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Voltage Regulator

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Todd

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I am using a LM338 to make a voltage regulator, this regulator is of the variable type.

The Input of the LM338 has the incoming voltage to be regulated and a 10uF capacitor going to ground. There is a small resistor from the Output of the LM338 to the Adj pin of the LM338. There is also a 100uF cap from the Output to ground. A 5K Reostat is attached to the Adj pin to ground, this allows me to adjust the output voltage to what ever I need.

Okay here is my problem, when I put a load on the output of the voltage regulator it causes my output voltage to drop drastically. I need to achieve 3.5 Volts and a maximum of 3.5 Amps from the voltage regulator, so this would mean that the load would have to be 1 Ohm. When I placed a 1 ohm resistor from the output of the LM338 to ground the volatage across the 1 ohm resistor dropped to .7V instead of 3.5V

Any suggestions on how to fix my problem are greatly welcome. Thankyou.

Todd
 
LM338 regulatro

The national data book recommends putting the resistor between the output and the adjust terminal right at the regulator pins. The also recommend a diode connected from the ouput back to the input. That is the anode on the output and the cathode on the input of the regulator.
Another thing that I have learned about the 338 is that you want to use a
1 point ground. That is to say the ground connection of the pot you are using to adjust the voltage needs to be at the load ground connection.
The data book I have lists the 138,238, and the 338. It says the resistor from the output to the adjust terminal on the 338 should be 120 ohms, but the standard of 240 ohms on the 138 and 238.
 
As i see, no problem with circuit. You wrote: 10uF cap on input, i hope this is not a filter cap for 3,5A? For 3,5V 3,5A need minimum 6V DC input, and 6800uF cap after bridge rectifier.
Check the input as Stevez and Nigel suggested...
 
You input source also needs to be able to supply the current you are trying to draw.........6volt at 2 amps just aint going to cut it.......you need a minimum of 4 amps....and put a heatsink on the device its going to get hot!
 
power supply

:D hi,

here's a simple rule when making a power supply and have its output regulated.

1. the voltage input to the fixed regulator should be higher by at least 2.5v to derive the rated voltage output of the regulator.
2. current need of the load should not be over the rated current output of the regulator and the transformer secondary
3. in case the current need is high the regulator should be properly heatsinked.

hope this helps :D
 
All of this is in case of linear regulator and all advices are right on.
You probably were trying to get 3-4 amp from the regulator which was
powered by cheap little wallwart. That's why you get only 0.7Amp.
Test the circuit with ca. 20-30 Ohm load first. See if it works etc.
an then try to get power from unit than provide it (4Amp sounds
very reasonable).
You CAN get higher output current than what source can provide
if the output voltage is much lower than input and you use switching
power supply.
This will also disipate much less heat and that's one of reasons
why switching power supplies have such popularity nowdays.

**broken link removed**
 
Guys!
Could I ask what is the difference between switching power supply and regualted power supply please?
could you provide some IC names or how the circuit is designed to be switch power supply which produces less heat?
Adnan
 
adnan_m_s said:
Guys!
Could I ask what is the difference between switching power supply and regualted power supply please?
could you provide some IC names or how the circuit is designed to be switch power supply which produces less heat?
Adnan

A simple way of thinking of a switching power supply is that it takes small bites of input power and integrates them (with a coil and a capacitor) into a constant voltage. Power disipation in this mode is very low as power is consumed only when taking a bite :)

Sorry... that analogy probably went too far. Suffice to say a switcher is active only part of the time, and hence the heat output is *greatly* reduced.

As an example, I recently needed 5v from a 21vdc source at just 200mA. A linear 7805 regulator gets very warm dissipating some 16v to get me 5v and a large heatsink was required despite the low current being drawn. I'm now using an LM2595T-5.0 from National Semiconductor and it requires no heatsink whatever and barely rises above room temp.

NatSemi has plenty of design tools on the web for helping you design a switchmode, and will even create a parts list for you, so they are *simple*!! My circuit has an LM2595, a schottky diode, an inductor, and a main capacitor, plus the usual assortment of small .1uF monolythic caps for removing high frequency switching noise from both the switcher and the other components.

Look here: **broken link removed** for the web tools to design a switching supply.

Switchmodes are also able to increase an input voltage to a higher one, albeit at lower current than the input side. A switchmode that drops voltage is called a "buck" convertor, one that increases the input voltage is called a "boost", and one that can do both (e.g. for running off batteries) is called a "buck/boost".

Hope that helps,
P.
 
The datasheet for the LM338 shows that it reduces its output current if there is more than 10V across it. So for an output of 3.5V at 5A then the input voltage must not exceed 13.5V.
 
High guys,
I just faced a streange thing,
I did a voltage regualtor using KIA78012 + a power pass transistor and it worked fine for 19 V In but for 30 Vin the regulator shut down so do you think I must keep to a voltage lower level? because I once read something like that somewhere??
 
adnan_m_s said:
High guys,
I just faced a streange thing,
I did a voltage regualtor using KIA78012 + a power pass transistor and it worked fine for 19 V In but for 30 Vin the regulator shut down so do you think I must keep to a voltage lower level? because I once read something like that somewhere??

At 30v in you're very close to the 35v specified max input for the 7812 regulator. Also, when you're dropping 18v (30-12) instead of 7 (19-12) you're going to be creating a lot more heat in the regulator. Finally, when driving a power transistor such as the 2n3055 it can require a drive current of up to 4A (!!) to drive it if you're trying to suck 15A out of it...

So was the input exactly 30v? How much current are you trying to suck from the power transistor? What's the application? You really need to consider a switchmode for what you're doing... even if it's just 3A or so given that you want 12v from 30v.

Hope this helps,
P.
 
because I tried it without a heat sink and it worked oK , it began to heat up but did not loose current , in addition I am using now two heat sinks taken from a PC case power supply
 
adnan_m_s said:
Well I am trying to drive 3A , but it turned off at 1.8 approximately ?
what do you think
I think the transistor was connected wrong and did nothing. Then the 7812 had 18v across it and 1.8A which is too much power dissipation so it got hot and shut down.
 
no you did not understand me, although getting hot it did not shut down,
it only shut down when i used the 30 V DC for IN !
 
adnan_m_s said:
no you did not understand me, although getting hot it did not shut down,
it only shut down when i used the 30 V DC for IN !
The regulator needs a heatsink when the input voltage is 30V.
A 7812 regulator can dissipate a max of only about 2W when it does not have a heatsink.

If the transistor is working properly and the circuit is designed properly then the regulator would get warm on a small heatsink and the transistor would be pretty hot on a big heatsink.
 
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