DIY project for constant voltage 5v, 2amp

[bitman]

Holy moly ... I found a TIP42C (sometimes you're lucky that way) and used to create Figure 13 (https://www.fairchildsemi.com/datasheets/LM/LM7805.pdf) - I still have no clue how the formulas work, all I know is my 5W 3ohm resistor got VERY VERY hot generating about 1.7amps but the TIP and LM7805 both stayed around 50-60 centigrades with the small heatsink that I had. So I think I'm on the right track (I know - 5watt is too small but it's what I had).

What strikes me is that I get about almost 7 volts out when there's no load. Only when applying a load does the voltage drop to around 5.3/5.4 volts. Still a bit high but better than the 1 and 2 volts I had on the other circuits under load.

Thanks for all the suggestions. Now to figure out how it works

[bitman]

Well, to reply to myself here. I figured out why the voltage was too high - that fixed, I now have the reverse issue. Under load the output voltage decreases "dramatically". Input is still the same - AMP just goes up. I measure about the same amp on the output - but much lower voltage so there's one heck of a loss. Input is 6v showing about 1.5amps (two 3 ohm resistors in parallel is the load) and output shows around 3.4 volts and just below 1.5amps. So I'm feeding 12W and measuring around 4.6W - that's quite a loss of ~7W ... yes, things are getting REALLY hot on the resistors (and since I corrected my mistake the TIP is getting hotter).

Not sure I understand why the voltage is dropping. If I was getting 5v the power loss would be a lot less.

(it's it's a very basic circuit and I still managed to screw up E/C on the PNP *sigh*).

[Ian.M]

You are starving it - it needs enough headroom for the regulator, and that's somewhere between 2V and 3V above the intended output voltage, plus the voltage drop across R1,which will be around 0.7V - Q1 Vbe - if Q1 is passing current.

3R||3R is 1.5R.  At 3.4V it *should* be drawing 2.27A.  Your measured 1.5A is bogus - probably because the load is increased by wiring resistance of 0.77R, but it could be by your Ammeter's burden resistance.   I suggest that to prevent stuff meting you use a single 3R resistor.  That should be 1.67A if you get the input up to where the 7805 can regulate, and sort out your wiring resistance problem, which is enough to prove the pass transistor is doing its job.   If you want to test it properly, you'll need a high current variable load.

[bitman]

You are starving it - it needs enough headroom for the regulator, and that's somewhere between 2V and 3V above the intended output voltage, plus the voltage drop across R1,which will be around 0.7V - Q1 Vbe - if Q1 is passing current.

Hmmm - I've set my bench supply to allow for up to 2 amps, and I've applied up to 10 volts with no significant change in the voltage (the output amp changes as expected). My "bet" is on a wrong setup (by me) more than anything else - I've gone over the simple circuit about a dozen times and I think it's correct now but it sure took me several tries to identify issues like reversing C/E on the TIP42C (I keep forgetting they work in reverse from NPNs - one day I'll learn).

Currently I've got the input set at 7V and it's still below 4 volts on the output.

3R||3R is 1.5R.  At 3.4V it *should* be drawing 2.27A.  Your measured 1.5A is bogus - probably because the load is increased by wiring resistance of 0.77R, but it could be by your Ammeter's burden resistance.   I suggest that to prevent stuff meting you use a single 3R resistor.  That should be 1.67A if you get the input up to where the 7805 can regulate, and sort out your wiring resistance problem, which is enough to prove the pass transistor is doing its job.   If you want to test it properly, you'll need a high current variable load.

I've measured right after the 3ohm resistor (going to the base) and when feeding 7V I see 7V - granted my current cheap meter is just short of 4 digits so it's not exactly seeing details (hoping for better multi-meter by christmas). I've measured on each side of the resistor and the meter shows exactly the same voltage. I presumed it was seeing voltage from the base on the TIP ... I've been wrong before.  I'll be happy to post a picture of the breadboard - I kept most wires very short except for the supply wires and the meter wires, everything is pretty short. I measure straight on the two parallel resistors.

