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4.5 V power supply using 7805 regulator - question

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Nepalien

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Hello all,

I have a short wave receiver that requires 4.5V power. Usually I use AC adapter but during power outage (which is frequent in Kathmandu), I use 3 D sized batteries. Current drain by the radio is around 120 mA for comfortable listening (AM mode with Sync detector ON). Listening to FM consumes around 100 mA. However it does vary but the maximum is not more than 150 mA.

I am thinking of using 12V and 7805 regulator to power this puppy. But I want to design the power supply that accomplish following things:
1. Voltage output should be 4.5V
2. Current should be limited to 150 mA
3. Any other protection that is required (for instance what if input polarity is reversed by mistake)

I have thought of putting a 5Ω 5W resistor to drop 0.5V at 100 mA in 7805 output. And for current protection, I want to put 200 mA fuse also. Finally I want to house this simple circuit in good old film roll plastic box with few holes for ventilation.

I would appreciate any inputs for this project. I really don't want to make one simple mistake and destroy the radio. Thanks for reading and Namaste from Nepal!
 
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Just use 5V it'll be fine. If you measure the output voltage of the mains adaptor you'll probably find it's more than 4.5V, I wouldn't be surprised if it's as high as 6V or more.

Using a resistor is a really bad idea because the current drawn continuously varies and can lead to instability.

If you're really paranoid, go for the LM317 with the component values listed below.
**broken link removed**
R1 = 150R
R2 = 390R
 
Agree with Hero.

However, if you really want to get 4,5V output from a 7805, you have to feed the center pin with -0,5V. Then you'll have a stable 4,5V supply. The center pin draws almost no current, so you could always use a battery with a simple voltage divider (just don't let the battery die).
 
A very easy way to get approximately 4.5V (e.g. 4.35V) from a 7805 is to put a diode in series with the 7805 output. There should be the normal capacitor on the 7805 output pin, plus another after the diode.

Beware a 7805 dropping (12-5=7)V at 150mA is going to be pretty warm.
 
Thanks for the tips so far.
Yes the AC adapter does produce around 5V. I guess it compensates when load is connected. With 7805, the radio will get 5V all the time. I guess it won't do too harm with 0.5V extra.
Regarding the diode, do I put it in output or input? I saw following thread (#9 comment) that indicated I should put it in input.
LINK
"Protecting against this can easily be done, as mentioned, with a 1N400X diode in the input circuit to the regulator (before the input caps, not between them and the 7805)"
 
Yes, the extra 0.5V won't hurt so 5V is find.

There are two reasons for adding a diode:

1) To provide reverse polarity protection, in case someone connects the battery backwards. In this case the diode needs to go in series with the input to the LM7805.

2) To prevent a large capacitor connected to the output from discharging back in to the regulator, if the input power supply is removed and short circuits, see the circuit on the bottom of page 8 of the datasheet linked below.
https://www.electro-tech-online.com/custompdfs/2010/03/LM7805.pdf

In this case I don't think there's any need for #2 but you might want #1, besides #1 will provide protection against #2.
 
4.5v

The three parameters were:
* 4.5V
* 100ma
* reverse polarity protection
The LM7805 will give stable voltage and a full bridge rectifier at the radio will provide the needed voltage drop and the reverse polarity protection (with the bridge rectifier right at the radio the POS & NEG can be hooked up either way and still work with no difference at the radio. Be sure that the rectifier is installed right at the radio!). There should be no need for the 100mw limit with a regulated supply as the load will not effect the voltage.
Abie
 
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