DIY project for constant voltage 5v, 2amp

[bitman]

If you don't have a constant current DC load, build one! 

I do? My bench supply provides constant voltage just fine. With over-current protection etc. without which I'm sure I would have burned down the house by now (just kidding but it is handy). I'm trying to NOT use my bench supply to power this setup.

A bench supply is *NOT* a DC load.  A DC load replaces those paralleled 3R resistors that were getting very hot, and lets you smoothly vary the load current from near zero up to a limit set by the power transistor used.  Look at the S.O.A and derate for heatsink temperature: e.g with a heatsink that can keep it below 70 deg C a 2N3055 based load would be good for 10A  at up to 7.5V across it dropping linearly to 2A at 40V.

DUH!! Sorry - I was looking into a DC load setup a few weeks ago, but that's way out of my price range. I did find a nice youtube video series to create one that I may look into (not a Eevblog one though) - yeah I found that looking for how to create a simple DC power supply and yes it took me a bit before I realized my mistake. Sorry for the misunderstanding.

[bitman]

Good idea. Whatever your favourite method is.
If the plastic breadboard was giving you problems now, the connections would have tended to get worse and worse, in coming months and years. With oxidation, dust etc. The wires can easily get knocked out as well.
So perfboard (verostrip), or the methods suggested by Ian.M, or any other soldered solution. Should be miles better than (plastic) breadboards.

Good luck with it! 

Thanks - I managed to get things soldered on a perfboard - not pretty but it'll do for now. The same test no longer shows a drop of output voltage beyond the 0.,03 volts so clearly the breadboard was the cullpret at higher amps. Important lesson learned (for me).  Using a 3ohm for load, pulls the AMP up to around 1.65. So things seems to match the numbers in this thread. What I need to do now is figure out what size heatsinks I need. I ran it at .2 amps for a few hours and the TIP42C and 7805 gets warm - not hot. I didn't see them get even measurably hot on the bread board at that load, so there's definitely NOT enough cooling as is. But it will be the weekend before I get a chance to look at it again.

Thanks again for all the input/help and more important showing a pathway to improving myself.

[MK14]

That's very promising, that the voltage drops are very small now.

As regards the heatsink, let's work out how much heat the TIP42C will have to cope with. The circuit I think you are using, will not send much current through the LM7805, so the current (small) heatsink, is probably enough, for now.
LM7805 current = 0.7 V (approx base voltage of PNP/TIP42C) across the 3  resistor.
Ohms law = I = V/R = 0.7/3 = about 230 mA's.

For these calculations, I am going to assume the incoming supply voltage is 10 Volts.

LM7805 disipation = (input voltage - output voltage) x current through it = (10 - 5) x 0.23 = 1.1 Watts
So it should run fairly cool. EDIT: Less than 1.1 Watts in practice, as some of the heat will be dissipated in the 3  resistor.

Current passing through TIP42C = 2 Amps (from title of thread, and ignoring the relatively small current through the 3    resistor)
Voltage across TIP42C = input voltage - output voltage = 5 volts
Ignoring the slight heating effects of the base current
TIP42C disipation = voltage across it x current passing through it = 5 x 2 = 10 Watts

Thermal resistance of TIP42C http://www.mouser.com/ds/2/149/TIP42C-890174.pdf
= about 2 deg per Watt (apparently not directly specified in datasheet)
Max junction temperature = 150 deg C

It is not a good idea, to go too close to the maximum junction temperature, as above it can rapidly break the transistor, and very high temperatures, although allowed, tend to significantly shorten the life expectancy of the transistor.

So a heatsink of around 5 deg C (as a minimum), is probably what you want. Let's check.

Total thermal resistance (ignoring thermal compound/interface part) = 5 + 2 = 7 deg C per watt
Thermal compound will add about 0.8 deg C per watt or so, so call it 1, final total = 8 deg C per Watt
Total dissipation = 10 Watts
Room temp = Ambient = 25 deg C

Expected final temperature = 25 + (8 x 10) = 105 deg C TIP42C junction temperature with a 5 deg C heatsink + thermal compound

Heatsinks can get a bit pricey, as they get bigger. So it can sometimes pay to get a more expensive power transistor (with lower thermal resistance) and/or use multiple, power transistors (best to refer to schematics designed for this, which usually use emitter resistors, e.g. 0.1  or more, to allow paralleling of transistors), to share the heat load. Often done, for high current (e.g. 5 to 20 Amp) linear power supplies.

The following example (5 deg C) one, is probably less than a dollar (£0.62).

https://www.digikey.co.uk/product-detail/en/assmann-wsw-components/V8813X/AE10770-ND/3511421



To get the best life and reliability. I prefer to use an even bigger heatsink (higher rated one). But peoples opinions on this can vary.