Btw. I put them in parallel because it's just 5W and it got way too hot when being the single load.  I'd calculated I needed around 3 ohms to test the load, got the wrong W though. Story of my life. 10W are being searched for in the mean time.  But for short usages I can use the 5W 3ohm resistors.

I measure the amps in series with the two 3ohm power resistors. If I try to measure amp straight on the +/- on the output it acts like a short and I stopped trying to do that. The amp I see on the meter is within a few tens of the amps on the power supply (as I stated). So I didn't see it as being off.

How did you come up with the 0.7V voltage drop btw? 

Well, I'm off for more testing and probing ....

[bitman]

I'm at this point feeding 8volts (according to your numbers that should be enough?) - I'm still just seeing above 4 volts on the output - if I'm lucky it gets to around 4.6volts.  On the pictures you can see me measure before and after R1 .. the numbers are pretty much identical, and at times I can see higher voltage (by 0.01) on the junction between the resister and the base. Yes, clearly the meter is the issue here. You don't get much for $25 but for almost a year it's been doing plenty good as I've needed things (considering I started with a $5 old radioshack one this is "delux").

Measuring before the resistor (straight on the Vin): https://photos.app.goo.gl/CKJW4lYVqHjWiCWo1
Measuring after the resistor:  https://photos.app.goo.gl/b6miXXMSuuX9Mxds1

This is during load btw (3ohm).

I definitely got the voltage up above 4v by increasing the Vin to 9 or 10. Anything higher than 10 doesn't seem to make a difference. But 8V on the Vin seems to produce just about 4volts out where as 9V produces around 4.6V (depending on temperature it will decrease as things get hotter). I definitely need to find something to keep the load going for a while to see what the transistor and regulator do heat wise. I think I may need bigger heat sinks.

[MK14]

I could easily be wrong and/or misunderstanding you, but ...

It sounds like you have broken the PNP transistor, and it has gone short circuit, between (at least) the base and emitter.

When you incorrectly wired it up, previously, you could have broken it, because they don't like reverse voltages. Especially as regards "wrong" polarity on the base, at significantly high voltages (see datasheet, from memory limit is 5V in reverse direction).

If it was me, doing this, I might pull out the PNP from the breadboard, and use the diode test mode of a multimeter, and check the emitter/base junction, is still around 0.6/0.7 volts. If it is short circuited, then the PNP is toast.

Also, breadboards like that, are not particularly useful at higher currents, like the 2 amps you are using. They can easily be damaged and give poor results. They don't like hot components either (it can melt the plastic).

[bitman]

I could easily be wrong and/or misunderstanding you, but ...

It sounds like you have broken the PNP transistor, and it has gone short circuit, between (at least) the base and emitter.

Interesting idea. See below (from my testing it seems to be ok). At no point even during my bad wiring did the PNP get very hot. It does get up to around 50-60 degrees centigrade now - eventually - under load. I haven't had it on for long so I don't know if it platoes there.

When you incorrectly wired it up, previously, you could have broken it, because they don't like reverse voltages. Especially as regards "wrong" polarity on the base, at significantly high voltages (see datasheet, from memory limit is 5V in reverse direction).

Wouldn't be the first time and you provided a good reminder to me not to take the state of affairs for granted. I never thought about testing the component as part of the diagnostics. Bad me ....

If it was me, doing this, I might pull out the PNP from the breadboard, and use the diode test mode of a multimeter, and check the emitter/base junction, is still around 0.6/0.7 volts. If it is short circuited, then the PNP is toast.

I think it's ok. When I put negative on the base, both the collector and emitter shows 0.6 volts or a little shy of that. I'm not sure if it's supposed to show this on BOTH but if I recall correctly that seems to make sense given the design of the chip.

I wonder if the 7805 could be hosed.

Also, breadboards like that, are not particularly useful at higher currents, like the 2 amps you are using. They can easily be damaged and give poor results. They don't like hot components either (it can melt the plastic).

