Hi,
I have purchased a 240W DC-DC Converter 24V@10A SKU: FIT0172
Is this product meant to have isolated input and output or do they share a common negative connection for both input and output?
Regards,
Ross
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Hi Ross
I think someone is going to have to check that.
The info is extremely skinny isn’t it.
DfRobot web site is not much better.
Cheers Bob
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Hi Ross,
It’s been a little while since I did my university project into switchmode power supply design but I think I can shed some insight here.
That DC-DC converter is almost certainly a boost converter under the hood and that switchmode topology by design will not have isolation between the input and output of the device.
This article is heavy on the technical jargon but goes to show there are a lot of different topologies, each with strengths and weaknesses.
Why do you need a power supply output that is isolated from the input? Is there a particular device you are trying to power that has this requirement or is particularly susceptible to power supply EM noise?
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Hi Trent
I would have thought that a simple resistance test between the 2 black wires would have given some clue without going into the design.
Some applications may need a positive and negative supply. Can do easily with this sort of converter if the outputs are isolated from the input. You can achieve a pos and neg supply with a single input supply. If not another solution is required.
Cheers Bob
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Hi Trent
Another way of determining if the output is isolated.
Connect input to power source.
Connect a load to output, 25V or 2 x 12V auto globes would be OK.
Measure output, should be 24V.
Now measure from input negative to output positive. If NOT isolated you should read 24V. If it IS isolated you should read nothing.
Cheers Bob
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Hi Ross,
I can see from the email you sent alongside this forum post that you were trying to connect a Dual Digital Display DC Voltmeter & Ammeter. These panel meters require a power supply that is electrically isolated from the measurement circuit and has minimal noise.
The simplest solution to this is to power the panel meter from a battery circuit that’s independent of your circuit under measurement.
If you need to power it from the same power supply that feeds the circuit you are measuring (like measuring a car battery) then you will need an isolator capable of supplying 4.5-30V DC at 20mA to power the panel.
A DC-DC converter that has an isolated output will normally reflect the isolation it the specs. I don’t believe any of the DC-DC converters we sell are isolated as it’s typically not required in maker electronics projects. You might need to search for a more industrialised isolated DC supply if a battery is not an option.
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James at Core Electronics also provided me a link “Boost converter - Wikipedia” showing that boost converters by design would share a common GND between input and output.
I was fault finding an issue and was surprised to detect some leakage between the input and output of the DC-DC converter. I then measured the resistance between the input negative wire and the output negative wire and found it to be zero. I had assumed incorrectly that the DC-DC converter inputs and outputs were isolated, so at this point I wasn’t sure if my assumption was wrong or if the DC-DC converter was faulty. It turns out to be the former.
The impact on my design is that I cannot connect a Core Electronics CE5132 dual panel meter that I ordered to measure the Boost Converter current input, since this sets up a varying differential voltage between the GND on the input to the Boost Converter (more specifically the input to the CE5132 shunt resistor) and the GND on the output of the Boost Converter. This would not be a problem if the DC-DC converter input & output were isolatred. A current meter shunt resistor that could be placed at the 8-18VDC positive input to the Boost Converter would not exhibit this grounding problem. An analog current meter in the positive rail would be a solution.
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Trent - I’ve ordered one of these which I can use with my multimeter.
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Hi Ross
I think this is an AC device. Says as much in the description.
Cheers Bob
EDIT Add on.
Clamp meter style current measuring devices suffer one problem when measuring DC. Every time a measurement is taken the residual magnetism has to be zeroed out or the reading can be so far in error that is practically meaningless. This is not a problem with AC as the current is rapidly reversing with no real residual effect.
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Hi Ross
You are right. There is still a case sometimes for an analog meter that does not have any external electronics that have to be powered.
You could use a Hall Effect device in the high side but you still have to power this externally, usually 5V.
Cheers Bob
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Woops. Well at least the input and output of the project I’m building are AC, so I can use it to measure the currents there.
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Trent,
I need to place the shunt in series with the postive input wire of the Boost Converter so that the GND voltages on eaither side of the Boost Converter are at the same potential. Are you saying that if I use an isolated battery to power the dual-meter, that I can do this?
Re: Dual Digital Display DC Voltmeter & Ammeter 0-100V 0-100A | Core Electronics Australia
There are no application diagrams which illustrate this option. Can you reveal what the wiring diagram would look like. I assume that the battery connects to the +/- Supply Connector and that the shunt connects to the IN+ and COM wires. What does the PW+ wire connect to in this case, or is this left floating? If PW+ is just used for the Voltmeter part of the display, then by not connecting it, the Voltmeter part of the display will not be used.
Below is a normal use-case.
Ross
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Hi Ross,
I’ve done the quick resistance test Bob suggested on the panel meter and can confirm that the power supply negative and the measurement circuit common are both tied together internally.
By connecting the panel meter power supply wires to a battery you are effectively using wiring option 3 with the annotations below (the black wire added is internal to the device)
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Hi Trent
I have just re-read this reply.
I originally suggested this test be done on the DC-DC converter.
Cheers Bob
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Hi Bob,
Sorry I should have been clearer, I did that check on the DC-DC converter, but we had already established it shared a common ground from Ross’s testing. Out of curiosity I then applied the same test procedure to the panel meter.
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Robert - I ended up buying a 1A analog panel meter with a resistance of 50 mOhm and a new 50A shunt with a resistance of 1 mOhm. The current at the input to the 8-18V to 24VDC/240W Boost Converter will be up to 30 Amps DC.
I will just read the 1A Meter value and multiply it by 50 (2% error is close enough). I did it this way since I don’t want to have to divert my high (30A) current cabling to/from the front panel of the prototype that I’m building. Instead, the most that will flow to the front panel is 0.6A, so a much smaller flexible cable can be used to connect the Meter to the Shunt.
Ross
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Hi Ross
If you are just looking for a close approximation then that is the way to go. I might add here you are not the first one to run into difficulty implementing one of these fancy meters. There was quite a long and drawn out discussion some time ago on just the meter you are trying to use. Like you say the 2% error is really nothing when compared to the simplicity of the analog counterpart. You will also find that sudden variations may show up better with this one as digital meters tend to take a while to settle.
Cheers Bob
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