Specific features for ECG cables safety in case of ... - Integral Process

Safety specific features for ECG cables in case of a defi shock.In case of a defi-shock, safety is provided in two ways.. The performance of the electrical shock has to beguaranteed by a limitation of the energy loss delivered to the patient and the ECG equipment must be protected.Integral Process offers two types of ECG cables:a) - One-piece cables (the derivation leads cannot be removed from the trunk cable)b) - Detachable type (the derivation leads can be removed from the trunk cable)All cable types can include (or not) a protection of the inputs of the electro medical device against defi shocks.The presence or lack of protection inside the cable is decided by the ECG device manufacturer, who can chose tooplace the protection inside the device.When the protection system is integrated to the cable, it is generally made of a wiring of resistors in serie and sparkgaps in parallel. It can be partially integrated to the cable (resistors) and the spark gaps can be placed inside theelectro medical device.The safety features of an ECG cable are not only related to the protection against defibrillation shocks, but also:- protection against electrical noise of the cable conductors- insensitivity to surrounding electromagnetic fields- limitation of the leakage current 50Hz those authorized and defined by the highest class of the electromedical device to which the cable can be connected- use of the cable in continuous way- effects of a high alternating voltage in a continuous way- effects of an early damage of the insulantsElectrodes plugsConnexion boxPlug to the deviceDetachable or attachedderivation leadsTrunkECG cable exampleMulti-conductors cableProtection against defibrillation shock efffectsApplying a defibrillation shock to a patient during a procedure (or in emergency) generates derivated currentsbetween the ECG electrodes, which can damage the electro medical device to which one the cable is connected, incertain conditions.Defibrillator400 joules 5kvVVPatientVElectrodeSignal inputElectrode Defibrillation de pad défibrillationSimplified example of the defibrillation voltage spread

The voltage generated by the defibrillation shock can provide high level tension (several hundred volts) to the inputterminal of the medical device. These tensions are of course dangerous for the medical device. A protectionagainst the damaging effects of these tensions has to be anticipated in order to protect the inputs.In addition the maximum of the energy provided by the defibrillator has to be delivered to the patient and it isnecessary to minimize a too high leakage of this energy. The resistors minimize this energy leak and guaranteethat the energy is correctly delivered to the patient.1r1,r2,r3 = protection resistorsr1E1, E2, E3 = device inputsE1v1e1,e2,e3 = protection spark gapse12r2E2v21,2,3 = patient electrodese23v3r3e3E3v1,v2,v3 = defi shock voltageFloating massProtection system exampleThe energy of a defibrillation shock applied to a patient can be over 400 joules, generating high tensions andvoltages (respectively 5kv and 50Amp)during a very short time (a few ms).The patient impedance when applying the shock is considered as being of 100ohms.In the worse conditions (direct contact of the defibrillation pads with the ECG electrodes) a 5kv tension can beapplied to the input terminals of the electro medical device.The resistors and spark gaps safety system reduces the tension to an acceptable level, protecting the device inputsif the components have been defined in a way to withstand the currents effect..The instantaneous currents spread in the protection circuits can reach very high levels. For example if theprotection resistor chosen has a low value (1kom) the instantaneous power at its terminals can reach 2 500 w,however the leading time being quite short (a few ms) the energy provided in this resistor decreases sensibly,which allows the use of resistors of a reduced continuous power.Table of the absorbed energy, power intensity and heat spread accoring to the tension30,0025,0020,0015,00R 1000,00t 0,00U 1000,00 2000,00 3000,00 4000,00 5000,00P=U²/R 1000,00 4000,00 9000,00 16000,00 25000,00P=RI² 1000,00 4000,00 9000,00 16000,00 25000,00W=Pt 1,00 4,00 9,00 16,00 25,00I=U/R 1,00 2,00 3,00 4,00 5,00Qµth=0,24*W 0,24 0,96 2,16 3,84 6,0010,005,000,001000,00 2000,00 3000,00 4000,00 5000,00W=Pt I=U/R Qµth=0,24*W