Best way to drop power supply output voltage by 1-2V

[shumifan50]

I have a 18.5V/6.5A power supply that I would like to use with a LiPo charger that needs 11-17V input voltage at max 50watt.
I was thinking of sticking a diode in the one lead to drop the voltage, but the ones that I have are rated at 1A (1N400x). Thing is at 5A and a 1V drop it needs to dissipate 5watt. Is there a better way to get a 1-2V drop on the output at 5A max. Typically I would imagine I would not charge above 1300ma, so I guess it would have to dissipate about 1.5watt and the 1N400x might do the job.
Any suggestions appreciated.

 

[Arouse1973]

I think unless you use a switching regulator your going to be stuck with getting rid of a lot of energy as heat. I think I would go for a diode on a heat sink it's a simple approach. Some of the other guys might have something better up their sleeves.
Adam

 

[shumifan50]

Thanks Arouse1973.

Hi people, anybody got some ideas, I dpn't want to blow up the LiPo charger.

I have tested the idea with 2 x 1N5004 diode and the drop is enough, but I have not actually hooked up a battery to charge yet. The current battery I need to charge is 1300mah so at 1C needs to be charged at 1300mA. Will the diodes cope without a heatsink or must I heatsink them - Y know I could just try but would like to know how to calculate how much heat it will have to dissipate.

Is there a better diode to use?
I saw some LMxxx regulators that can handle 5A. Would this be a better solution? I am trying to keep it simple and safe.

 

[Kiwi]

What about using the diode plate out of an old car alternator?

 

[Arouse1973]

The 1N5004 seems to be rated at 1A and you want 1.3A. I would look for another diode. The thermal resistance of the diode give you the rise in centigrade per watt. Times this by the watts and add ambient and that's how hot it will get. So say the diode had a 20deg/watt thermal resistance and you only wanted a 10 degrees centigrade rise then you would need (diode voltage drop at current say 0.8*1.3A=1.04W*20degrees=20.8+20 degrees centigrade ambient=40.8 degrees centigrade diode temp. If it needs a heat sink then choose one with say 10 degrees/watt thermal resistance and the temp will only rise 10 degrees + ambient. But all that said you should be good to run the diode at 40 degrees without a heat sink.
Adam

 

[shumifan50]

Thanks for the answers.
I went ahead and tried the diode. I charged the LiPo (11.1V, 1300mah) at 1300ma and it completed the charge cycle in about 30 minutes(it was only discharged to storage voltage). The diode got slightly warm, but not even hot, so I am still no closer to understanding heat dissipation.

As a note: It is a balancing charger, so presumably it does not charge at 1300ma all the time, which might explain why the diode in the input power supply does not get hot. I can now make it a more permanent installation.
I had a look inside the PS, but could see no easy way of dropping the PS output voltage, so I will just add this diode inside.

 

[Arouse1973]

That seems quick. At 0.7c which is recomended it should take about 2.5 hours from flat to charged. The first cc charge charges to about 55% and then the rest is cv for another 2hours ish until about 3% of the initial 0.7c current. I would say your battery was not flat and it was not putting 1.3amps in. Battery chargers that charge in 1 hour are not cherging the battery fully. About 75% is it for 1 hour charge.
Cheers
Adam

 

[shumifan50]

No the battery was not fully discharged: it was at storage voltage(new) which is about 3.85V per cell rather than 4.2V*charged) per cell or 3.3V per cell(discharged).

It is supposed to be quite a good charger: Turnighy acucell 6, which is a balance charger capable of charging Lithium, NiCad and LeadAcid batteries, depending on selection. It is microprocessor driven with a display for setup and status. You select number of cells(and it checks the balance plug), and charge rate and battery type. Note that this battery is a 1C charge rate, newer more expensive LiPos are up to 25C charge rate, they say.

I will see how it goes, but it is looking promising as this is the highest mah battery I want to charge. I will keep the diode outside until I have run down the battery and recharged it, monitoring the diode.

Thanks again for your input.