h i am making this circuit i have altered it from the original one could any one tell me that i could use as Q1 to Q10 for 500 MA or above say 750MA and what the value of resistors i need for the Q1 to Q10
thanks

 

What is the output voltage of U2 and how much current each output can source?

 

4.95 volts x 500 MW if this helps

 

The maximum output current of U2 when it is power with 5V is 0.36 mA. Thus you will need a 12K resistor for each transistor. Also, you have to use darlington transistors which have a gain greater than 1500 at saturation.

 

will a TIP23 do the job or can you recomend something else i have in stock some IRF530's

 

The IRF530 is not a logic level MOS. It needs 10V to fully turn on however you may use it because the current through it will be only 500mA. Try one and if it gets very hot then you have to use another one or power U2 with 10V.

I can't find TIP23.

 

sorry tip32

 

sorry tip32

Its gain its too low, you can't use it.

 

You are going to need to use a different regulator than the 78L08 since it is only rated for 100 milliamps maximum.

hgmjr

 

i have the 78l08 so i will replace it

 

Yeah. I see 5 amps through the total of 10 transistors. You do realize that (15V-5)*5A = 50 Watts. You will need atl east a TO-3 package and one whale of a heatsink to dissipate 50 Watts or you will cook the regulator.

BTW a TIP120 is a TO-220 NPN Darlington

 

Marko,
You'd be better off using MOSFETs.
Order some of these IRLD014's: http://www.mouser.com/ProductDetail...=sGAEpiMZZMveMCOqFR6qCPeAjNEHCd78YZDGJGvck3Y=
[eta] Actually, Jameco is considerably less expensive: http://www.jameco.com/webapp/wcs/st...amecoall&ddkey=http:StoreCatalogDrillDownView
Get some extra. They're logic-level N-channel MOSFETs rated for 60v 1.7A in a 4-pin DIP package; really slick. You will use them for lots of projects.

The 78L08 won't work, as has already been mentioned. A 7808 might, but will require at least 2v higher input.
D1 is a 1N4148; those are rated for 100mA maximum. At 500mA, it would have a very short life indeed (maybe about 5 seconds) before it turned into a miniature volcano. A 1N4000 series or 1N5400 series rectifier diode would last far longer.

You are not showing Vdd and GND connections for the 4017.
You don't show a 0.1uF cap across the power/GND pins of the 555 and the 4017. Don't forget them.

In the lower right corner, you have connected your 7808 output to GND. I'm sure you didn't mean to abuse your regulator in that fashion.

 

Yeah. I see 5 amps through the total of 10 transistors. You do realize that (15V-5)*5A = 50 Watts.

Papabravo,
Look again; he's driving the bases/gates using a 4017 Johnson counter. Only one transistor/MOSFET will be on at any given time.

 

@Sarge. As I recollect a Johnson Counter aka a twisted ring counter does in fact have a state where all outputs are on and another state where all outputs are off. The lengthof the sequence is 2*n where n is the number of stages. Isn't that right?

 

Papabravo,
The 4017 is a 5-stage Johnson counter that only has one output (Q0-Q9) high at any given time. If desired, an early reset can be triggered by bringing the reset input high - but no, this particular counter will never have more than one output driven high at a time.

 

I did miss the decoded output part of the chip specication. I stopped at "5 stage Johnson Counter"