Ahh yes - I'm not claiming it's the best method, and once I have something that looks like it works I'll definitely get a perfboard out and get soldering to make longer tests. For now, I'm thinking my wiring is bad (or components) and soldering them seems kinda pointless until I get that right. Btw. no components touches the breadboard - and the leads haven't so far showed anywhere near the same heat, but yeah it's plastic. It will melt. This is a throw-away bread board. The very wide leads for the power resistors have made the holes so wide nothing else will "stick" in there. So I'll call this my "power tester" and leave it for future projects like this. But no argument here - for anything that needs to be running for more than a few minutes, this is a BAD THING.  I don't know if you can see, but to the very far left is a perf-board with a transformer on it - something I "rescued" from a piece of equipment. I didn't even TRY to put that on a breadboard for this exact reason  

Thanks for the suggestion - I hadn't thought of testing the PNP at all.

[MK14]

On the 7805, what is the voltage difference (ideally with a load on the output), between its input and output (pins 1 and 3) ?

It should be around 2 volts (EDIT: more is fine, but 0V probably means there is a short or the LM7805 is faulty), assuming the incoming supply is at least 8 volts (a bit less might work as well).

Your results are confusing me here. Because you also seem to be saying the PNP's collector is at around 0.6V, implying that there is no voltage across the voltage regulator. Yet you also seem to be saying the input voltage is up to 10V, yet the output is only about 4.6V.
tl;dr
Something is a bit out with your results or my interpretation of them.

[MK14]

Summary:
So you have 10V coming in to the circuit.
The PNP's base is around 9.3V
The output voltage SHOULD be at around 5V
The PNP's collector should also be at around 5V
The PNP's emitter should be at 10V
Hence there is around 5V 4.3V between the input and output of the LM7805 (which is fine).

[Ian.M]

A 7805 *REQUIRES* the minimum capacitors given in the datasheet application circuits: 0.33uF on the input and 0.1uF on the output.   They should be as close as possible to the regulator.  Without them, it can go unstable and oscillate (more likely if the layout is poor, e.g. when breadboarding).  The oscillation tends to depress the DC average of the output voltage and also, most multimeters don't do well when you try to measure voltages round a RF power oscillator!

Re: The measurement of high currents in low voltage circuits.
Cheap meters tend to have a high burden voltage, i.e. their current ranges do NOT look like a perfect piece of wire from the circuit's point of view but instead look like a resistor, which on low current ranges may go as high as hundreds of ohms.  The exception is their 10A range that is usually a solid wire shunt (possibly fused) between the 10A and common terminals.  Even good quality precision DMMs can have quite significant burden voltages.  In addition you've got the possible voltage drop in the test leads and their contacts.

Its therefore preferable to measure the voltage drop across a known resistor, and using Ohm's Law to calculate the current.  Preferably the resistor should be one you have directly measured the resistance of to 1% accuracy, and know doesn't change value significantly when hot, but even assuming the nominal value of a commercial grade low-Ohm 10% resistor is likely to be more accurate than adding a cheap multimeter reading current in series with it.  For low-Ohm resistors, the multimeter leads should be clipped directly to the resistor leads next to its body, so you dont include the voltage drop across the breadboard contacts (or other connections) in the measurement.

[bitman]

Summary:
So you have 10V coming in to the circuit.
The PNP's base is around 9.3V
The output voltage SHOULD be at around 5V
The PNP's collector should also be at around 5V
The PNP's emitter should be at 10V
Hence there is around 5V 4.3V between the input and output of the LM7805 (which is fine).

I think one detail is lost in this now much longer than I ever thought it would be thread. I see 4.99v on the output WHEN THERE IS NO LOAD. Once I apply about 1.5amp load, the voltage drops to about 4.5 +/- 0.1 (which is much better than when I started where it was close to 3v).

[MK14]

I think one detail is lost in this now much longer than I ever thought it would be thread. I see 4.99v on the output WHEN THERE IS NO LOAD. Once I apply about 1.5amp load, the voltage drops to about 4.5 +/- 0.1 (which is much better than when I started where it was close to 3v).

