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+5V voltage regulator

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Maverickmax

Maverickmax

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Hi

I have a small problem with my below schematic because I could not get 5V as I keep getting 3.5V to 4V.

My battery supplies under 8V at the moment.

I welcome to hear your suggestion

Maverick Max
 

assuming that all the conecction and calculkations are correct and ig the increment of the input voltaje do not coause anything, you may think for a malfuction of the devise, however if you are looking for a voltaje regulator i recomend you to use LM7805 it worked for me everytime and is very easy to implement
 

The quiescent current for this device is 6mA max, and have a dropout voltage of 2.5V max, your battery can source this 6mA more your load current, and when your battery is connected the voltage is more than 7.5V. If not try to change the components...
 

Maverickmax,

I think I saw your problem here. You need a larger input capacitance, preferably 0.33-µF to 0.47-µF, than the output capacitance. Not making both the same of 0.1-µF because should your input is switched off, back-injection can kill 7805 in your design.

For 7805 to operate properly, you need to have at least 6.7Vrms to 7Vrms DC at the input of voltage regulator, preferably between 8 to 10Vrms to save on the use of a heatsink.

You also need to have at least 100mA to 250mA input current from your supply. Surprisingly, this is not specified in the datasheet. I personally tested a 7805 before and I remember clearly that you need to have input current above 100mA. If your battery is producing less than 250mAH, you are likely to have a problem with getting 5Vrms DC.
 

Capacitors, currents and all that stuff doesn't matter much.
What really does matter is the input (batery) voltage. For standard 7805 you should have Vin=Vout+3V, and if you don't have it the output voltage can be below 5V.
With battery supply you should consider a Low Dropout voltage regulator such as LM4940-5 or similar. This device will produce 5V of anything above 5.5V ..
Regards,
IanP
 

IanP, are you sure LM4940-5? It doesn't exist.
I checked LM4940 and it is an audio amplifier.

As the name has already implied, 7805 is a linear voltage regulator, not a controlled voltage source. You need a minimum amount of current for it to operate because it taps its power from the input, unlike an amplifier that gets its power from the power rails.

I have personally tested a 7805 before and I am sure without sufficient input current, you will not get regulated voltage. That's assuming if you have already 8V. Try to go to 7V or 6.7V although it is the minimum. Usually it is recommended to maintain 20% higher than minimum in any electronic design to offer you better reliability.
 

The correct part number is LM2940CT-5.0..
Regards,
IanP
 

and another LM2575-5

good luck.
 

Hmm

Interesting, since I don't get 5V and my batter is going die soon, Im going to get a 12V battery from my local store. Do you think is it a good idea?

At the moment I use Panasonic 9V battery

Maverick Max
 

7805 voltage regulator without load will draw less than 5mA current and you should be able to measure 5V at its output pin.
12V or 9V should not make much difference.
As was said before for battery operation better option would be a lowdrop voltage regulator.
DrWhoF
 

*groan* I am very confused with your all messages. I need to iron this problem out asap. Please give me a clear instruction


Maverick Max
 

With or without a resistor, even a capacitor, so long you probe the voltage across the regulator's output with reference to ground, your digital multimeter DMM should still read about 5Vrms anyway beause your DMM has internat impedance afterall.

It's not the 9V or 12V battery that's your concern. These batteries certainly pump above 7V well.

What you can do is to conduct a simple test. Make a potential divider of R1=1K, R2=10K and the node in between be A. Connect A to the input of voltage regulator. The potential divider connects to the battery. Best is... Make R2 a potentiometer or trim pot. With this, you can check if voltage is the issue.

You can also use this bridge to check if the battery suffers from low mAH that causes insufficient current pumped into the regulator.
 

The story with batteries is like this: if the voltage of a 9V battery drops to 8V it indicates that the battery is already dead and you can't squize anything more from it.
For a 5V voltage regulator ,such as 7805, all what you need from a battery is >8V.
Using a 12V battery will not alter this situation at all.
So, if I were you I would use 9V battery.
Make sure you supply the battery voltage to the input pin, and you take the measurement at the ouput pin (not the other way around - confirm this with the data sheet), this voltage regulator, if not faulty, will give you ≈5V.
Regards,
IanP
 

basic reg out = input - 3

then to get 5 u need to input at least 8
 

hi

u can try this modification.

A 4007 diode is connected to L7805 regulator middle pin(common ground)
diode anode is connected to 7805 middle pin and cathode is connect to ground.

best of luck.
satya
 

IC could have problem too.
Have to trial and error.

Just got started using DC-DC, and getting to like them now.
some reference for LM2576, LM2575.
for 5V dc-dc regulation.

**broken link removed**
 

If you want to be able to generate 5V until your 9V battery has WAY low output voltage then you must use a DC-DC converter. If such a converter is properly designed, it will have decent efficiency when the battery voltage is low and it will have superior efficiency (compared to a linear regulator such as 7805) when the battery voltage is high.
 

Most probably your device might be functioning eratticaly .I have tried the same with NiMH batteries and Iam get exactly 5V withLM7805 from National semiconductor
 

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