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# What happens to Voltage source if i take out both V and I with separate circuit?

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#### gwarming82

##### Newbie level 4
I have 3.8V and 1.2A battery.I will use a step up circuit that gives 5v and some current output using 2v in(from my battery).Then i will use diode(say zener) to take voltage only(5v).I will also use a step down circuit that gives 1v and 2A output using 3.5v in(from my battery).Then i will use another blocking diode to take current(2A) only.So Can i get 5v and 2A output in this way from a 3.8V and 1.2A battery?And Now what will happen with my battery?Will the battery be empty after producing separately stepped volt and current or can i get more volt(5v) and current(2A) greater than my battery(3.8V,1.2A) in this way?My step up and step down circuits are as following:

https://www.alibaba.com/product-detail/DC-DC-2-0-5-v_1547670640.html?spm=a2700.7724838.102.7.vDaE3Z

https://www.amazon.com/DROK-Convert...1436885493&sr=8-16&keywords=current+regulator

Your understanding of electricity is confused. A power source, whether battery or from mains AC, produces a voltage. The voltage can be sustained as more current is drawn from it but you can't use the voltage and the current independantly. The battery isn't 3.8V and 1.2A, it is probably 3.8V with a 1.2A limit. Batteries are usually specified as having an 'Amp Hour' (AH) rating rather than just Amps which also introduces the concept of how many hours the current can be drawn for. If your battery is really 3.8V with 1.2AH rating it means you can draw 1.2A for one hour or 0.6A for two hours or 0.12A for 10 hours before the voltage drops to an end point.

Also consider what you are asking your circuits to produce. The actual usage of power is measured in Watts, that is calculated by the voltage multiplied by the current. I'm afraid no circuit with an input power of 3.8 * 1.2 = 4.56W will give an output of 5.0 * 2.0 = 10W out. That would imply an efficiency of over 200% !! Even well designed regulator circuits can only achieve ~95%.

Brian.

gwarming82

### gwarming82

Points: 2
The voltage booster product is used in a cell phone charging bank. The charging bank uses a 3.7V Li-ion battery and its output is 5V at maybe 1A to feed the charging circuit inside a cell phone.
With a 3.7V input the voltage booster you show has a maximum output current of 1A and the current drained from your 3.7V battery will be 1.15A.

I also think your "3.8V" battery has a capacity of 1.2Ah so it can fully charge a cell phone with 5V/1A only one time.

Re: What happens to Voltage source if i take out both volt and current?

So first off is your battery 1.2A or 1.2Ah? I rarely see batteries state the maximum current they can source in Amps as opposed to a multiple of their capacity in Amp-hours. It sounds like you may be using a 1 cell LiPo, which should have a max current rating of (#)C. If this is the case the maximum current your battery can source in # times the capacity of your battery. 1C it can source 1.2A, 10C -> 12A. Do you have a link to the battery?

Other than that I'm not really sure what you mean by "take voltage only" or "current only", do you have any links to the concept you're describing that we could see?

If the battery can really only provide 3.8V @ 1.2A, or about 4.5W than it won't be capable of providing even 1A @5V without an additional power source.

The charging current of a cell phone with a Lithium battery is not maximum for the entire charging duration. The charging current slowly drops during the last maybe 1/3rd of the time until the charging stops when the charging current becomes very low. Then the battery is fully charged. So a 3.8V/1.2Ah battery in an 85% efficient voltage stepup circuit can charge a cell phone with "1A" one time until it is fully charged.

Wait a minute. I looked up a Samsung Galaxy battery and its capacity is 2100mAh! The 1.2Ah battery in this thread is too small to stepup the voltage and fully charge the phone one time.
The power banks listed in Google have capacities of thousands of mAh.

Your understanding of electricity is confused. A power source, whether battery or from mains AC, produces a voltage. The voltage can be sustained as more current is drawn from it but you can't use the voltage and the current independantly. The battery isn't 3.8V and 1.2A, it is probably 3.8V with a 1.2A limit. Batteries are usually specified as having an 'Amp Hour' (AH) rating rather than just Amps which also introduces the concept of how many hours the current can be drawn for. If your battery is really 3.8V with 1.2AH rating it means you can draw 1.2A for one hour or 0.6A for two hours or 0.12A for 10 hours before the voltage drops to an end point.

