Question before I try this

Thread Starter

diggerdugger2022

Joined Feb 6, 2024
11
Hello all, I am still learning here so bad with me.

I would like to use the following laptop adapter as a breadboard power supply for multiple projects for testing. The adapter is 19.5v 3.33A 65W.
I removed a 5v regulator from another board that I tested with a multimeter. I have 1000uf 25v capacitors and some 3v voltage regulators alond with some 100uf ceramic capacitors. If i connect the adapter to the breadboard and use the 5v regulator with 1 x 1000uf cap and a 100uf cap, would it be safe for a 5v device?
 

WBahn

Joined Mar 31, 2012
30,065
Depends on a lot of details you haven't provided.

What is the 5 V regulator? If it's a regulator that isn't designed to have an input voltage of 19.5 V, it's not going to be safe.

If you are trying to power something that is going to draw 3.33 A, that poor regulator is going to be dissipating nearly 50 W, which it is unlikely to be happy about.
 

Thread Starter

diggerdugger2022

Joined Feb 6, 2024
11
Depends on a lot of details you haven't provided.

What is the 5 V regulator? If it's a regulator that isn't designed to have an input voltage of 19.5 V, it's not going to be safe.

If you are trying to power something that is going to draw 3.33 A, that poor regulator is going to be dissipating nearly 50 W, which it is unlikely to be happy about.
1708723108307.png this is the 5v voltage regultor. I wanted to use the 19.5v laptop charger as a power source for testing 5v microcontrollers with sensors on a breadboard. I wanted to use the voltage regulator to only grant the device I am trying to power 5v.
 

WBahn

Joined Mar 31, 2012
30,065
View attachment 316011 this is the 5v voltage regultor. I wanted to use the 19.5v laptop charger as a power source for testing 5v microcontrollers with sensors on a breadboard. I wanted to use the voltage regulator to only grant the device I am trying to power 5v.
That's just a generic 3D rendering of a particular case style (TO-220, if I'm not mistaken).

DigiKey is currently showing 90 active 5 V regulators in a TO-220 package with maximum input voltages ranging from 6 V to 40 V.

What is the part number on the actual part you want to use?
 

Thread Starter

diggerdugger2022

Joined Feb 6, 2024
11
That's just a generic 3D rendering of a particular case style (TO-220, if I'm not mistaken).

DigiKey is currently showing 90 active 5 V regulators in a TO-220 package with maximum input voltages ranging from 6 V to 40 V.

What is the part number on the actual part you want to use?
I can't read it. I tried to zoom in on it but it is unlegiable. I measured it with a multimeter and got 5.5v. It has 6a stamped on it as well.
 

KeithWalker

Joined Jul 10, 2017
3,096
If that is a LM7805 5V regulator then it should be good for circuits that use less than an amp. It is really OK up to 1.5A but it will need to be mounted on a good heatsink to dissipate the heat generated. Don't connect a 1,000uF capacitor across the 5V output to the circuits. On power up, the charge current will probably activate the regulator's over-current protection.
Here is the recommended circuit from the LM7805 datasheet:
1708725753436.png
 

MrChips

Joined Oct 2, 2009
30,819
Think about power dissipation.
If the power supply output is 20V, then the 5V linear voltage regulator has to dissipate 15V x ?A.
The regulator will get VERY HOT and likely shut down by the thermal protection circuit.
 

MisterBill2

Joined Jan 23, 2018
18,538
Both the power rating and those capacitors are very important. You can use a series resistor to reduce the heat in the regulator, possibly quite a bit. BUT then the capacitors on the input side become even more important. Keep the 0.33 cap, ADD a 0.1 Mfdcap, and one of those 100Mfd caps across the input. If you drop 10 volts across the resistor that will leave 9 volts for the regulator and that will be good.
 

Jon Chandler

Joined Jun 12, 2008
1,051
For any and all linear regulators (this applies to ALL variants of LINEAR REGULATORS): POWER DISSIPATION is the key.

Pd = (Vin – Vout) × I,

Where

Pd = power dissipation in watts

Vin = regulator input voltage in volts

Vout = regulator output voltage in volts

I = output current in amps.

This power dissipation is released as heat.

So let's say you want 500mA output current.

Pd = (19.5 – 5) x 0.5 = 7.5 WATTS dissipated by the voltage regulator. Ever burnt your finger on an incandescent night light bulb? That's 4 watts. 7.5 watts is a lot of heat to radiate from a TO-220 package.

A voltage regulator is rated to supply some maximum current. But the ultimate limitation is the power dissipation of the regulator package. For a small difference between input and output voltage, the regulator can deliver the max rated current. As the voltage difference increases, power dissipation is the limiting factor.

Solutions:

1. Reduce input voltage.

2. Add heatsinks.

3. Adding series resistors or diodes to the input will reduce power dissipation of the regulator by spending it out across the resistors or diodes....but in total, the same amount of power is released as heat.
 

WBahn

Joined Mar 31, 2012
30,065
I can't read it. I tried to zoom in on it but it is unlegiable. I measured it with a multimeter and got 5.5v. It has 6a stamped on it as well.
If it's outputting 5.5 V, then don't use it -- it is almost certainly way out of spec, indicating that it's either damaged, or possibly a counterfeit part to begin with. There's no telling what the 6a means.

Considering that a brand new regular, even at DigiKey's exorbitant onsey prices, is less than 50 cents, you've already spent a lot more time trying to salvage a part than it's worth. Admittedly, buying a single part from DigiKey is also going to expose you to shipping costs (~$5), but that still makes it hard to justify more than a fraction of an hour.

Try to find an electronics store in your area, particularly a surplus place. They are getting few and far between in the U.S., but some do still exist.
 
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