I wonder if that is because of the voltage drops across the breadboard connections, and breadboard connection leads, which may not be 100% copper, and/or not rated for such high currents. I.e. it is more meant for currents in the hundreds of milliamps range, rather than 1.5A .

tl;dr
It does not take much resistance (at 1.5 Amps) to drop a few hundred millivolts, if it is made on a breadboard with possibly weak (electrically) leads. Some of the cheap Chinese wires are NOT 100% copper or similar. Some are copper coated aluminium, or some other kind of higher resistance configuration.

[bitman]

A 7805 *REQUIRES* the minimum capacitors given in the datasheet application circuits: 0.33uF on the input and 0.1uF on the output.   They should be as close as possible to the regulator.  Without them, it can go unstable and oscillate (more likely if the layout is poor, e.g. when breadboarding).  The oscillation tends to depress the DC average of the output voltage and also, most multimeters don't do well when you try to measure voltages round a RF power oscillator!

I have a 0.33uF on the input and a 0.1uF on the output. They're not ceramic but the cheap ones you see in the picture. I read somewhere that you had to use ceramic caps for this - I spent 2 hours going through a bunch of recovered caps of which maybe a dozen were ceramic and well, not much luck finding anything at those levels. But I'll keep searching.

(note, the 0.033uF is on the base on the TIR which is connected straight to the IN (pin 1) on the 7805). According to the diagram I follow I can use the same cap for both the base and the input of the 7805?

[bitman]

Summary:
So you have 10V coming in to the circuit.
The PNP's base is around 9.3V
The output voltage SHOULD be at around 5V
The PNP's collector should also be at around 5V
The PNP's emitter should be at 10V
Hence there is around 5V 4.3V between the input and output of the LM7805 (which is fine).

Hmmm - this is WAY far from what I'm seeing.
Input voltage of the 7805 is 10v. I do NOT see a voltage drop what-so-ever (at least not that I can measure) through the 3ohm resistor/base/7805-in.
When there's no load, I see about 4volts between input and output (somehow it shows up negative when I put the multimeter zero on in, and positive on output). That looks like what you're describing?
When there's a load, this drops to about 2 volts.
 
When there's no load, I see 5v on the output. When there's a load (approx 1.5amp) that drops to 4.5 +/- 0.1 volts.

[bitman]

I think one detail is lost in this now much longer than I ever thought it would be thread. I see 4.99v on the output WHEN THERE IS NO LOAD. Once I apply about 1.5amp load, the voltage drops to about 4.5 +/- 0.1 (which is much better than when I started where it was close to 3v).

I wonder if that is because of the voltage drops across the breadboard connections, and breadboard connection leads, which may not be 100% copper, and/or not rated for such high currents. I.e. it is more meant for currents in the hundreds of milliamps range, rather than 1.5A .

tl;dr
It does not take much resistance (at 1.5 Amps) to drop a few hundred millivolts, if it is made on a breadboard with possibly weak (electrically) leads. Some of the cheap Chinese wires are NOT 100% copper or similar. Some are copper coated aluminium, or some other kind of higher resistance configuration.

I have no idea - however, the stuff I want to run is on breadboards too (it's large - lots of boards). It draws up to 1.3amps at full implementation and so far I haven't seen anything indicating a problem of this caliber. Granted, this is a single board and the others is well about 15 or so - so much longer leads and maybe that makes a difference. It just seems odd to me that a few mm of bread-board metal would make this big of a difference. At least I would expect to see plastic melting or something if the connectivity is having issues.

Anyway - I'll check again tonight if everything is hooked up right. Double check the datasheets to be sure the pins are right etc. And then update.

[MK14]

Summary:
So you have 10V coming in to the circuit.
The PNP's base is around 9.3V
The output voltage SHOULD be at around 5V
The PNP's collector should also be at around 5V
The PNP's emitter should be at 10V
Hence there is around 5V 4.3V between the input and output of the LM7805 (which is fine).