Also consider what you are asking your circuits to produce. The actual usage of power is measured in Watts, that is calculated by the voltage multiplied by the current. I'm afraid no circuit with an input power of 3.8 * 1.2 = 4.56W will give an output of 5.0 * 2.0 = 10W out. That would imply an efficiency of over 200% !! Even well designed regulator circuits can only achieve ~95%.

Brian.

Dear sir,
Say you said,"This is fine as far as it goes, but why you are stepping down from 3.8 to 2 volts before
stepping up to 5 volts makes no real sense to me."
Well i use both step up and step down because the step up circuit only increase voltage from 2v to 5v. Say here i found my desired voltage like 5v.Again i use step down that decrease voltage from 3.5v to 1v and increase current like 2A output.So here i found desired current 2A.Now i want to combine both this circuit to get 5v and 2A as output in combination.Then i want to use this 5v@2A to recharge the battery again inversely.

Say you again said,"Blocking diodes will stop current. You most certainly need voltage; current will not flow
without some. Assuming for the moment that 1 volt is your output voltage and 2 amps current,
then your power output is 2 watts. A 3.8 volt battery with 1.2 amp-hours capacity has a total
available energy content of 4.56 watt-hours, so your battery will be exhausted in 2.28 hours."
Well why i have 1v with 2A(2 watts)? I already combined 5v from step up and 2A from step down so will not i get 5v@2A as output in combination?

​Pls tell me one thing.If i combine the step up and step down circuit separately(using blocking diode) can not i get 5v at one end and 3A at other end of circuit?​Here small voltage and current is ignored after blocking with diode and i will use only biased volt and current.​​Why i will get 1v@3A?The step down circuit gives me 1v and 3A output using 3.5v in(from my battery).Then i will use another blocking diode to take current(3A) only and i dont need ​small ​voltage from this circuit.So can not i get 3A here(voltage blocked) and get 5v out​(current blocked)​ from ​both circuit and combine them to get 5v@3A?

​My main purpose is to recharge the same battery using back the stepped energy of the battery itself.If i feedback the stepped energy 5v@2A (using step up and step down​​) to the battery to recharge the same battery how long will my circuit work? Will my design really recharge the battery using the same battery energy(3.8v@1.2Ah) as input and then recharging the battery again using the combined stepped voltage and current(5v@2A)?
Hope you understood my project.

All Step converters have efficiency losses and at same time transform source impedance, so stepping up will always be higher impedance causing more regulation losses if you cascade step up then down. They are not ideal voltage sources.

Even neglecting the above, if you start with 3.8V @1.2Ah or 4.5Wh you cannot get 5Vx2Ah= 10Wh out.

V
Points: 2
Your understanding of electricity is confused. A power source, whether battery or from mains AC, produces a voltage. The voltage can be sustained as more current is drawn from it but you can't use the voltage and the current independantly. The battery isn't 3.8V and 1.2A, it is probably 3.8V with a 1.2A limit. Batteries are usually specified as having an 'Amp Hour' (AH) rating rather than just Amps which also introduces the concept of how many hours the current can be drawn for. If your battery is really 3.8V with 1.2AH rating it means you can draw 1.2A for one hour or 0.6A for two hours or 0.12A for 10 hours before the voltage drops to an end point.

Also consider what you are asking your circuits to produce. The actual usage of power is measured in Watts, that is calculated by the voltage multiplied by the current. I'm afraid no circuit with an input power of 3.8 * 1.2 = 4.56W will give an output of 5.0 * 2.0 = 10W out. That would imply an efficiency of over 200% !! Even well designed regulator circuits can only achieve ~95%.

Brian.

Say you said,"This is fine as far as it goes, but why you are stepping down from 3.8 to 2 volts before
stepping up to 5 volts makes no real sense to me."
Well i use both step up and step down because the step up circuit only increase voltage from 2v to 5v. Say here i found my desired voltage like 5v.Again i use step down that decrease voltage from 3.5v to 1v and increase current like 2A output.So here i found desired current 2A.Now i want to combine both this circuit to get 5v and 2A as output in combination.Then i want to use this 5v@2A to recharge the battery again inversely.

Say you again said,"Blocking diodes will stop current. You most certainly need voltage; current will not flow
without some. Assuming for the moment that 1 volt is your output voltage and 2 amps current,
then your power output is 2 watts. A 3.8 volt battery with 1.2 amp-hours capacity has a total
available energy content of 4.56 watt-hours, so your battery will be exhausted in 2.28 hours."
Well why i have 1v with 2A(2 watts)? I already combined 5v from step up and 2A from step down so will not i get 5v@2A as output in combination?

