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I have a question about the V7805-2000R DC-DC converter.
I am designing a custom PCB that uses a 12VDC 3.3Amp Class 2 power supply.
On the PCB I am planning to use 4 x V7805-2000R to convert the 12V to 5V.
Each V7805-2000R will supply 4 audio players with 5V DC voltage. Each audio player uses about 0.4 Amp.
In order to for the PCB to pass local CSA electrical certification the inspector is asking me if the V7805-2000R is a “kind of diode with SCR that will reduce the voltage but doesn’t have property to increase the current by conversion. If it is like a power supply 12vdc that convert voltage to 5VDC it could be an issue because it won’t limit the current under a class 2 level. But if it is only a voltage reducer like a kind of diode where the nominal current could not be higher than what the power supply can provide. This shall be acceptable.”
If I’m not mistaken, class 2 supply limits are 60V and 100VA, with no constraint on current per-se. As long as your apparatus is powered by a class 2 supply and cannot generate an aggregate output in excess of these limits, there should be no issue regardless of any DC-DC conversion performed.
If the quote in the original post is verbatim, it would be a strong indication to me that the correspondent lacks the knowledge necessary to function appropriately in their apparent role. SCRs are used primarily for manipulation of AC power at line voltage levels, and describing diodes as a “voltage reducer” is, well, not the sort of thing that folks familiar with diodes would do…
So, no. The DC-DC converter in question is not an SCR device, but there’s also no rational basis for questioning the presence of a DC-DC converter in the system.
Politely asking your inspector to point out where in statue/code such conversion is prohibited in a class 2 system might be a useful approach to take.
Thank you for your fast reply. And thank you for sharing the same confused look that I have when trying to untangle the inspectors comments.
At some point he also said:
" Depend how does this device work.
If it is like a power supply 12vdc that convert voltage to 5VDC it could be an issue because it won’t limit the current under a class 2 level. But if it is only a voltage reducer like a kind of diode where the nominal current could not be higher than what the power supply can provide. This shall be acceptable.
What we are expecting is that there is no risk that voltage and current could exceed value from the section 16-200 of the CEC part 1.
In this case 0 to 20V 5amps.
12V X 3.3A = 39.6watts / 5v = 7.9amps"
My feeling is that this DcDc converter is a kind of “voltage reducer”.
I found this on the web when searching for the documents he refers to:
" Class 2 Circuits
This subsection outlines the requirements for installation of Class 2 circuits.
Rule 16-200 describes the limitations of a Class 2 circuit to prevent a shock or fire hazard under normal circumstances by:
limiting the current of Class 2 circuits by various methods depending on the voltage;
energy limiting through use of suitable rated series resistors or similar devices; and
preventing a transformer or other power supply device having a Class 2 output to be connected in series or parallel with another Class 2 power source. (Note: a device having a Class 2 output must be marked as having energy-limiting characteristics and as suitable for the purpose)."
I am adding an image so you can see the PCB I am talking about. But I understand that from just looking you will not gain much more info.
The V7805-2000R is a switching DC-DC converter. These differ from the alternative, a linear regulator, in that it pulls enough power from the input to supply the necessary power to its load. A linear regulator will draw enough current to supply to the load and burn off the excessive voltage as heat. Whenever a switching DC-DC converter switches from a higher voltage to a lower voltage, the current draw from the input voltage source will always be lower than the output current.
For example, in your application, the load for the V7805-2000R is 2 Watts (5V x 0.4A = 2W). The datasheet states that at full load (2.00A for this part) it is about 91% efficient. It does not clarify its efficiency at lower loads, but the efficiency of most DC-DC converters does typically drop somewhat at lower loads.
So, just for erring on the worse case side, we’ll assume an efficiency of 50% at 0.4A output current (in reality, it’s probably closer to 70%). If the DC-DC converter was 50% efficient, the input power would be 4W (2W / 50% = 4W). Since the input voltage is 12V, this means that the input current to the DC-DC converter would be 0.333A (4W / 12V = 0.333A). Notice that even with an assumption of very poor efficiency, the input current is still lower than the output current.
If you add four of these together, you get a pessimistic max current draw of 1.333A at 12V for a total power of 16W. This would not seem to be a problem for your system.
As long as your power supply is a class 2 supply, there should be no problem.
Commenting further on the info you gave regarding the Class 2 requirements, those are all specifications for the Class 2 power supply itself, not the specifications of the circuit that the Class 2 supply powers. The Class 2 specifications allow one to be somewhat less concerned about the design of the circuit powered by the Class 2 supply, because it will do much of the protection itself.
David_1528
Thank you for this great explanation.
I will share this with the inspector and hope it will convince him that my device and the use of 4 x V7805-2000R is worry free.
I shared your valuable information with the CSA inspector. It did not to help much. I feel I probably need to add a fuse to make him happy.
The DC-DC converter (12V to 5V) is rated for 2 Amp. So I might need to use a 283-PTSLR12066V110TR-ND which has a trip current of 2.2Amp.
Here is his message, please let me know what you think.
I am not sure they understood what I was asking for.
I was not looking if you might overload your power supply.
I was trying to confirm the way the voltage is converted into the converter.
It shall have no risk that the current can be exceeding the class II level.
I am more looking on the way it could fail then the functionality.
The fact of having a none certified power supply 12-5vdc doesn’t show evidence that it will actually act properly under failure.
We can consider that a 60watts sources 12vdc 5amps through a power supply can provide a 60watts 5vdc 12amps (not considering the efficiency here, it is only for the understanding)
Having this said, now how could this append?
Device do have internal protection but cannot rely on them.
So the approach will be more a justification on how it could become a possible situation and what is in place to prevent it.
What make it impossible like having already a constant load other than the one for those power supply that will limit the available current to those item.
With a class II power supply, if you draw too much current it will shut down the power supply. Maybe you may have some acceptable justification here that will demonstrate it is not a possible situation because the system pull out to much current under normal operation to permit such current to go through the none certified power supply. We are looking for a maximum of 2amps current draw on the primary side of the 12-5vdc power supply .
If one’s 5HP lawn mower engine throws a connecting rod, does it become a 50HP engine? Or scrap metal?
The V7805-2000R is a non-isolated buck converter of the common sort, and the output current protection is there as much for the sake of the converter as anything else. Output currents substantially beyond the device’s design limits will cause the internal inductor to saturate, precipitating failure of the switching element. This commonly manifests as a short circuit for a brief period of time (during which your upstream supply would provide appropriate limitations) and then to an open circuit once the switching element has been converted to a vapor. It would appear that your inspector is postulating a state where, in spite of a fault, the converter continues to operate in a sustained and otherwise-normal manner at more than 2.5x its rated capacity. Such a state is quite unlikely.
IIRC, your max anticipated load per-converter would be 1.6A; this is in the gray area between the hold and trip currents for the PTC device suggested. Depending on use case and circumstances, nuisance tripping is likely to be a problem. A one-time fuse should be substantially cheaper, and in the unlikely case of porcine aviation your board would by definition be jacked up anyway, so the lack of reset shouldn’t pose any additional burden.
