Current produced from LM7805 !

[MEERA]

I feeding 6v 1300mAh to LM7805 ( 5v voltage regulator)
i need to know what is the current value in output of the regulator

According to OHMS law,
if voltage limited(decreased) , Current will be Maximized (Increases)

also tell me how to check the Current from output of LM7805 using multimeter !

 

[Harald Kapp]

The LM7805 is a volateg regulator. It will provide 5V as long as the current limit (see datasheet for the specific variant of LM7805 you are using) is not reached. If the current limit is reached, the 7805 will decrease the output voltage to keep the current at the limit.

Note: The 7805 will not work correctly with a 6V input. It has a so called dropout voltage of at least 2 V. This is the minimum voltage difference between input and output for the regulator to operate. this means Vin >= 7V is required.

You will need a so called low drop regulator which perates with a dropout voltage below 1V. Here's an example.

The current from the regulator is measured by inserting an ammeter (or multimeter set to measuring amperes) in series with the load.

 

[MEERA]

is LP2954 (ACTIVE) 5V Micropower Low-Dropout Voltage Regulator
suitable for producing constant current source of 500mA and voltage source of 5v from a battery of capacity 6v 1300mah !

 

[(*steve*)]

Please! don't! use! exclamation! marks! inappropriately! unless! you! work! for! Yahoo!

is LP2954 (ACTIVE) 5V Micropower Low-Dropout Voltage Regulator
suitable for producing constant current source of 500mA and voltage source of 5v from a battery of capacity 6v 1300mah !

Perhaps you had better tell us what you need this for.

I'm not sure that you understand what "constant current source" or "1300mAh" actually mean.

If you can tell us what you're trying to do we may be able to give you a correct solution rather than trying to answer very poorly phrased (and potentially very poorly understood) questions.

 

[MEERA]

Please! don't! use! exclamation! marks! inappropriately! unless! you! work! for! Yahoo!



Perhaps you had better tell us what you need this for.

I'm not sure that you understand what "constant current source" or "1300mAh" actually mean.

If you can tell us what you're trying to do we may be able to give you a correct solution rather than trying to answer very poorly phrased (and potentially very poorly understood) questions.

im going to design a portable mobile charger (Power bank)
im having 6v battery
i need to get constant output, as im using it to charge a mobile phone which needs 5v, 500ma

 

[BASICFreak]

USB is 5V -/+ 0.2 so if the battery is 6V a silicon diode would drop 0.7V get you 5.3V and a germanium diode would drop 0.3V which will get ~5V

Depending on the phone it will regulate the charger itself as the battery inside is 3.7V

I've measured my phone's wall charger at 5.15V (I guess under load it would drop down some..?)

or a DC-to-DC converter
http://www.us.tdk-lambda.com/ftp/Specs/cc-e.pdf
CC3-0505SF-E should work nicely for you with 600mA

 

[(*steve*)]

At 500mA, you'll be running up against the limits for the dropout voltage. As the battery goes flat the problem will become worse.

Have you actually measured the voltage of the battery when fully charged? What type of battery is it? (lead acid or several MiMH cells I presume).

Deriving a voltage at a relatively high current where the desired output voltage is only a little less than the input voltage is tricky to do with regulators like these.

There are options with switchmode regulators, especially buck/boost designs where the input voltage can fall below as well as above the desired output voltage, but these are somewhat more complex and may be best purchased as modules.

 

[(*steve*)]

USB is 5V -/+ 0.2 so if the battery is 6V a silicon diode would drop 0.7V get you 5.3V and a germanium diode would drop 0.3V which will get ~5V

Firstly it's 5V +/- 0.5V, and secondly this may not be a reliable method, especially as the battery discharges.

If the current is low, the voltage drop of the diodes will be less, possibly exceeding the 5.5 maximum spec for power, and under high current the voltage drop across the silicon diode could exceed 1V.

This can be an acceptable solution where the load and the inut voltage is very close to constant.

Also, you wouldn't use a germanium diode.

Depending on the phone it will regulate the charger itself as the battery inside is 3.7V

And it may work for input voltages up to 7V, ot 12V, but you don't know so it's better to go with the specs.