Hmmm - this is WAY far from what I'm seeing.
Input voltage of the 7805 is 10v. I do NOT see a voltage drop what-so-ever (at least not that I can measure) through the 3ohm resistor/base/7805-in.
When there's no load, I see about 4volts between input and output (somehow it shows up negative when I put the multimeter zero on in, and positive on output). That looks like what you're describing?
When there's a load, this drops to about 2 volts.
 
When there's no load, I see 5v on the output. When there's a load (approx 1.5amp) that drops to 4.5 +/- 0.1 volts.

Sorry, I was mistaken. When I wrote it, I had tried to remember the circuit, but couldn't see it while typing the response.
R1 3 is what was confusing me. That is why you are only getting around 2V across the LM7805, which is fine/correct and how that circuit intends to work.

When there is no load, the PNP pass transistor is inactive, so the LM7805 works as normal. But with a load, there is a significant voltage drop across R1 3 , which activates the PNP pass transistor, into supplying the bulk of the output current (asssuming the output load current is high enough).

[MK14]

I have no idea - however, the stuff I want to run is on breadboards too (it's large - lots of boards). It draws up to 1.3amps at full implementation and so far I haven't seen anything indicating a problem of this caliber. Granted, this is a single board and the others is well about 15 or so - so much longer leads and maybe that makes a difference. It just seems odd to me that a few mm of bread-board metal would make this big of a difference. At least I would expect to see plastic melting or something if the connectivity is having issues.

Anyway - I'll check again tonight if everything is hooked up right. Double check the datasheets to be sure the pins are right etc. And then update.

Assuming you give it a big test load (1.5 A ?), and the output has dropped to 4.6V (say).
You could connect (briefly touch) the multimeter probes (very carefully), (to the pins of the LM7805) middle pin (Gnd) and the output pin (3), directly. Is it now exactly 5V ? (4.99V or whatever).

If it is almost exactly 5V, you can then try and find out why the final output has dropped to 4.6V. E.g. voltage drops across the grounds, etc etc.

[Ian.M]

I wonder if that is because of the voltage drops across the breadboard connections, and breadboard connection leads, which may not be 100% copper, and/or not rated for such high currents. I.e. it is more meant for currents in the hundreds of milliamps range, rather than 1.5A .

tl;dr
It does not take much resistance (at 1.5 Amps) to drop a few hundred millivolts, if it is made on a breadboard with possibly weak (electrically) leads. Some of the cheap Chinese wires are NOT 100% copper or similar. Some are copper coated aluminium, or some other kind of higher resistance configuration.

Yep.  Simple high current low voltage circuits like this are best *NOT* built on breadboard.  Tack solder them together, component to component, wire to wire (use insulated solid wire), over a sketch of the physical layout (in biro or marker, not pencil, so its non-conducive), flat on your bench.  If they need more support, knock brass panel pins into a thin plank of softwood  (which should ideally be impregnated with paraffin wax to keep it dry to minimise leakage) and solder to them as support points)  This is the original 'Breadboard' construction.  If you use soft enough wood and a panel pin setting tool with a drop of tacky flux up it to retain the pin, you don't even need to hammer anything.


For more complex circuits, use a bare un-etched single sided PCB as a ground plane, copper side up, Jim Williams style.  For even more complex or critical circuits you can transition into Manhatten style construction, with glued on isolated PCB 'islands' as connection points or even whole breakout PCBs.  If the breakout boards are double-sided, use foam tape (3M VHB for a permanent build, the cheap white stuff lets go with age) or saturate a slip of cartridge paper in superglue to insulate them from the board ground plane.  Also see https://hackaday.com/2016/05/04/getting-ugly-dead-bugs-and-going-to-manhattan/ for some nice photos of Jim Williams and Manhattan style construction.

However if you retrieve components from such a build for solderless breadboading, you *MUST* make sure their leads are cleaned of all excess solder, especially spikes or lumps, and wipe off any flux residue with IPA, or they'll trash your breadboard.  This generally isn't worth doing for smaller passives so keep the used ones separate for future solder breadboarding.

