Originally Posted by Morgen

I suppose this is a gambling site. I've reread the original post and have no further comment.

Poor loser? :lol:

Did I goof with my maths?

Hi Audio,

About the value of R1, I sort of reckon the 7805 should reach its full load current of 1A when the excess is taken up by the boost transistor.

So if I design a 4.5A supply, the current distribution would be 1A via 7805 and 3.5A via boost transistor. For a 10A output, 9A via the boost transistors.

With the 7805 mounted along with the transistors, its thermal protection would take care of my boost transistors too. That's my view. Any comment.

Hi Ron,
It looks like your sim is using 0.228 ohms for R2{R1} and R3. Then with 1A through the 7805 regulator the voltage drop across R2{R1} will be only 0.228V and the booster transistor Q4 won't be anywhere near being turned on.
The 7805 must pass 0.6V/0.228 ohms = 2.63A for the booster transistor to begin conducting, and when the booster transistor reaches a collector current of 4A with a Vbe of 1V, the 7805 will have 1.912V/0.228 ohms = 8.39A through it!

Hi Ron,
Don't you argee that if {R1} is 3 ohms, and if the regulator has 1A thru it, then the voltage drop across {R1} is slightly less than 3V and your booster transistor Q4 would be turned-on so hard that the other transistor Q3 would current-limit? Then the Rsc would have about 0.9V across it and wouldn't you know, have nearly 4A through it! :shock:

But it doesn't work that way because Vbe of Q4 plus the voltage across Rsc reduces the voltage across the 3 ohms resistor, reducing the current in the regulator to about only 633mA. :lol:

Hi Ron,
I changed the file type of your sim file before downloading it and got SwCADIII to open your schematic. I watched it run a few things but it is different with R1 = 2.2 ohms.

Do you think the Current limiter transistor is active with a 5A load?
If this is the case then I am giving my appologies to Russ and Morgen. ops:

Do you think it looks like this:

Originally Posted by amazing_ang

hi
i need help with designing a power supply of 5v-5a(high voltage,high current).
cant get the current to boost by using a current booster.if u have the circuit diagram even that will do...but not by using a MJ power transistor only current booster

Sounds like you need a switching regulator.

Hi there,


Here's how it all works......


Since Ir1 depends on the two Vbe drops of the transistors just before cutoff:

R1=(Vbe1+Vbe2)/(Ir1-Ic1/B1)

but since Vbe changes we want to allow a 50 percent
change in the two Vbe so we would get:

R1=1.5*(Vbe1+Vbe2)/(Ir1-Ic1/B1)

or

R1=1.5*(Vbe1+Vbe2)/(RegIlimit-Io/B1+RegIlimit/B1)

Since Q2 begins conducting at Vbe2:

Rsc=Vbe2/((Io-RegIlimit)*(1+1/B1))

where

Vbe1 is base emitter voltage of Q1, and
Vbe2 is base emitter voltage of Q2, and
RegIlimit=Regulator I limit
Ic1=collector current of Q1
Io is total output current
B1 is beta of Q1
Rsc is Q2's base emitter resistor
Ir1 is current through R1


Not too difficult?


Numerical example 1:

Io=5 amps
RegIlimit=1 amp
B1=100
Vbe1=1 volt
Vbe2=1 volt

Calculate R1:
R1=1.5*(Vbe1+Vbe2)/(RegIlimit-Io/B1+RegIlimit/B1)
so, R1=3.125 ohms.

Calculate Rsc:
Rsc=Vbe2/((Io-RegIlimit)*(1+1/B1))
so, Rsc=0.248 ohms


Numerical example 2:

Io=5 amps
RegIlimit=1 amp
B1=10 (much lower than before)
Vbe1=1 volt
Vbe2=1 volt

Calculate R1:
R1=1.5*(Vbe1+Vbe2)/(RegIlimit-Io/B1+RegIlimit/B1)
so, R1=5 ohms.

Calculate Rsc:
Rsc=Vbe2/((Io-RegIlimit)*(1+1/B1))
so, Rsc=0.227 ohms


Another point of view is that if R1 goes too low the regulator will cut back too soon, and if R1 is too high Q2 has to do more work.
A good value would be one that always allows Q2 to work, but as little as possible.