First of all, good afternoon to all my fellow forum members!!!
I came here by chance but I thought it was wonderful to find this space
I have a question about a medical device to see if you can give me a hand or at least come up with a solution.
Context, medical device type BF, floating applicable part. The device works with an internal battery (16V) or with an external certified and isolated power supply (24V). When operating in battery mode, without a power supply, the power supply connector is uncovered and accessible by the test finger. In battery mode, there is insulation between the battery poles of 2xMOOP. There is no voltage at the exposed connector.
In the patient current leakage test, FIG 16, the current is slightly higher, because an external source is applied with the main device voltage 240+10% volts = 264 volts. This implies that the clearance on the connector pin must be at least 2.5mm.
The questions are:
If it is battery powered, why is a voltage of 264 volts applied?
Is there any way to justify the use by risk analysis?
Is there a solution that does not involve a redesign?
Any suggestions, maybe there is something I am not understanding correctly.
Notes: I can't post images because I'm new. But I refer to the following points in IEC 60601, Figure 16 Patient leakage current via the patient connection, Table 12 creepage and clearance for MOPP, paragraph 3.29 F-type applied part, subclause 8.7.3 Allowable values.
Of course, thanks in advance for simply taking the time to read.
I came here by chance but I thought it was wonderful to find this space
I have a question about a medical device to see if you can give me a hand or at least come up with a solution.
Context, medical device type BF, floating applicable part. The device works with an internal battery (16V) or with an external certified and isolated power supply (24V). When operating in battery mode, without a power supply, the power supply connector is uncovered and accessible by the test finger. In battery mode, there is insulation between the battery poles of 2xMOOP. There is no voltage at the exposed connector.
In the patient current leakage test, FIG 16, the current is slightly higher, because an external source is applied with the main device voltage 240+10% volts = 264 volts. This implies that the clearance on the connector pin must be at least 2.5mm.
The questions are:
If it is battery powered, why is a voltage of 264 volts applied?
Is there any way to justify the use by risk analysis?
Is there a solution that does not involve a redesign?
Any suggestions, maybe there is something I am not understanding correctly.
Notes: I can't post images because I'm new. But I refer to the following points in IEC 60601, Figure 16 Patient leakage current via the patient connection, Table 12 creepage and clearance for MOPP, paragraph 3.29 F-type applied part, subclause 8.7.3 Allowable values.
Of course, thanks in advance for simply taking the time to read.