 5mA maximum. V=180k*5m= 900V.

Using standard TVS may spoil the measurement due to leakage currents.

Read the document:

INPUT PROTECTION All terminals of the AD8232 are protected against ESD. In addition, the input structure allows for dc overload conditions that are a diode drop above the positive supply and a diode drop below the negative supply. Voltages beyond a diode drop of the supplies cause the ESD diodes to conduct and enable current to flow through the diode. Therefore, use an external resistor in series with each of the inputs to limit current for voltages beyond the supplies. In either scenario, the AD8232 safely handles a continuous 5 mA current at room temperature. For applications where the AD8232 encounters extreme overload voltages, such as in cardiac defibrillators, use external series resistors and gas discharge tubes (GDT). Neon lamps are commonly used as an inexpensive alternative to GDTs. These devices can handle the application of large voltages but do not maintain the voltage below the absolute maximum ratings for the AD8232. A complete solution includes further clamping to either supply using additional resistors and low leakage diode clamps, such as BAV199 or FJH1100. As a safety measure, place a resistor between the input pin and the electrode that is connected to the subject to ensure that the current flow never exceeds 10 µA. Calculate the value of this resistor to be equal to the supply voltage across the AD8232 divided by 10 µA.

Answered by Marko Buršič on October 29, 2021

Add your own answers!

Ask a Question

Get help from others!

© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP