hi
I am after using a 9v battery to power a digital circuit that needs only 5v
My supplier is out of 9v to 5v regulators

I had a couple of questions

1. Can i accomplish the same regulation using resistors and capacitors? Is it better to use a voltage regulator?

2.It seems that there are lots of 5v regulators but they are minimum 10v. I presume this is no good if I am using a 9v battery, or is it acceptable?

There seem to be quite a few of these types
I wonder if this would be ok
http://cpc.farnell.com/texas-instruments/lm2594n-5-0/switching-reg-0-5a-5-0v-2594-dip8/dp/SC08553

Thanks

 

Whats your location.?

There are lots of regulators that will work with a 9V battery and give 5V output.

What is the 5V current load.?

E

 

To be totally honest, impatient as I am, I have not got really paid lots of attention to the current, I was just buying parts off a crib sheet and was hoping to learn about the current as I got started. My location is UK

There is a mention in document about the circuit consuming 60mA at 5.0Volt

 

hi dux,

This supplier is a low cost source, I buy from them sometimes.
http://www.bitsbox.co.uk/

http://www.bitsbox.co.uk/index.php?main_page=index&cPath=140_171

E

 

Thanks I will look on here.

I was reading an article where the chap used a
LM7805CT
whilst using 9v dc battery

Is that ok?

 

A 7805 will work down to +7.5V input for currents upto ~ 250mA

Always add a 100nF cap on both the input and output of the 7805.

 

Just wondered why capacitors are needed either side?

 

How much current does your circuit require?
If the current requirement is low enough you can use a 78L05 or even a 5.1V zener diode (with a current limiting resistor).

Note that the LM7805CT while readily available and inexpensive, it is not the only solution.

Voltage regulators are servo feedback mechanisms. If you get a perturbation it is possible that the regulator could oscillate. Sometimes the datasheets will call for a 1 to 10μF tantalum electrolytic capacitor on the output.

 

One more great question

http://www.fairchildsemi.com/ds/LM/LM7805.pdf

when I look here I see it says input voltage as low as 5v

Am I reading that correctly?

Where would I find a datasheet that might state capacitance?

Thanks

Just trying to educate myself so I don't have to keep asking the same questions

 

You are not reading the specs correctly.

It says typical dropout voltage is 2V. Hence the input voltage > 7V.

 

Where would I find a datasheet that might state capacitance?

The datasheet you linked to in post #9! See Fig 9, page 19.

 

Ok Thanks

I have (a little) more understanding about dropout voltage now.

but....

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I wonder why the datasheet has

Input Voltage​

VO = 5 V to 18 V
mentioned if the output is 5v and the dropout is 2v

Unless the output can possibly be 3v at anytime

 

The datasheet is for the LM78XX regulator series.
It comes in variants with output voltages of 5, 6, 8, 9, 10, 12, 15, 18 and 24 volts

The data sheets shows the maximum input voltage for regulators with outputs from 5-18 volts is 35 volts, and for the 24 volt type, 40 volts.

For any of them the dropout voltage is 2 volts, so the input must be at least 2 volts more than the output and never more than the maximum input voltage.

For the LM7805 that is an input voltage in the range 7-35 volts

 

Of course, you know that three AA or AAA batteries would supply a good working voltage with no other components, and would last many times longer than a 9V battery.

http://forum.allaboutcircuits.com/showthread.php?t=46378&highlight=holder

 

I wonder why the datasheet has "Input VoltageVO = 5 V to 18 V " mentioned if the output is 5v and the dropout is 2v

You can input 5V if you want, but that won't give 5V out.
Here's a sim of output versus input: