• Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

1,2v to 3v dc/dc converter

Status
Not open for further replies.

kalaman

New Member
hi,
i have charged one ni-mh battery and used 3v led. I'd like to make 1,2v to 3v dc/dc converter which is not include any IC. Because I couldn't find that type ICs where i live.

Actually i found a circuit and it worked nicely but it waste source when i don't connect load.
 

Hero999

Banned
That circuit is fine for powering LEDs, I don't understand why you should want anything better.

For a simpler circuit Google Joul thief.
 

kalaman

New Member
Firstly, thanks you for help.

It works properly but I measure 60mA standby situation. It spends 60mA when i disconnect LED. Is it normal ?

By the way my English so bad. I'm sorry.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
kalaman said:
Firstly, thanks you for help.

It works properly but I measure 60mA standby situation. It spends 60mA when i disconnect LED. Is it normal ?
Then don't disconnect the LED - and yes, I would imagine that is perfectly normal.
 

Hero999

Banned
Does the capacitro blow up when you remove the LED too?
 

Hero999

Banned
Yes, there's nothing to control the maximum output voltage from this thing so fi you're not careful the capacitor will explode as it's maximum voltage rating will be exceeded.
 

kalaman

New Member
Hero999 said:
Yes, there's nothing to control the maximum output voltage from this thing so fi you're not careful the capacitor will explode as it's maximum voltage rating will be exceeded.
Yes absolutely right. Because i remove the led electrolitic cap's voltage goes 11-12V.

Allright. this circuit eat 60mA from my poor battery without led. any idea alternative circuit?

thanks for replies
 

kalaman

New Member
Hi Nigel. You are angry with me. I'm so sorry. But I supposed that something is wrong in my circuit. Maybe there might be an alternative circuit.
Anyway thank you.
Best regards
 

Nigel Goodwin

Super Moderator
Most Helpful Member
kalaman said:
Hi Nigel. You are angry with me. I'm so sorry. But I supposed that something is wrong in my circuit. Maybe there might be an alternative circuit.
You still haven't answered the question?.

There's nothing wrong with the circuit, just don't disconnect the LED - and why would you want to?.
 

kalaman

New Member
My electronic knowledge is a little bad. So according me any circuit spends more watts when you use load. When you remove the load every circuit's standby current decreases. But this circuit surprised me in this case.

As a result i'm wrong. I will use this circuit with solar energy and any mA will be important for me.

This was just curiosity.

I won't remove the led forever. :)

Best regards
 

Hero999

Banned
I haven't analysed the circuit in detail but from a quick glance:

This circuit is a flyback step up converter.

It works by suddenly interrupting the current through an inductor so the large back EMF charges a capacitor.

When you disconnect the LED the current in the inductor decays faster so the gap between the inductor being turned on and off is shorter causing it to draw more current.
 

justDIY

Active Member
come on guys, lets not be too dense here

the OP talks about disconnecting the led. perhaps they mean they're turning off the led, upstream of the converter. not exactly a good solution, since it leaves the converter running.

I don't know if you're interested in making things more complex, but the TPS61040 from Texas Inst is a very small boost converter with some nice features. 28 uA quiescent current with no load and 1uA current in shutdown.

I built a flashlight based on one of these long ago, powering eight white leds (in series) off a pair of aaa batteries. My ammeter can't measure the current draw from the batteries when it is "off", it doesn't show up on the 2mA (lowest) scale.
 

iONic

Member
If you use the circuit switch to "turn off" the LED it will disconnect the battery from the rest of the circuit and NO current should be discarded.
 

audioguru

Well-Known Member
Most Helpful Member
Why do so many people who live on the moon come here asking for circuits then say they can't get ordinary common ICs on the moon???
Why don't they ask the other people on the moon???
 

Hero999

Banned
Looking back, I think we've all misunderstood what the origional posted wanted. I think we got tied to the idea of this circuit and the LED. I think, that all he wanted was a DC-DC converter to boost 1.2V to 3V without the LED. He mistakenly thought that he could just remove the LED from this circuit and it would work. Not that this matters now, the origional poster has gone but it's still a good idea to take not of this for future reference.
 

mvs sarma

Well-Known Member
kalaman said:
My electronic knowledge is a little bad. So according me any circuit spends more watts when you use load. When you remove the load every circuit's standby current decreases. But this circuit surprised me in this case.

As a result i'm wrong. I will use this circuit with solar energy and any mA will be important for me.

This was just curiosity.

I won't remove the led forever. :)

Best regards
Imagine - 'energy can neither be created .....'

100% efficiency doesn't exist. whether you load or dont load you are asking the circuit to do some work -- 'produce higher voltage from 1.5V cell.' YES it will at the cost of some current-- Nigel tried to explain the same and he said it is NORMAL.
you expect output voltage to remain ata point you imagine-- imean- without realizing a circuit element for your spec. naturall in the abscence of any feedback network from the output DC, the capacitoe will encounter more that the voltage you have planned for. some might get HOT some might EXPLODE.

IF YOU WANT TO MOD THE CIRCUIT FOR ANOTHER USE - NOT FOR THE LED PROVIDED BY DESIGNER-- YOU DESIGN PRIOPERLY WITH A DC FEEDBACK AND A NOMINAL DUMMY LOAD.

please don't feel some one is angry with you -- to be angry - as you know it costs us.
 
Status
Not open for further replies.

Latest threads

EE World Online Articles

Loading

 
Top