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[SOLVED] Dual 5V output from one 5V input circuit design, help.

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Rickooo

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Hi, I am working on some circuit. I am using 220V to 5V 5W converter https://www.electronicscomp.com/hlk...XWfRMe9VBNYrs6X75c5rFFSSsTbXCBOYaAkCGEALw_wcB board. But, it have just one 5V output. How can I convert one 5V to two 5V output?

I tried using circuit of LM317 to get fix 5V output. but, I am not sure if we can use same volt as input and output for this ic or not. I selected R2 = 715 ohm and R1 = 240 ohm. I want 5V 1A as one output and 5V 1A as other output. What do you suggest?
 

HI,

please be more clear.
What does "dual 5V output" mean?
* two completely isolated 5V outputs. If so, why two? Why not simply use one supply?
* or a combined +5V/-5V output. If so, then a voltage inverter could
Do you need isolation from input to output?

Klaus
 
What does "dual 5V output" mean?
* two completely isolated 5V outputs. If so, why two? Why not simply use one supply?
* or a combined +5V/-5V output. If so, then a voltage inverter could
@KlausST The dual output mean. When input volt is 5V, I want two isolated output supply of 5V each, like 5V 1A and 5V 1A and these are isolated supply.
I want two, because, one supply goes to the main thermometer circuit and other 5V supply goes to USB connector to which camera 5V will be inserted.
If we use one supply for all, it may give non stable 5V output and which may effect the functioning of other circuit.
 

Hi,

I am not sure if we can use same volt as input and output for this ic or not.

for me it sounds you are trying to create a 5 V output voltage by the use of a 5 V input voltage, which also corresponds to your stated resistor values (V_out = 1.25 V \[\cdot\] (1 + 715 \[\Omega\]/240 \[\Omega\]) = 4.974 V). That does not work, a voltage regulator always needs a higher input voltage (drop-out voltage) to be operated, and for the LM317 a voltage difference between input and output (drop-out voltage) up to 3 V is required (DS page 10 [1]).

What are you aiming to realize exactly?


[1] https://www.ti.com/lit/ds/symlink/lm317.pdf

BR
--- Updated ---

When input volt is 5V, I want two isolated output supply of 5V each, like 5V 1A and 5V 1A and these are isolated supply.

By using a linear regulator like the LM317, the output voltage is not GALVANIC isolated. Therefore, you would nid a SMPS.
 
Last edited:

When input volt is 5V, I want two isolated output supply of 5V each, like 5V 1A and 5V 1A and these are isolated supply.
I want two, because, one supply goes to the main thermometer circuit and other 5V supply goes to USB connector to which camera 5V will be inserted.
If we use one supply for all, it may give non stable 5V output and which
Doesn't sound like a substantiation for isolation. Please explain the application more detailed. How are both circuit parts connected?

In any case, you can't have dual 5V/1A (10W) from single 5V/1A (5W).
 

@KlausST @FvM @stenzer As I said previously, I am using this board https://www.electronicscomp.com/hlk...XWfRMe9VBNYrs6X75c5rFFSSsTbXCBOYaAkCGEALw_wcB
Via this board I am getting 5V 1A. Now, I wants to convert this 5V 1A to two isolated 5V supply. One 5V is given to camera and another 5V given to thermometer. Now, I want you to suggest me a ic/circuit in which I can use to get two 5V supply (output) from one 5V supply (input). For more details, you can look to this board, https://www.tindie.com/products/armtronix/mi002power-supply-5vdc5w/
Awaiting response.
 

Hi,

I´d use an isolated 5V/5V DCDC converter. Cheap. ready to buy. Many manufacturers, just do a search.

Klaus
 

Hi,

in post #1 you asked for:
" I want 5V 1A as one output and 5V 1A as other output. "

Thus: NO.
200mA is not 1A.

Klaus
 
Why do the camera and thermometer need to be isolated from each other?
The power supply module that you have says its output is 5V at 1A. It cannot produce the 2A that you want.
 

    Akanimo

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Hi,

Just like FvM has stated, the AC-DC converter you linked to in post #1 is a 5W converter. Using it as supply to a DC-DC converter, you can't you can't get two 5V 1A outputs from such converter. You can't even get 5W from such converter because of losses.
 

Hi, @FvM @KlausST @Audioguru As, it seems a but confusing to all. I have made a rough sketch to make you understand what I actually want. Their is 220V to 5V converter circuit. Now, I want to use this 5V to operate two different circuit. One is camera, which will attached to USB. Second is thermometer circuit, which is main circuit and which is also work as power supply for whole circuit. I am using now, 220V to 5V 10W converter circuit. Now my question to you is, If I use diode to direct the path in both circuit as shown in sketch 1, will id difflect the end voltage. Or I should use sketch 2? Looking for the response.
 

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Hi,

I still don´t see a need for an isolated power supply.
I also see no need for the diodes.

Maybe I´d add separte filters and polyfuses for each power output.

Klaus
 
The AC/DC described in post #1 has 5W output, why does it appear as 10W converter in the schematic? If you think about changing the converter, why not a dual output AC/DC module?
 

Why are you using diodes that reduce the voltage to about 4V to 4.4V? If the 5V power supply can produce enough current then it can power hundreds of 5V devices without using any diodes.

A car has a 12VDC power supply that powers many lights and things without using any diodes.

A diode might be needed in parallel with an inductive load to clamp the high voltage flyback produced when it is turned off.
 

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Hi,

why do you want to separate those two branches? If you are afraid of noise generated in one branch propagating to the other one, use a filter as klaus suggested. E.g. a dedicated PI filter for each brach and an additional ferrite bead.

BR
 

Okay, Thank you all for the suggestions. I will use Pi filter. Let me back with the filter. As I am new to this hardware. Can you suggest/direct me for pi filter calculation? As it include frequency, so what frequency do I use? How do I began calculation for this? If I use capacitor, won't it drop the volt? Thanks
 

Hi,

usually you would start with the switching frequency \[{f}_{sw}\] of your SMPS. If you are using a second order PI-filter the cut-off frequency can be determined by

\[{f}_{c }\]=\[\frac{1}{2 \cdot \pi \cdot\sqrt{L\cdot C}}\].

As a start point set \[{f}_{c }\] = \[\frac{10}{{f}_{sw} }\]. Choose reasonable values by means of their physical size. Keep the current rating of your inductor in mind. If you have enough space you can use high values which results in a low cut-off frequency (-> high attenuation). High valued capacitors and inductors have usually are larger physical size.

This gives you the values in one direction, from the SMPS towards your load i.e. USB. If your attached load generates noise as well use the expected frequency to calculate the cut-off frequency from the load towards the SMPS. Therefore, your inductance is already determined (see above), thus only the capacitor on the SMPS side has to be determined with the equation above.


BR
 
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