​Pls tell me one thing.If i combine the step up and step down circuit separately(using blocking diode) can not i get 5v at one end and 3A at other end of circuit?​Here small voltage and current is ignored after blocking with diode and i will use only biased volt and current.​​Why i will get 1v@3A?The step down circuit gives me 1v and 3A output using 3.5v in(from my battery).Then i will use another blocking diode to take current(3A) only and i dont need ​small ​voltage from this circuit.So can not i get 3A here(voltage blocked) and get 5v out​(current blocked)​ from ​both circuit and combine them to get 5v@3A?

​My main purpose is to recharge the same battery using back the stepped energy of the battery itself.If i feedback the stepped energy 5v@2A (using step up and step down​​) to the battery to recharge the same battery how long will my circuit work? Will my design really recharge the battery using the same battery energy(3.8v@1.2Ah) as input and then recharging the battery again using the combined stepped voltage and current(5v@2A)?
Hope you understood my project.

If you start with 3.8V/1.2Ah and step it up to 5V then its current for one hour will be less than 5V/(3.8V x 1.2Ah)= 1.1A but probably 0.8A.
If you start with 3.8V/1.2Ah and step it down to 1V then you get less than 1V/3.8V x 1.2A)= 4.6A but probably 3.4A for one hour.
If you stepup the 1V/3.4Ah to 5V then you get less than (1V x 3.4Ah)/5V= 0.7A but probably 0.5A for one hour.

If you want 5A at 3A for one hour then you need a battery that is 3.8V/5.3Ah.

V
Points: 2
If you start with 3.8V/1.2Ah and step it up to 5V then its current for one hour will be less than 5V/(3.8V x 1.2Ah)= 1.1A but probably 0.8A.
If you start with 3.8V/1.2Ah and step it down to 1V then you get less than 1V/3.8V x 1.2A)= 4.6A but probably 3.4A for one hour.
If you stepup the 1V/3.4Ah to 5V then you get less than (1V x 3.4Ah)/5V= 0.7A but probably 0.5A for one hour.

If you want 5A at 3A for one hour then you need a battery that is 3.8V/5.3Ah.

So do you mean If i want 5V at 3A for one hour then i need a battery that is 3.8V/5.3Ah? If so then can i recharge the same battery again using the stepped voltage and current? And can not i recharge the battery not only for one hour but again and again when i need in this way?

Sorry gwarming82, what you propose is nonsense. Even if it was possible for a battery to recharge itself, what would be achieved? If it was possible to do it electrically and the circuit was 100% efficient all you would end up with is a battery that didn't lose charge if you didn't use it. Taking energy out for any other purpose would still drain it. I have lots of batteries that hold charge if I don't use them already!

As I already stated (and although you quoted my message, it wasn't the one you replied to) you can't use the voltage and current as separate entities. The current is what you draw from the power source while it is producing a voltage, one is the 'push' and the other is the 'flow'.

So do you mean If i want 5V at 3A for one hour then i need a battery that is 3.8V/5.3Ah? If so then can i recharge the same battery again using the stepped voltage and current? And can not i recharge the battery not only for one hour but again and again when i need in this way?
If you want 5V at 3A you need 15W and a battery that can provide it for one hour would be a 5V and 3Ah. That means it provides 5V across it's terminals and can sustain 3 Amps load for one hour. In practice you will not find any battery that can hold constant voltage under load then suddenly 'die', the voltage gradually drops so the rating might for example be specified as 5V 3Ah with a 4V end voltage. In other words it tells you the end voltage under the specified load current after the specified time.

Brian.

If you try to charge a Lithium battery cell at the much too high voltage of 5V then it will explode or catch on fire. Lithium, Magnesium (white hot flares) and Titanium (turbine blades in jet aircraft) all burn with a VERY hot fire. 4.20V with the charging current limited to the spec's of the battery is the maximum charging voltage.

Use another fully charged 3.8V/5.3Ah battery to drive a step up voltage booster. Then they can provide 5V at 3A for one hour to feed a battery charger circuit. That is what a "Power Bank" does. The 5.3Ah battery will need to have its own charger.

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