I've measured my phone's wall charger at 5.15V (I guess under load it would drop down some..?)

It might, but it could even go up. (it depends on how it is regulated)

or a DC-to-DC converter
http://www.us.tdk-lambda.com/ftp/Specs/cc-e.pdf
CC3-0505SF-E should work nicely for you with 600mA

Possibly. Have you priced one?

Something like this might work down to about 4.5V. When the input voltage is less than the desired output voltage it essentially passes straight through.

 

[MEERA]

Im using Ni-Mh battery 6v 1.3A
can u guide me what is current and voltage required to charge this battery

 

[BASICFreak]

Firstly it's 5V +/- 0.5V, and secondly this may not be a reliable method, especially as the battery discharges.

I'll admit I was wrong there I misread my facts

Also, you wouldn't use a germanium diode.

True, especially if its +/- 0.5V a silicone diode would work in a pinch where you needed the phone charged right then and there, yes as the battery drains and more current is drawn the output will drop down as well.

Possibly. Have you priced one?

Something like this might work down to about 4.5V. When the input voltage is less than the desired output voltage it essentially passes straight through.

Yes, actually I was looking at the 2A one for a project of mine, and I have also seen similar ones like that on eBay but I rather not buy a cheap china made board I have had very bad luck with parts like this.

 

[BASICFreak]

Im using Ni-Mh battery 6v 1.3A
can u guide me what is current and voltage required to charge this battery

To charge the battery I would use around 6.3-6.9V and a slow charge about 15-45mA to avoid over charging it.
The faster you charge (more current) the hotter the battery will get and if too hot explode.
My thought is charge it over the course of hours, if you want a quicker charge I personally would not put more than 150mA into the battery.

 

[(*steve*)]

Im using Ni-Mh battery 6v 1.3A

Firstly, the battery is (presumenably) 4 AA NiMH cells. They are each rated at 1300mAh

mAh is a measure of capacity, not current, but it often gives a guide to the charge/discharge current. In this case 1C would be a current of 1300mA (1.3A) and 0,1C would be 130mA (C is the theoretical 1 hour discharge rate)

can u guide me what is current and voltage required to charge this battery

NiMH cells are among the hardest to correctly charge. I would recommend you get a charger that can reliably sense either the temperature rise when the charge is almost complete, or the small voltage dip which occurs at this point.

It is a common misunderstanding that a particular voltage is required. The fact is that all that is required is that current is flowing into the battery (i.e in the opposite direction to discharge).

The battery will respond to this by having its voltage rise by a combination of achieving a higher charge level and due to resistive losses in the battery. When the charging current falls to zero, the battery voltage is simply that voltage it would have at the current state of charge.

Chargers typically (although not universally) require a higher voltage than the battery (typically at least a couple of volts higher) and a current a little greater than the charge current. (There are exceptions, but let's pretend they don't exist -- the conditions I mention here must exist within the charger in any case).

A simple charger might charge at 0.1C for 14 hours. Note that this means you are charging the battery with 140% of its capacity. This is because charging is not 100% efficient and you need to put more power into the battery than you take out (resistive losses are one reason for this).

A simple charger will overcharge any battery that is not fully discharged, or which is left in for more than 14 hours. NiCad batteries were often charged this way because they were reasonably tolerant to this level of overcharge. NiMH are far less tolerant.

Fast chargers (4 hour or even 1 hour) require some means of detecting the end of charge because they can very easily damage the cells. You often find the batteries have a third contact. This connects to a thermistor that the charger uses to detect the end of charge. The batteries begin to heat up because the power is no longer going in to reversing a chemical change that stores energy, but is being dissipated resistively, heating the battery. This can cause the liquids in the battery to boil, or electrolysis might create hydrogen, or other things (depending on the battery type)

 

[MEERA]

First of all thanks for your valuable information !
This battery has 4 cells per-packed !
Its total capacity is 6v, 1300mAh


Im using 8v 500mA for charging thos battery
is it permissible or any changes should i do it for a better performance !