title was "High current PSU's and chargers using phone charger" caught by bug.

When building linear PSU's or chargers getting a suitable transformer can be difficult/expensive. Some expensive units have transformers with an extra winding for powering the stabilser circuit or simply have 16v or more secondaries to overcome the dropout. (12v case)

Today I just split open a switch mode, 4.9v phone charger and rescued a neat small pcb with an isolated output up to 500Ma. I'm thinking that if that was connected to add to the main rectified DC, it could supply an LM7812, which in turn feeds the multiple power (pass) transistors.

The question then is would a 12v transformer suffice ? I'm assuming capacitors @ 2000uf per amp and a bridge rectifier - which reduces the drop out to 2v total off the main DC.

 

How about some more words and a circuit or block diagram?

I'm unclear how a 4.9 volts supply could power a 7812?

 

I think he's talking about using a 12 volt rated transformer and stacking a 4.9 volt switching supply on it to run a 7812. This is because he doesn't know that the capacitors will charge to 12√2 volts minus 1.4 volts for the rectifiers. That's 15.57 volts which leaves 3.57 volts for the necessary drop-out voltage of the 7812 chip under full load (on the transformer) conditions.

 

No circuits or block diagrams because it's an idea for comment rather than a project under way....
I do know that the main DC line will rise x 1.4 but only off load. The large ripple current at say 10 amp may cause the regulator to drop out at 100 Hz depending on the size of the capacitor(s) and the actual load.
Are you saying you can make a 10 amp supply with a 7812 plus pass transistors that would work ok from a 12ac transformer ?

 

No circuits or block diagrams because it's an idea for comment rather than a project under way

Shrugs and goes back to creosoting fence.

 

OK. It is a dumb idea. That's the idea that is dumb, not the OP. Either make a linear supply or make a switching supply. For high current applications there is no substitute for an appropriately sized transformer -- period full stop, regardless of the type of power supply.

 

done some research..It would seem that I had an exaggerated idea of the ripple voltage that might be on the raw DC line. I found an formula online to calculate it and with the values I quoted (10 x 2000uf) and 10 amps draw the ripple voltage would only be 10Mv. so safe to ignore..Anyway, as you say, it's dumb idea....
Further thought are that the total minimum overhead would be 2.5(7812) + 1.4(bridge) +.7pass transistor(s) = 4.6v.
12 + 4.6 = 16.6. 16.6/1.4 = 11.85 ac secondary required. But I'm thinkingthat even if the transformer wattage is ok it's own winding resistance can't be ignored because of the extremely high charging spikes into the capacitors.
I'm going back to cutting the grass....

 

10 mV ripple for 10 amps from 20000 uF? No way.

i = C dv/dt

dv = i dt / C

dt = 1 / (2 fmains)
dt = 0.01 sec (for 50 Hz)
i = 10 Amps
C = 0.020000 Farads

dv = 10 (0.01) / 0.020000

dv = 5 Volts ripple p-p

 

10 mV ripple for 10 amps from 20000 uF? No way.

i = C dv/dt

dv = i dt / C

dt = 1 / (2 fmains)
dt = 0.01 sec (for 50 Hz)
i = 10 Amps
C = 0.020000 Farads

dv = 10 (0.01) / 0.020000

dv = 5 Volts ripple p-p

Hi..I came back to this thread and saw your post. I too had begun to doubt that figure. So much so that earlier today I did an experiment. I grabbed a 12v 200VA torroidal transformer, a standard bridge, a 34.000uf cap and set a 100 watt car bulb as a load. Current draw was about 9 amps.

Result..transformer no load 12.3v ac with above load 11.75v ac
and ... across cap no load 16v DC. With the load there was 1.5v ripple and the bottom of the ripple was at 11.9v DC. !

I'll try and find the webpage where I found the formula that gave me 10mV but summat was wrong somewhere !
Anyway I think it's clear that for high currents real world losses are higher than I bargained for. And there is not enough overhead for even one pass transistor. I'm may repeat the experiment with shottkey diodes but I havn't got any right now..

 

The formula I use is:√2 C Er F = I
where Er is the peak to peak ripple
and F is the rectified frequency, 100 Hz for a 50Hz main with full wave bridge.
This gives you the voltage loss at the bottom of the sag compared to the peak voltage. Very accurate. I have bench tested this to within 1% accuracy. Original source was a National Semiconductor Audio Handbook.

 

The RSGB publishes ripple charts for many values of capacitance, load, frequency and also inductive and other filter arrangements, along with tables of voltage and current requirements for sundry rectifier configurations.

They are pretty comprehensive.

RSGB Radio Data Reference Book
G R Jessop

pages 117 through 123 in my 5th edition.

 

title was "High current PSU's and chargers using phone charger" caught by bug.

When building linear PSU's or chargers getting a suitable transformer can be difficult/expensive. Some expensive units have transformers with an extra winding for powering the stabilser circuit or simply have 16v or more secondaries to overcome the dropout. (12v case)

Today I just split open a switch mode, 4.9v phone charger and rescued a neat small pcb with an isolated output up to 500Ma. I'm thinking that if that was connected to add to the main rectified DC, it could supply an LM7812, which in turn feeds the multiple power (pass) transistors.

The question then is would a 12v transformer suffice ? I'm assuming capacitors @ 2000uf per amp and a bridge rectifier - which reduces the drop out to 2v total off the main DC.

Keep an eye out for surplus open-frame switchers - especially ones with a TL431 programmable zener driving the feedback opto-coupler.

The TL431 is programmed with a 2x resistor driver, you can adjust the values or put a preset pot in place of the bottom one - but watch you don't set a voltage higher than the electrolytics can take - or its party-popper time!

Surplus open frame switchers are still plentiful with +5V & +12V outputs, you have to load the 5V rail to make the 12V rail active - or you can re-calculate the sensing resistors for 12V and move the sensing point to the 12V rail.

Another source of switchers is some TV set-top boxes - you have to trace out the secondary side to find out which rail is sensed, then re-calculate the divider resistors and move them to the rail you want to use.

 

http://electronics-diy.com/electronic_schematic.php?id=657

Returning to my my original post...I have just come across this circuit on the internet. The circuit is shown for 5v @ 10amps which puzzles me. With the 160,000uf caps shown, and 12.6ac from the TXfrmr I would have thought there was plenty of "overhead" without a need for the additional supply to the 317.

However the penultimate sentence seems to suggest that with that large smoothing cap. AND the separate supply for the 317 you can indeed get regulated 12dc from a 12v ac transformer.

He doesn't imply he has actually built it though....and my original phone charger at 500Ma may not supply enough current to bias those 2n3055s

edit..sorry..didn't notice thread had gone to page two. Reading the other now.