Question on How to Wire Parallel Circuits Properly

Thread Starter

Benjamin0904

Joined Oct 3, 2020
22
I've pretty new to parallel circuits and can't seem to understand them.
To Begin:
I have a 5v DC power source with barely any amperage, and the goal is to covert it to 5v 2A using no excess power. Just recently I bought (5) dc-dc 1v-5v step up converters. Apparently all of these are supposed to work like this:
When 1v DC is put in the input it will give you 5v DC; 2v in input = 5v; 3v in input =5v, etc. up to 5v input. Now when a full 5 volts is inputted it will still give 5v output, but increase amperage to 500ma. Therefore with the smallest 1v input it should give you 100ma.
If each converter, at full input gives output of 5v DC and 500ma then is it possible to wire all of the converters to keep the same voltage but increase amperage x5?
By the way, each converter has a input+, an output+ and a ground-
 

jpanhalt

Joined Jan 18, 2008
10,515
Welcome to AAC.

It's not parallel circuits that are the issue. You need to understand "power." The watt (W) is the unit of power. In DC circuits, it is easily calculated at voltage times current or current squared times resistance. Those relationships are know as Ohm's law.

No circuit is 100% efficient. That means there is always some loss of power. So, if you put 1V in and get 5V out, the current at 5V can be no more than 1/5 th the current at 1 V. Because of losses, it is probably considerably less.

Back to your goal, you simply cannot convert 5V at low current to 5V at 2A without adding power.
 

Thread Starter

Benjamin0904

Joined Oct 3, 2020
22
Welcome to AAC.

It's not parallel circuits that are the issue. You need to understand "power." The watt (W) is the unit of power. In DC circuits, it is easily calculated at voltage times current or current squared times resistance. Those relationships are know as Ohm's law.

No circuit is 100% efficient. That means there is always some loss of power. So, if you put 1V in and get 5V out, the current at 5V can be no more than 1/5 th the current at 1 V. Because of losses, it is probably considerably less.

Back to your goal, you simply cannot convert 5V at low current to 5V at 2A without adding power.
Hi jpanhalt! Thanks for the info. I just want to make sure I've got something down. So for a single converter, if I put 5v 80ma into input and get 5v output with 85% conversion efficiency I wont get 500ma? So how many milliamps do I get or does it vary?
Thanks again,
Benjamin
 

djsfantasi

Joined Apr 11, 2010
6,962
Hi jpanhalt! Thanks for the info. I just want to make sure I've got something down. So for a single converter, if I put 5v 80ma into input and get 5v output with 85% conversion efficiency I wont get 500ma? So how many milliamps do I get or does it vary?
Thanks again,
Benjamin
Did you mean 5V in? Or 1V in? Like your proposed circuit.

I’ll run the numbers for 1V input for you. For 1V input and 5V output, your current out is 1/5th the current in. 80mA input becomes 16mA output. But due to losses, 16mA times 85% becomes 13.6mA output.
 

Thread Starter

Benjamin0904

Joined Oct 3, 2020
22
Did you mean 5V in? Or 1V in? Like your proposed circuit.

I’ll run the numbers for 1V input for you. For 1V input and 5V output, your current out is 1/5th the current in. 80mA input becomes 16mA output. But due to losses, 16mA times 85% becomes 13.6mA output.
So the input is 5v 80-90ma and the output will be 5v but I'm not sure how much amperage
Thanks!
 

Tonyr1084

Joined Sep 24, 2015
5,239
Think of a see-saw: At 5V you have 500mA. At 2.5V you have 1000mA (1A). Both come to the same wattage: 5V x 0.5A = 2.5W. And 2.5V x 1A = 2.5W. You can not exceed the wattage. As the voltage goes up the amperage goes down. And vice versa. Using a converter comes with losses, so you're giving up some power as wasted heat energy. If you need more amperage then you need a bigger supply. One capable of supplying the total current you need. Would suggest you multiply your needed current by 1.3 or 1.5 times higher and get the appropriate size supply.

If you're putting several devices in parallel, all running on 5V, assume you have three devices, each drawing 250mA. Your power supply will need to be able to supply 5V @ 750mA. In that case I'd opt for 975mA to 1125mA. In that case a 5V 1A supply will be able to drive all three of those devices.
 

Thread Starter

Benjamin0904

Joined Oct 3, 2020
22
Think of a see-saw: At 5V you have 500mA. At 2.5V you have 1000mA (1A). Both come to the same wattage: 5V x 0.5A = 2.5W. And 2.5V x 1A = 2.5W. You can not exceed the wattage. As the voltage goes up the amperage goes down. And vice versa. Using a converter comes with losses, so you're giving up some power as wasted heat energy. If you need more amperage then you need a bigger supply. One capable of supplying the total current you need. Would suggest you multiply your needed current by 1.3 or 1.5 times higher and get the appropriate size supply.

If you're putting several devices in parallel, all running on 5V, assume you have three devices, each drawing 250mA. Your power supply will need to be able to supply 5V @ 750mA. In that case I'd opt for 975mA to 1125mA. In that case a 5V 1A supply will be able to drive all three of those devices.
I'm beginning to think that this Amazon listing lied to me, It is stating what is impossible according to all of the comments. This is where I bought the 1-5v converters, here is a picture of the statistics: I would be in the category of more than 3v input.
Amazon Circuit 1.jpg
 

Tonyr1084

Joined Sep 24, 2015
5,239
Maximum output current is just how much current it can handle. Doesn't claim it will increase the current to that level. If you have a pitcher of water you can fill only so many glasses with water. You can double the number of glasses if you only fill each half way. But you can't give more water than you have.
 

jpanhalt

Joined Jan 18, 2008
10,515
Those numbers are probably correct and do not "lie."

Example 1: 1.5 V in -->5V @ 100 mA out. Out = 0.5 W. In = 333mA + inefficiency. Let's say 500 mA in.
Example 2: 2 V in , 5V @ 150 mA out = 0.75 W. In = 375 mA + inefficiency.
Example 3: 3 V in , 5V @ 380 mA out = 1.9 W. In = 633 mA + inefficiency.

It looks like the input current must be below 500 mA or just a little more if the input voltage is higher.

One way around your need for higher current is to charge batteries. If it takes 4 hours to charge to charge 4.2 V, then you can draw twice the charge current for half the charge time (actually considerably less because of inefficiency).
 

Veracohr

Joined Jan 3, 2011
726
Seems like an awfully complicated way of getting a 5V 10W power supply. Those should be pretty cheap to buy new, or you can probably find one at your local thrift shop.
 
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