I bought 2 or 3 of these http://www.sparkfun.com/commerce/pro...oducts_id=7816 with different frequencies. They work flawlessly on the mini PCB I made when powered by my power supply but nothing at all when powered by a 9 volt. I made a pcb with a receiver of one frequency and a transmitter of another frequency for 2 way communication. each receiver/transmitter pair is powered by a single 7805...Im not an expert reading data sheets so maybe someone can tell me whats wrong. Im assuming the 9 volt is incapable of powering a 7805 but maybe im wrong...please church me up if possible THANKS

 

Yes, a 9V batter will struggle to power an LM7805 for long because it requires an input voltage of at least 7.5V and the battery's voltage will quickly drop below that.

Use a low dropout reuglator like the LM2931A 5.0 which will work down to 5.6V at which point the battery will be totally dead.

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You could also use a 7 cell pack of AA batteries in series. That will give you 10.5 volts and plenty of voltage overhead. A 9 volt R/C car battery pack will also do you just fine (and is pretty cheap) if you don't want to build your own pack.

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To understand the problem, first estimate what your load is. The current draw for the receiver is only about 3 mA or so at 5V according to the receiver data sheet. The transmitter takes about 22 mA at 5V according to its data sheet. So the regulator must provide 25 mA at 5V to operate one of each at the same time. Assuming the 7805 is operating normally it will need some current to operate, about 3.2 mA typically for an MC7805 (from OnSemi). So we are pulling about 28.2 mA from the battery.

Next, have a look at the specs for a typical 9V alkaline battery:
http://data.energizer.com/PDFs/522.pdf
scroll down and you will see the characteristic of output voltage vs time for the TOY load at 270 ohms which is quite close to your circuit resistance. As you can see, the voltage drops from 9V fairly quickly and gets to about 7 volts in 15 hours, assuming you are starting with a new fresh battery. The data sheet for the MC7805 shows, on page 22, that the regulator stops regulating when the input to output voltage differential drops to about 1.6 volts, when you are drawing less than 200 mA (with an IC temperature of 40 deg). So, your regulator should work fine until the battery voltage comes down to 6.6 volts. This implies you should get about 17 hours of operation.

So, on the surface, it should work if you don't have any circuit faults and you are using a good fresh battery. That the exact same pc board works ok with an external supply rules out most circuit faults. I wonder what the actual current drain is. Did you measure it?

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RadioRon

 

Ummm. Should't the tx and rx both be the same freq?

Edit: Scratch that. I see wht you're saying.

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Another great regulator for most any 5V battery application is the LP2954AIT, which has comparable Vdropout of the above unit (300mV at 100mA load), but also only a third the quiescent current, and an Imax of 250mA instead of 100mA. I've used these many times and they're great.

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Thanks for all the input...I guess I failed to mention the other components assosiated with this project there are also two 8 pin pic's which encode/decode 4 bits. I will have to find out which pic's they are to figure out what current load they have. Thanks for all the help...I will try some of your idea's and keep you updated. THANKS AGAIN!

 

I can't find a datasheet with that information on.

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The NSC datasheet for the LM7805CT I'm looking at shows roughly the same as what Radio Ron's datasheet shows: Vdropout of 1.5V at 250mA at 25C:

http://www.national.com/ds/LM/LM7512C.pdf

(Don't mind the 7512 part number, the 7805 datasheet indeed comes up. The plot in question is on page 6.)

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Here's the data sheet from On Semi. Look at Figure 14 on page 22.

http://www.onsemi.com/pub_link/Collateral/MC7800-D.PDF

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RadioRon

 

Datasheets for the 7805 have it spec'd with a minimum input voltage of 7V even if its load is olly 5mA.
You cannot rely on "typical" numbers because you don't know if your device is a weak but still passing one. You can "gamble" or "hope" that your device is typical.
Or design the circuit with the worst case numbers so that every passing IC will work perfectly.

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Uncle $crooge

 

Not quite. Both referenced datasheets (On Semi and NSC) specify a Vdropout of 7V and 7.5V typical, respectively, at 1A.

Yep.

Checking the parts by hand is always useful, just like going through a bin of 12k resistors to find one that's really 12k. It's tedious but I'm sure many of us have done that many, many times.

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Passing parts are a lot worse than typical.
When a regulator is in dropout then it is not regulating.
I was talking about a regulator meeting its specs and regulating properly.

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Uncle $crooge

 

Right AG, I agree with you. That's why I added the statement about hand sorting parts. If I can find a reg that has a lower Vdropout than the rest in the bin, by all means, I might need to use it. Typical values are hopeful, as you say, and the best way to go is with worst case; we're on the same page.

(My only objection was, well, the part I objected to. The two datasheets only specified *typical* Vdropout, not *min,* and at 1A, as opposed to 5mA, as you stated. It would have been nice if they spec'ed min, but they didn't. That was my point, and why I went to the device curves, because they showed some actual data. I presume the curves represent behavior over some considerable sample size, which is a good place to start...)

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