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Stable 5 Volt supply

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adwnis123

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Hello, I want to have stable 5 Volt supply for my circuit. What IC to use and what voltage should I give. The current should be around 500mA -1A
 

At this current, you probably want a switching power supply.

What input voltage you want to convert to 5 volts?
 
shaiko said:
At this current, you probably want a switching power supply.

What input voltage you want to convert to 5 volts?
Click to expand...

What would be a good supply to take 5 Volts on the output???
 

You can use a LM7805 to get constant 5V supply. The input to 7805 should be 3V higher than the output. So, output is 5V and input should be higher than 5 + 3 = 8V. Provide 9V or 12V to LM7805 input and you get 5V output.
 
milan.rajik said:
You can use a LM7805 to get constant 5V supply. The input to 7805 should be 3V higher than the output. So, output is 5V and input should be higher than 5 + 3 = 8V. Provide 9V or 12V to LM7805 input and you get 5V output.
Click to expand...
For a current of 500 mA, an input of 9V or 12V would result in a power dissipation of respectively 2W and 3.5W in the LM7805. This will require a rather bulky heat sink.

I agree with shaiko's opinion, that a switched mode buck converter is much better suited for this application. They have relatively high efficiencies and won't get hot at 500 mA load current. A type I have good experiences with is the LM2596.
 
milan.rajik said:
You can use a LM7805 to get constant 5V supply. The input to 7805 should be 3V higher than the output. So, output is 5V and input should be higher than 5 + 3 = 8V. Provide 9V or 12V to LM7805 input and you get 5V output.
Click to expand...

And I use the "Fixed-Output Regulator" circuit on page 19 here:
https://www.fairchildsemi.com/datasheets/LM/LM7805.pdf

???

- - - Updated - - -

ArticCynda said:
For a current of 500 mA, an input of 9V or 12V would result in a power dissipation of respectively 2W and 3.5W in the LM7805. This will require a rather bulky heat sink.

I agree with shaiko's opinion, that a switched mode buck converter is much better suited for this application. They have relatively high efficiencies and won't get hot at 500 mA load current. A type I have good experiences with is the LM2596.
Click to expand...

And I use the "Typical Application" circuit on page 1 here
https://www.ti.com/lit/ds/symlink/lm2596.pdf
??

Do electronic stores sell coils, where can I find the 33μΗ coil???
 

I presume you need the stable 5v as a part of a larger project. If that is the case, why not just pop in a readymade cellphone charger rated at 1A? They are cheap and abundantly available.
 
ard said:
I presume you need the stable 5v as a part of a larger project. If that is the case, why not just pop in a readymade cellphone charger rated at 1A? They are cheap and abundantly available.
Click to expand...

I used a 5 Voltage supply, but the voltage drops down to 4.23 Volt and the circuit does not work correctly
 

Are you sure it was rated at 1A?

Is your circuit drawing way too much current?
 
ard said:
Are you sure it was rated at 1A?

Is your circuit drawing way too much current?
Click to expand...

The circuit's current is maximum 500mA now, but in case I put some extra components I want to have "space".

A PIC18 what current does it draw?
 

I use an LM7805 because I didn't find the LM2596. I give 10 Volt Input and I get 4.5 Volt Output, and the circuit does not work properly. What should I do???
 

adwnis123 said:
I use an LM7805 because I didn't find the LM2596. I give 10 Volt Input and I get 4.5 Volt Output, and the circuit does not work properly. What should I do???
Click to expand...
The power you're dissipating into the LM7805 is probably too high, forcing the IC into a thermal shutdown mode and causing it to throttle down the output current.

The LM2596 can be purchased online, and **broken link removed** are also available at a low cost.
 

ArticCynda said:
The power you're dissipating into the LM7805 is probably too high, forcing the IC into a thermal shutdown mode and causing it to throttle down the output current.

The LM2596 can be purchased online, and **broken link removed** are also available at a low cost.
Click to expand...

I touch it. It is not hot.
 

Strange. Perhaps a feedback/stability issue? Have you included an input and output capacitor in your design?
 

Do you have another LM7805 available? The only other cause of the problem I can think of is that the regulator is either defective or wrong connected in the circuit.
 

If you connect a 10 volt at the input, and you get 4.5v at the output (Presuming your meter to be perfect) your 7805 is faulty OR the Input drops below 6 volts. If you are sticking to perfect 1 ampere output, first confrm the input source has a amphere capacity more than you need in the output. Next you may get a 7805K which is a TO3 metal package with 3 amps capacity. On the cheaper alternative you may fix 2 nos 7805 TO220 packages in parellel which should deliver a stable 1Amp current.
 

pranam77 said:
If you connect a 10 volt at the input, and you get 4.5v at the output (Presuming your meter to be perfect) your 7805 is faulty OR the Input drops below 6 volts. If you are sticking to perfect 1 ampere output, first confrm the input source has a amphere capacity more than you need in the output. Next you may get a 7805K which is a TO3 metal package with 3 amps capacity. On the cheaper alternative you may fix 2 nos 7805 TO220 packages in parellel which should deliver a stable 1Amp current.
Click to expand...

The current at this stage (because I haven't connect all the ICs ) is around 350mA. The problem is that even with a voltage supply of 5Volt (without the LM7805), I meter 4,23-4,5 Volt! In the circuit I use LD1117 which takes the 5 Volts and gives 3.3 Volts as output. On the input of LD1117 I use 100nF ceramic capacitor and on the output 10uF electrolytic and in parallel 0.1uF electrolytic. The 0.1 uF it is not on the schematic of the datasheet. I put it because I was advised. Could this cause any problems to the voltage of 4.5 Volts (instead of having 5 Volts)??
 

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In the case of LD1117 there are two versions. One is for 3.3 volts and other for 5volts. In both cases it will require at least 8 volts in the input side to work efficiently. 104 or 100nf has nothing to do with the voltage drop unless it has a leakage. You are using LD1117 but please let me know what is written except that on the device. 3.3 or 5.0?
Also let me know what voltage is your circuit built for? 3.3 or 5 volts?
 

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