DIY project for constant voltage 5v, 2amp - Page 1
Forum section: electronics / beginners

author: bitman
author_rank: Unknown
timestamp: November 25, 2017, 08:33:02 pm
content: I've watched eevblog on you tube for quite a while, and one of the series I've watched more than once is when Dave creates a bench power supply from scratch. Very informative and an inspiration to repeat. HOWEVER I'm trying to find something "small" I can build to generate 2amps 5volts to power some TTL. According to the bench supply it currently draws between 1.15 and 1.35amps depending on "state" so I think 2amps would be a safer thing than 1.5. That leaves out the basic 7805 circuits - or rather I'm experimenting with two 7805 in series. The problem is I need to calculate cooling requirements and a few other ares I don't feel comfortable in right now.So I'm looking to see if there's a DIY project spec out there creating a basic 5v PSU (from 120v) which protects against shorts and reverse polarity. It doesn't have to be a Dave's level of teaching but something that explains the circuit would be great since I'm on a learning path here. It should include a gerber or similar file so I can get the circuit created permanently.I've created several LM317 and LM7805 circuits from different templates I've found - but they all seem to crack around .5 -> 0.7 amps. I then found an old CPU heat-sink drilled a few holes in it and used it with the LM7805 - that got me closer to 1.5AMP but not all the way to 2AMP. It seems odd to have to keep adding 7805s - in particular because the voltage seems to drop to around 4.8 volts under load.I know the problem - it's meI don't know which components are best for these kind of efforts. Initially I didn't think I needed a simple arduino or other micro-processor but the more I see small projects using a simple LCD the more it appears to me that it's the right way to go. Anyway - I can find a lot of projects out there but well, they show their age. Some are even malware sites today - most have pictures of circuits not the actual data I can use with software like KiCad. I am NOT planning to etch out a PCB at home - those chemicals don't belong in my closed work-space, so being able to get it printed somewhere is important.Anyway - the short version: Looking for a project site to create simple PSUs 5v and at least 2amp output (from 120v).

author: Benta
author_rank: Super Contributor
timestamp: November 25, 2017, 08:50:38 pm
content: Quite frankly - for 5 V, 2 A from 120 VAC I'd simply buy a USB wall-wart. You won't get anything easier or cheaper. 10...12 W types are plentiful.

author: bitman
author_rank: Unknown
timestamp: November 25, 2017, 09:12:51 pm
content: Quote from: Benta on November 25, 2017, 08:50:38 pmAhh yes. But you see, part of this is to learn. Ie. figuring out how much thermal cooling I need etc. I'm not doing it because I cannot find a PSU that just plugs in (if fact I have plenty lying around).

author: Ian.M
author_rank: Super Contributor
timestamp: November 25, 2017, 09:26:36 pm
content: Simple linear regulators above about 1.5A are a P.I.T.A.and best avoided unless you cant tolerate the noise of a switcher, or *really* need the learning experience!Perhaps the easiest option is simply put a separate 7805 on each board, or if its all one big board break it up into sections, such that each 7805 load current is under 500mA, and distribute a higher voltage between 8V and 9V to all the regulators. That splits up the dissipation into reasonable 2W 'chunks' which can be handled by a small heatsink on each regulator.However, if you must go above 1.5A:L78S05is a 7805 that's rated for 2A load current. Its input voltage and heatsinking are *CRITICAL* - like any other linear series pass regulator, the heat dissipated is equal to the product of the load current and the voltage differential, and if you don't get it out of the die quickly enough it goes into thermal limiting and the output collapses. The dropout voltage *should* be under 3V differential, but it rises at low temperatures, so figure on 8V in at the troughs of the ripple at 2A load current. If you give it a smooth 8V in, you've got 6W to get rid of. That's a 30 deg C junction temperature rise above the heatsink temperature. However 4V pk-pk ripple (not unusual for an unregulated rectified supply) will give you an average of 10V in and you've now got 10W to get rid of and a 50 deg C junction temperature rise. That's on-top of whatever temperature rise the heat-sink is running at, and ideally you want to run the die at no more than 100 deg C with 25 deg C ambient to give you enough margin to cope with elevated ambient temperatures and still get a long life out of the regulator.If the transformer output voltage is a bit too high, a power resistor in the feed to the L78S05, calculated to drop enough voltage to just get the troughs down to 8.5V at 2A load current will help considerably. The imput decoupling cap specced in the datasheet goes after the resistor.Also look at theLM3503A adjustable regulator. That will let you trim the output to get it closer to 5.0V. Its got a lower thermal resistance so can run its junction cooler than a 78S05 (by about a factor of two) with the same input and output voltages, load current and heatsink. Otherwise all comments above about the L78S05 also apply to the LM350. N.B. it has a worst-case *MINIMUM* load current of 10mA, so either make the top resistor in the voltage adjustment divider no more than 120R (the datasheet shows 240R, and still needs an extra 5mA load to guarantee it always regulates), or add a power LED off its output with a series resistor chosen to make up the difference with a little margin.There are more complicated circuits that pre-regulate the voltage to a L78S05 or LM350, or use a LM337 with an external pass transistor to split the dissipation between the regulator and another power semiconductor to make heatsinking easier, or one can go for a full-up LM723 based PSU design, with multiple external pass transistors if you have ridiculous load current requirements and want to be able to preset a current limit (or even more complex circuits to add an adjustable current limit) but that's getting into serious project territory and is over-kill for just running TTL logic.

