why does the voltage of 7805 drop?

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when i connected a led directly to the 7805 regulator the voltage output of the 7805 droped to 3.8v.
please explain .

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1. Which is the full code of the 7805?
2. Do you have any resistance in series with the LED?
3. Do you have connected all the capacitors needed (as the data sheet tells tou?

If all the above are ok, please take a measurement at - exactly - the pins GND and Vout of the 7805 and post again.

Michael.

Michael.

User of:
IAR Embedded Workbench C/C++ Compiler
Altium Designer

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Oh I forgot it,

If the power supply befor the 7805 is not stong enough and the LED needs current that produces a drop voltage (at the power supply) which means lower than 7,5V at the Vin pin of the 7805 then you will have problems like this.

Please check my two posts.

There is no magic.

Michael.

Michael.

User of:
IAR Embedded Workbench C/C++ Compiler
Altium Designer

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vicky3413 wrote:
when i connected a led directly to the 7805 regulator the voltage output of the 7805 droped to 3.8v.
please explain .

LEDs do not like 5V directly, they work with lower voltages. Read the Wikipedia article on LEDs.

Actually you should not put a voltage over a LED, you should think it like putting a current through a LED. So at least you need a series resistor to limit current to a safe value - You don't say if it is a regular LED or high power white LED or something.

Actually, there are LEDs that can be used with for example 5V directly. They just have the current limiting resistor built-in and are very expensive compared to normal LEDs.

- Jani

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Output voltage on 7805 drops for one of three reasons:

(1) Output current limiter has turned on because load current is too high

(2) Temperature limiter has turned on because the internal temperature is too high. This happens because of the product of load current and voltage difference (Input minus output) and insufficient heat sink

(3) Input voltage is too low. It must be at least 1.7V higher than the output voltage.

Jim

 

Until Black Lives Matter, we do not have "All Lives Matter"!

 

 

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Must be a VERY BRIGHT led with 1A. :?

John Samperi

Ampertronics Pty. Ltd.

www.ampertronics.com.au

* Electronic Design * Custom Products * Contract Assembly

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Can a schematic of your hookup be posted? Sounds like there could be a few reasons.

the other Jim

I would rather attempt something great and fail, than attempt nothing and succeed - Fortune Cookie

 

"The critical shortage here is not stuff, but time." - Johan Ekdahl

 

"Step N is required before you can do step N+1!" - ka7ehk

 

"If you want a career with a known path - become an undertaker. Dead people don't sue!" - Kartman

"Why is there a "Highway to Hell" and only a "Stairway to Heaven"? A prediction of the expected traffic load?"  - Lee "theusch"

 

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Atmel Studio6.2/AS7, DipTrace, Quartus, MPLAB, RSLogix user

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Hello to all,

Before starting this is my first post, just signed up :))) quite exciting!!

I'm currently having a similar issue with my circuit. Trying to make a 5V phone charger from a 12V Battery which is also controlled by a PIC and in order to make electronic switch i think i need to use a transistor. But when i try to connect the transistor through the  LM7805 Regulator my output voltage drops to 4.2V (I think it is because of the voltage divider) also the current is an another problem. In this case i have some questions to ask;

 

1-) Do i need exactly 5V as an output in order to charge a smartphone?
2-) How do i increase the amount of current? For more quick charging i think the current should be 1A referring to its adaptor :)
3-) How to choose the correct component (transistors, regulators) for these kind of circuits?

4-) How do i calculate the voltage through the transistor to determine the output voltage??

 

I'm on the simulation part of the project using Multisim and the circuit is below;

 

 

 

Many thanks,

Kaan

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Can you please explain more about what you're trying to do?
What is the purpose of controlling the charger with an MCU?

 

I suggest reading up on how transistors work.  You need to

limit the current going into the base or it will burn up. When

saturated, the voltage at the emitter will be about 0.7 volts

less than the base voltage due to the BE-junction diode which

explains why you're seeing 4.2V in the simulator.

 

--Mike

 

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Greetings and welcome to AVR Freaks -

 

Not sure what this has to do with AVRs, since you mention PIC. But, as far as your circuit is concerned, what point do you call "output"? If it is the emitter of the bipolar transistor, then I would expect about the voltage you quote. An NPN transistor that conducts will have its emitter voltage close to 0.7V less than the base voltage. Your base voltage is 5V, so I would expect the emitter to be about 4V3.

 

Be aware that, in a real circuit, your 7805 will get VERY hot at this current. Vin = 12V. Vout = 5V. So, Vdrop = Vin =- Vout = 7V. Power = Iload * Vdrop = 7W. That is a lot of power to dissipate in a 7805, even with a very good heat sink. You are lucky, at this point, because this is only a virtual circuit so you are only releasing virtual smoke. 

 

Most smart phones will probably work with 4V3 or 4V2 because the specified voltage range for USB is quite wide. On the need for a switch transistor, that is unclear to me, at this point. The smart charger inside the phone should take care of that. After all, you can leave it  plugged into USB for a long time after it is fully charged.

 

Jim

 

Until Black Lives Matter, we do not have "All Lives Matter"!

 

 

Last Edited: Thu. Sep 6, 2018 - 12:13 AM
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Welcome to the Forum.

 

Your questions are good ones, but you really need to take a course in electronics.

There are no short and simple answers to your questions on component selection.

 

The USB spec is 5V, (+0.25 V, - 0.6 V), IIRC.

Google and Wiki can help in this regard.

 

An option, of course, is to purchase a commercial car USB power adapter.

WalMart sells them for under $2 USD.

They are reasonably well designed, and include the voltage regulator and the female USB cord connector.

You can't make one for this low price.

