How to decrease the output current and voltage from this DC to DC converter?

[Willen]

OK then all can feel happy with electronics because they all are creative!! Because they are engineers and electricians yes! I feel these advanced parts of the world like a heaven for all. But 'money' is key thing there to get all.

AG, good news. I believe that happiness is first medicine for us.

While talking about happiness, I remembered an old 'misunderstanding' from a very old people of here towards American white people. It was so funny so laughed lot yesterday---

More than 100 years ago there were no advanced technology around the world and no more communication. Education presented here very late around 20th century. That time just few white Americans had here for research. Here most of the peoples were uneducated. So they used to talk about this Americans like this way (I heard from an old person)- "Very years ago their parents were big giants and used to live inside (down) of the earth. So now they have a tail at back of their body near head. They are so much rich so they drink milk always instead of water so they are amazingly white." Hehe how narrow world was exist!

And now we feel entire world as a room!! All credits goes to electronics frist!!!! Want to salute those inventors!!!

 

[audioguru]

I used to drink a lot of beer and rum but my doctor and my bank said it was bad for me. Now I drink a lot of coffee, hot chocolate and chocolate milk but my skin is not brown. Sunshine in summer makes my skin red then it turns brown for a couple of months.

Some people around the world have more money than brains. They own cars like Lamborghini, Ferrari, Bentley and Rolls Royce that are extremely expensive then they rust away in a few years.
Some people travel around the world in their own jet airplanes.

 

[Willen]

Here I found an Chinese Li-ion cell charger circuit. I made this circuit by seeing the real charger device, hope I didn't make any mistake. Here LED 1 blinks so fast while battery is charging, How it blinks?

LED2 is a simplle power LED to show ON state. How amazing bias around LM358, is it possible? Another amazing is how battery gets negative supply? (there were in PCB marking same 'B-' for Pin1 and 'B+' for pin7 I did same in schematic.

 

[audioguru]

You forgot to post the schematic.
The datasheet for the LM358 shows that pin 1 and pin 7 are the outputs of its two opamps.

 

[Willen]

Yes! You became so fast than my upload hehe Now I edited the post by adding a schematic (see my previous post).

 

[audioguru]

The top opamp will do nothing with its input pin 2 connected to nothing. Its input pin 3 powers the green LED with such a small current that this LED will not be seen. The green LED might be a poor attempt to make a 4.20V voltage reference with a different voltage for each different green LED.
The lower opamp will blow up since its input pin 6 is more negative than its pin 4. The lower opamp has the entire power supply voltage between its inputs so it will limit the current.

 

[MrAl]

Hi,

It looks like there is something wrong with that circuit drawing. For one thing the op amp is missing one input. That means there is probably other problems too with the drawing. Maybe check over the circuit one more time with a good magnifying glass. It also helps to take pictures of the board top and bottom, then look at those to try to determine the correct connections.

 

[Ashish Bhandari]

hi, i have been trying to charge my cell phone using a solar powered battery...its output is 12 v and 5A ...i used 7805 to reduce the voltage to 5 v but dnt knw how to reduce the current to 500mA.. can u help me???

 

[audioguru]

The charger circuit in your phone will take only as much current as it needs. It will probably take less than 500mA.
As the battery reaches full charge the current continues to drop until it is very low then the charging stops and the charging circuit will take NO current from your 7805 regulator.

 

[Willen]

hi, i have been trying to charge my cell phone using a solar powered battery...its output is 12 v and 5A ...i used 7805 to reduce the voltage to 5 v but dnt knw how to reduce the current to 500mA.. can u help me???

If you did it then 7805 will dissipate 3.5watts which produce more than 200 degrees celcious temperature which might burn your finger. May be you need extremely large good heatsink. But there are lots of 'DC Charger 12V in to 5Vout' found in Kathmandu to charge cell phone without wasting such lots of powers (heat).

 

[MrAl]

Hello,

Yes so the 7805 will require a heat sink. 3.5 watts isnt that much when there is a heat sink so it should work. But without the heatsink it's goodbye 7805 hello smoke.

 

[Ashish Bhandari]

Thank u guyz so much!!!

 

[Willen]

Hi MrAl,

Hope you are fine. Today as an experiment I made little silly modification to your Li-ion charger. I made it 'Auto termination charger' hehe but I know nothing what I am doing. I think my mistake is 'proper base current for Q2 BD135'. Um... As a whole what it is? (Ops! 1Meg resistor on comparator connected mistakenly, please forget about it for now.)

