Dear All,
Regarding the defibrillator protection circuit on the input stage of the ECG, i have read all the thread and its answers on the forum,
but still i have to discuss some points with all of you.
The input stage is as follows
Where YOKE_RA is the Patient Cable Side, And ELEC_RA is the connection to ADS1298
.
From the previous topics i discovered that,
"The standard tests for compliance is a 5000V discharge at 360 joules across a 100 Ohm resistor that simulates the body resistance,
therefore to assure less than 10% shunt current the resistors used in the ECG electrodes should be no less than 1K ohm."
Q1) what about the amount of current will be created from the charging of the capacitor?
I have read some values ranges about 50A.
So only the amount of current entering the input stage will approximately equal to ((100)/ (100+1000)) * 50 = 4.5A
and by using a resistance with high value on the input stage as 30K, the current may reach to ((100)/ (100+1000+30000)) * 50 = 0.16A
so my question here
Q2) What is the optimum wattage for the that 30kohm resistance to withstand a higher current without any wattage problem?, and is the 0603 with 1/10 watt is efficient?
We reduced the current, but there's still a high voltage, so we'll used a diode for clamping.
From previous posts i discovered that
"The first step is to clamp the defibrillator discharge voltage down to a manageable level, using neon lamps or high power TVS diodes such as SMBJ-14 or SMCJ-14."
These two diode are with low voltage breakdown (14 voltage reverse standoff) and (23.2 Max voltage clamping), the difference between them is only the Max. power (SMBJ with max 600 Watt, while SMCJ with 1.5KWatt)
Q3) Is it okay to use diodes with small voltage ratings?
As you have discovered from my questions, i have some issues regarding the calculations of the Voltage, Current, Power.
I need to make some circuit analysis for every point to know my circuit is save.
Anyone want to help me because I'm so confused