 

You mentioned having a micro control the current / On/off of the charger?

If so, having a micro turn on and off a relay, to turn on/off power to the car charger might be an option.

 

Note that you don't have to plug a car charger into a cigarette lighter / accessory socket.

You could just solder your wires to the device's connectors.

 

JC

 

 

 

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In the U.S., you can get phone chargers that plug into a 12V "cigarette lighter" outlet on a vehicle. These chargers cost around $5, US. You cannot afford to even make your own at that price.

 

Jim

 

 

Until Black Lives Matter, we do not have "All Lives Matter"!

 

 

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Q2 is dropping 0.7 volts in a common-base configuration as shown.   In the real-world, there would never be a 6 ohm resistor connected to a 7805 voltage regulator, because it draw almost 1 amp.  The 7805 would get so hot that it would shut off. 

  In a real-world power supply, it would be better to use a $0.75 switching power-supply module board like https://www.ebay.com/itm/DC-DC-B... instead of a $0.75 linear regulator like the 7805.   The 7805 turns all the voltage between V input and 5 Volts into wasted heat, while the module board switches Vinput so fast that the output voltage never rises above +5V across the inductor that is located after the (internal) switching transistor.

 

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Whats the purpose of the 7905 (U1)?  THats useless in this circuit.

 

E.Kaan222 wrote:
which is also controlled by a PIC

You are in the wrong community for PIC's, even though the Evil Empire owns Atmel now.

 

Pic Questions are better serviced here:

http://www.microchip.com/forums/

 

General electronics questions are not a problem. smiley

 

Welcome to the Freak show

 

JIm

I would rather attempt something great and fail, than attempt nothing and succeed - Fortune Cookie

 

"The critical shortage here is not stuff, but time." - Johan Ekdahl

 

"Step N is required before you can do step N+1!" - ka7ehk

 

"If you want a career with a known path - become an undertaker. Dead people don't sue!" - Kartman

"Why is there a "Highway to Hell" and only a "Stairway to Heaven"? A prediction of the expected traffic load?"  - Lee "theusch"

 

Speak sweetly. It makes your words easier to digest when at a later date you have to eat them ;-)  - Source Unknown

Please Read: Code-of-Conduct

Atmel Studio6.2/AS7, DipTrace, Quartus, MPLAB, RSLogix user

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I suppose you want to switch the power on/off with Q2, in that case it should be between the load (your 6 ohm resistor) and ground, that is, a low side switch. And, as already mentioned, you can't just power the base directly, you need to put a base resistor there to limit current, or use a MOSFET instead (but remember bipolar transistors are in general more robust than MOSFETS).

Last Edited: Thu. Sep 6, 2018 - 08:10 AM
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Thanks to all for giving hand,

 

USB car charger is out of option because this is used with a battery in any case. So adding a renewable battery to the usb charger price will be above 5$ anyways.

This circuit is a part of an automation system. When a coin insterted into the system it allows the charger to charge a smartphone for a while. I thought to do that i do need to use a microcontroller just haven't decided which company's product to use.

Relay seems reasonable btw thanks.

 

In the circuit the Vcc represents an output pin of a microcontroller. 

The heat is a huge problem that i know as far as i already burned a regulator down :)) I thought using a good heatsink would solve the problem but as i understand it won't. 

So what about connecting two regulators in series as 12V to 9V then 9V to 5V? According to the equation it would reduce the power dissipation assuming 3W and 4W respectly to the regulators. Furthermore where to read the datasheet to know its limit for power dissipation??

 

Thanks,

Kaan

 

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E.Kaan222 wrote:
USB car charger is out of option because this is used with a battery in any case.

 

I interpret "USB car charger" as just a figure of speech, meaning you should use a switching regulator instead of a linear regulator. See the link on #13.

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    To OP: Your approach looks like 30y old. These days there are switching regulators with controlled limited output current, short circuit protection and an input to enable it or shut it down. Or you can read the output voltage with one ADC input from your PIC so you know what is going on. You can also use a switch specially designed for this purpose; look at this Digikey selection: https://www.digikey.com/products...

 

    Even if your switching regulator is short circuit protected, the output capacitors makes it to react slow, hence the switch IC mentioned before. Some of them are designed especially for USB applications. You can also measure (no need for high accuracy) the output current, so when the phone is done taking power you turn on an LED to show it.

 

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Why use a micro at all? A 555 will do what you want.

#1 Hardware Problem? https://www.avrfreaks.net/forum/...

#2 Hardware Problem? Read AVR042.

#3 All grounds are not created equal

#4 Have you proved your chip is running at xxMHz?

#5 "If you think you need floating point to solve the problem then you don't understand the problem. If you really do need floating point then you have a problem you do not understand."

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Brian Fairchild wrote:
Why use a micro at all? A 555 will do what you want.

Considering this looks the OP is trying to make a commercial product, a simple 8 pin tiny without any external parts would be cheaper than a 555 with passives.

Jim

I would rather attempt something great and fail, than attempt nothing and succeed - Fortune Cookie

 

"The critical shortage here is not stuff, but time." - Johan Ekdahl

 

"Step N is required before you can do step N+1!" - ka7ehk

 

"If you want a career with a known path - become an undertaker. Dead people don't sue!" - Kartman

"Why is there a "Highway to Hell" and only a "Stairway to Heaven"? A prediction of the expected traffic load?"  - Lee "theusch"

 

Speak sweetly. It makes your words easier to digest when at a later date you have to eat them ;-)  - Source Unknown

Please Read: Code-of-Conduct

Atmel Studio6.2/AS7, DipTrace, Quartus, MPLAB, RSLogix user