 

[MrAl]

Hi Willen,

That's an interesting idea there. So you want to drive the opto coupler with the LED indicator output, which then drives the transistor which turns the charging on or off.

First problem is that the saturation voltage of the Q2 transistor may not be as constant as we would like as it heats up and cools down. That changes the voltage setting at the "4.15v" test points. You'll have to check that yourself. If it does not change too much or changes in a favorable way then it's ok, but also watch it over ambient temperature changes too and see if it is ok.
There might be a way we can work it's voltage into the regulation part of the existing circuit but it will take a little thought, or we might use a MOSFET instead of a bipolar.

Second problem is that the drive for the opto coupler LED is in parallel with the other LED, and that is not good. To drive the opto LED you might be able to place it in series with the other LED.

Another problem may be that the opto can not drive the 2.2k resistor properly, so a two transistor driver might be needed. That's not hard to do though.

Pretty good idea.

 

[Willen]

...Another problem may be that the opto can not drive the 2.2k resistor properly, so a two transistor driver might be needed.

Hi MrAl,
You know what I am trying to learn about electronics. I am interested on a sentence quoted above. What is mean by 'opto cannot drive 2.2k resistor'? Transistor inside the opto will not get proper current or will short or what? I am sure it's VERY important bias consideration for each circuit.

 

[MrAl]

Hi Willen,

Opto couplers have a certain spec called the "current transfer ratio" or CTR for short. It expresses the maximum output current with respect to the current used to drive the LED in the opto. Often this spec is given in percent.

For example, with a CTR of 100 percent if we drive the LED with 1ma then we expect to get 1ma at the collector of the transistor (roughly). If we drive the LED with 10ma then we can expect to get 10ma collector current.
With a CTR of 200 percent if we drive the LED with 1ma we can expect to get 2ma collector current.
With a CTR of only 50 percent however we need twice as much input current as what we need on the output, so 1ma output would required 2ma input.

So all you need to do is see if the current through the LED is enough to drive the collector resistor in a manner needed for the application. If it can then all is fine, but if not you need to use a higher input current or a different opto.
Higher input current means faster LED light output degradation over time, like a year or two, so it is best to stay as low as possible on the LED drive current. 5ma is a good place to run it (for a 20ma LED) and 10ma might not be too bad either.

Base biasing also effects the CTR though. With no base biasing the CTR is highest, but the switching speed is the slowest. If you dont need high speed switching though then you might not need a base bias resistor.

 

[Willen]

Hi,
It's nice but again, If opto has 10mA collector current (200% CTR/ input 5mA), then which value resistor needed as a collector resistor? (I mean how to calculate?) My main problem is I know every Ohm's formulae and can calculate but I cannot apply it to a circuit.

 

[MrAl]

Hi,

Oh i see, ok.

To calculate the resistor value when the resistor goes to +Vcc and to the transistor collector and the emitter goes to ground, when the transistor is 'on' we figure it as a short circuit for simplicity, which leads to Ohm's Law:
R=Vcc/I

So if the output has to supply 10ma with a 10v supply then the resistor value is 1k. Have to be a little careful though to make sure the opto can really pull the load close to ground if that is needed. 2k would be better for example unless you really need 10ma.

 

[Willen]

Hi,

That's pretty good really. Just one small thing, notice the other circuit had a 1k resistor on the output of the reference. So you have to move the 10uf cap to the left then install a 1k resistor from the top of the 10uf to the 1M resistor and non inverting input so it is just like the other circuit.

The 4.15v measurement is at point A common to the battery negative terminal ( -Battery ).
On the new schematic below, the 4.15v is between points A and B, and is measured when the battery is not yet fully charged.
Of course if you set that to 4.15v then you must set the other voltage setting to charge the battery to around 4.16v or the voltage monitor LED may never turn on.

You may not need the whole 10v, that's just there in order to have some value shown. You might be able to use 8v or 9v, but if the regulator gets hot then use a heat sink. The second LM317 can be a regular type in the original circuit but since you now use a TL431 you dont need it. You never mentioned that you also have TL431's on hand too.

Hi,

Now I got lot more dual comparator LM393, so can I use this one chip instead of LM358 here without any bias change? Or which thing do I need to change please.

 

[MrAl]

Hi,

It's been a while since we talked in this thread. What circuit is this you are using?

Comparators can often work as op amps, but their outputs are different than op amps and their spec's may not be as good for some things.

The output of the comparators often need a pullup resistor while the op amps do not, for one thing.