The following is my (I'm a manufacturer engineer) personal opinion, for reference only, the specific understanding of the engineers of different testing centers are not the same, to their point of view shall prevail (of course, if you are more familiar with the 9706, you can explain to them, they agree with or not don't know)
You mean is not the 3.7v lithium battery powered "internal power supply equipment", and this battery is a special charger, and the ECG machine is not to charge the battery (i.e., the ECG machine will not be connected to the external power grid). power supply device" powered by a 3.7v lithium battery, and that this battery has a special charger, and that the ECG machine does not have the ability to charge the battery (i.e., the ECG machine is not connected to the external power grid)?
If this is the case, please check "20.2 Requirements for equipment in the application section" in gb9706.1 "For all equipment in the application section, the strength of the electrolyte shall also be experimented with" "B-a "B-a" in the reference to "between the application part and the charged part", I understand the "charged part" is the battery (or battery contacts) and the application part (electrocardiogram limb clamps / suction bulb) loaded with 500V high voltage, that is You should measure it the way you said
I don't know what the purpose of your question is, is it because the isolation device to isolate the ECG part from the power part is quite big? In fact, it seems to use optocoupler isolation can be solved, and the device does not occupy too much space