Diode voltage drop

Link to your LM2596 module? Most of those small switching modules handle around 10W - 15W, over 10W use heatsinks. They will handle a max of 3A @ 5v or less.

http://www.aliexpress.com/item/Free-Shipping-20PCS-GW1584-Ultra-small-size-DC-DC-step-down-power-supply-module-3A-adjustable/1819402588.html

Link to your LM2596 module?

|500x388

What kind of battery were you charging?

LiPo 3S 2.2A

3 lipo cells charged @ 2,2A gives 27,72W. If you set the step down to 12V this means you have a current of 2,31A not including losses!. If i include losses and say the charger is +-80% efficient your up to 35W so @ 12V gives you 2,9A, very close to the max of the step down.

If you go to 18V output of the step down you only have 1,54A without or 1,94A with losses.

And now about the losses in the step down (chip only, losses in the coil not taken into account). At an input of 22V and an output of 12V to 18V it's 90% efficient max under full load. If it has to supply 35W it is still burning about 4W in the converter already which can get hot.

So set the step down to 18V and indeed cool the step down. Also don't charge faster then 2,2A or more then 3 LiPo cells with this setup because the step down is just to light for that.

So set the step down to 18V and indeed cool the step down. Also don't charge faster then 2,2A or more then 3 LiPo cells with this setup because the step down is just to light for that.

I charge the LiPo at 1C so it's normal for the lipo and B6 about the step down "let it burn" :D I'm joking, it's working fine, yes it heats up a lot but with a 5cm x 5cm fun it gets in a normal temperature (under 40C`).

btw the step down i got it at +12v or i can put it 12~18v i don't really care :D it works.

Yeay, I know it's normal. But that doesn't mean you don't blow the step down if you charge faster/more. @18V the step down will be cooler / capable of charging more LiPo's etc

Its an emitter-follower driven from a zener, so its basically a high-power zener in function.

Of course you have to waste all the power, whereas using the right power supply in the first place would avoid that.

Of course you have to waste all the power, whereas using the right power supply in the first place would avoid that.

I got as well a 12v 800mA power supply which i charge my battery drill but when i tested the output of that power supply i got over 19 volts so i can't understand why the hack do that.

is there any other kind of diode that i can drop it more but with a single diode not more ?

Well, not to state the obvious, but if it dropped 7V instead of 0.7, it wouldn't be a diode drop now would it ?

FYI, any linear power supply is going to consist of an input voltage and a regulator circuit. You already have your input voltage (19V). All you need is the regulator circuit. 12V regulator circuit

Of course if that's too complicated you can go with the 10 diodes in series...

You already have your input voltage (19V). All you need is the regulator circuit.

If you look few comments above you will see that i got one :D

Those LM2596 modules are cheap enough that you can put two of them in parallel. to eliminate your overheating problem.

Somebody is now going to prairie dog and say that is a bad idea!

Obviously paralleling regulated p.s. requires that BOTH P.S. be calibrated for the SAME output voltage. Other than that, there is no limit to how many p.s. you may parallel. It is best if they are identicsl make & models but that isn't mandatory. There may be some residual current differences but nothing worth worrying about. If you have any doubts about it yiu can try it yourself. No P.S. knows tge ithers exidt .It adjusts it's output until it sees the setpoint. It is not capable of outputting a negstive voltage so all it can do is turn maintain the setpoint by toggling the output device.

@Mark, you need to do some reading. The module he is using is a switch mode step down module. It's not linear nor a zener diode. And even a switch mode step down gets hot with this drop and current. Doing it linear would be a disaster....

And you, two (non fixed output) steps downs in parallel is a bad idea or at least not working that great... If the set voltage of one is set a little bit higher then the other the first one will do all the work and the second almost non.

But if you modify the modules so you use one potentiometer to adjust both it can work. Just desolder the pot on one and find out which pin goes off to the feedback (hint, it's pin4) and connect the feedback pins of both modules.

If the set voltage of one is set a little bit higher then the other the first one will do all the work and the second almost non.

That’s right. It’s only true if there is a large discrepancy in the setpoints because BOTH P.S. will SHARE the load if both see a difference between output voltage and setpoint. Using only one pot is a smart approach though

If you question the fact that paralleled P.S. with similar setpoints will share the load,
insert DMMs in current mode in series with each P.S. and you will see that BOTH have the same current until you get to the voltage range that corresponds with the difference in setpoints. If one has a setpoint of 11.80 V and the other a setpoint of 12.25 V then the former will basically turn off it’s output at 11.8V and the later will be the only P.S. delivering current in the range between 12.25 V and 11.80 V. Below 11.8V, both supplies will be at work and above 12.25, only the one.

Those LM2596 modules are cheap enough that you can put two of them in parallel. to eliminate your overheating problem.

It’s working really great as it is, i just added a 8 x 8cm fun on the side and it’s cooling down really well/fast.

Why would i need 2 of them to drop down the voltage if they work really good as it is.


As you see they can handle the voltage and the current I need, If i will need to charge more than 3A then that will be a problem. But right now I don’t charge more than 2A it’s getting hot but as i said the fun do great job, cooling it down from 80C` to room temperature.

Your call.

I got as well a 12v 800mA power supply which i charge my battery drill but when i tested the output of that power supply i got over 19 volts so i can't understand why the hack do that.

That's the "No Load" voltage.

The battery drill represents an 8.75 ohm load.

(19V -12V)/0.800 A = 8.75 ohms

septillion: @Mark, you need to do some reading. The module he is using is a switch mode step down module. It's not linear nor a zener diode. And even a switch mode step down gets hot with this drop and current. Doing it linear would be a disaster....

The OP asked for a way to drop many volts, I pointed out a common circuit for this function, so what's the issue? You can drop 7V from any voltage source, it doesn't matter what it is internally. I said it will dissipate a lot of power.

As for reading does The Art of Electronics editions 1, 2 and 3 count?

raschemmel:
Your call.

That’s the “No Load” voltage.

The battery drill represents an 8.75 ohm load.

(19V -12V)/0.800 A = 8.75 ohms

No… that would be the internal resistance of the power supply.

The drill charging resistance, if it were using 800mA while at 12V, would be 12/0.8 = 15 ohms

Of course, all this is an extreme oversimplification…

No... that would be the internal resistance of the power supply.

The drill charging resistance, if it were using 800mA while at 12V, would be 12/0.8 = 15 ohms

Yes of course you are correct. Actually that's what I meant but I misstated it. Thanks for the correction.

Of course, all this is an extreme oversimplification...

Yes, maybe so, but that's probably appropriate for this thread.

:)