[bitman]

Assuming you give it a big test load (1.5 A ?), and the output has dropped to 4.6V (say).
You could connect (briefly touch) the multimeter probes (very carefully), (to the pins of the LM7805) middle pin (Gnd) and the output pin (3), directly. Is it now exactly 5V ? (4.99V or whatever).

If it is almost exactly 5V, you can then try and find out why the final output has dropped to 4.6V. E.g. voltage drops across the grounds, etc etc.

Sure - I have exactly 5.00v when there's no load. And 4.97v with load. That seems within reason. I guess this shows the 7805 is ok.

[Ian.M]

Nope.  7805 (and the rest of the 78xx family) regulators have a tendency to die by going high resistance, so you don't know its good till you have swept its load current up to a couple of hundred mA while monitoring the output voltage to check it can deliver real current.  If you cant sweep a load current, at least spot check with a few load resistances to give currents in the range 5mA to 200mA.

If you don't have a constant current DC load, build one!   Beefy power MOSFET designed for linear mode operation on a salvaged CPU heatsink, 0.1R power resistor as the current shunt in series with its drain and an OPAMP and a few passives to close the loop, a pot and a Zener for the current setpoint, and a floating output 12V wallwart to power the fan and control circuit.  You can also use a beefy power transistor (e.g. the classic 2N3055), Darlington paired with a high gain driver, but it needs more voltage headroom than the MOSFET so is less useful if you are experimenting with stuff like plotting NiMH dscharge curves.

[MK14]

Sure - I have exactly 5.00v when there's no load. And 4.97v with load. That seems within reason. I guess this shows the 7805 is ok.

Then it does seem to be because of excessive resistances, due to the construction method.

The thing is, voltage drops are NORMAL (but unwanted), even in professional builds. E.g. a professional 5.00 Volt power supply, powering many boards, with long (solid copper) leads. At higher currents (many amps), and at longer lead lengths, voltages do drop.
That is why some power supplies have/use voltage sense terminals/connections. So that it can regulate the voltage AFTER any voltage drops, in the leads.

Which is why care is sometimes needed, to ensure that these voltage drops, do not affect the circuit operation. I.e. they are not so big, that the supply is too low or even unstable.

As Ian.M has suggested, as well. You may want to consider building it using some kind of soldered solution. Which will significantly reduce (but not fully eliminate), these voltage drops.

tl;dr
1.5 Amps is quite a lot of current, so reasonable care needs to be taken. Otherwise you will drop a fair amount of voltage in your wiring and connections.

[bitman]

Nope.  7805 (and the rest of the 78xx family) regulators have a tendency to die by going high resistance, so you don't know its good till you have swept its load current up to a couple of hundred mA while monitoring the output voltage to check it can deliver real current.  If you cant sweep a load current, at least spot check with a few load resistances to give currents in the range 5mA to 200mA.

I've run at much lower loads and at those I see the expected results.

If you don't have a constant current DC load, build one!   Beefy power MOSFET designed for linear mode operation on a salvaged CPU heatsink, 0.1R power resistor as the current shunt in series with its drain and an OPAMP and a few passives to close the loop, a pot and a Zener for the current setpoint, and a floating output 12V wallwart to power the fan and control circuit.  You can also use a beefy power transistor (e.g. the classic 2N3055), Darlington paired with a high gain driver, but it needs more voltage headroom than the MOSFET so is less useful if you are experimenting with stuff like plotting NiMH dscharge curves.

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.

[bitman]

Sure - I have exactly 5.00v when there's no load. And 4.97v with load. That seems within reason. I guess this shows the 7805 is ok.

Then it does seem to be because of excessive resistances, due to the construction method.

Sure - let me give the perfboard a roll then. I've tried pretty much everything else. It may be a day or actually a week before that's done (heading on the road early tomorrow for the rest of the week). But I'll get there.
[/quote]

[MK14]

Sure - let me give the perfboard a roll then. I've tried pretty much everything else. It may be a day or actually a week before that's done (heading on the road early tomorrow for the rest of the week). But I'll get there.

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! 

[Ian.M]

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.