author: bitman
author_rank: Unknown
timestamp: November 25, 2017, 09:34:22 pm
content: Thanks Ian - do you know if there are any projects out there where the 350 is used? I currently have two 7805s and I guess I may need a 3rd. I also have a few 34063AP lying around that may work better here.Btw. I guess I really goofed during testing. I'm feeling about 12-15v from the bench supply. Totally ignored the heat that generates and it may explain quite a lot.

author: iainwhite
author_rank: Unknown
timestamp: November 25, 2017, 09:40:51 pm
content: The 78xx datasheet shows a sample circuit which combines a power transistor with the 7805. Maybe give that a try?Look at fig 13 on page 20:https://www.fairchildsemi.com/datasheets/LM/LM7805.pdf

author: Ian.M
author_rank: Super Contributor
timestamp: November 25, 2017, 09:52:10 pm
content: LM350 is effectively just a LM317 on steroids!Take any LM317 project, fit a big-ass heatsink (well over double the size as you need half the thermal resistance), double the transformer VA rating, double any secondary fuses, and halve the value of any resistors in series with the input or load, and you've got a LM350 project.

author: bitman
author_rank: Unknown
timestamp: November 25, 2017, 10:13:00 pm
content: Quote from: iainwhite on November 25, 2017, 09:40:51 pmWell, I don't have any bd536sChecking the specs to see what I may have close to it.

author: Ian.M
author_rank: Super Contributor
timestamp: November 25, 2017, 10:43:10 pm
content: It doesn't even need to be close as long as it can handle the load current and input voltage and has a high enough power rating. Check the S.O.A. (safe operating area) plot (Ic vs Vce) in the transistor's datasheet, specifically the DC line, not any of the pulsed time lines.for what it can handle. Anywhere under the line with enough margin is good, after derating for temperature: The permissible current for all lines drops proportionately by multiplying by the derating factor(Tj-25°C)/(Tjmax-25°C). The factor is always less than one, and drops to zero at Tjmax. Below Tj=25°C. you dont derate. If there isn't a DC S.O.A line, in the most comprehensive datasheet you can find, you need to change transistors to one with a DC S.O.A ine.You can even use a NPN power transistor - but you need a lower current PNP to drive it to make it work like a PNP. Seehttps://en.wikipedia.org/wiki/Sziklai_pair.

author: bitman
author_rank: Unknown
timestamp: November 25, 2017, 11:42:08 pm
content: Ian.M - I'll pretend I understood the bulk of what you wrote there and look for a way to use my TIP31G (NPN) that I do have in a circuit like this. As I stated, I have a lot to learn which is why I was looking for a complete project that would cover the basics. Thanks though - I'll go back to this as a reference once I get "wiser".

author: mariush
author_rank: Super Contributor
timestamp: November 26, 2017, 12:07:05 am
content: LM1085 can do 3A and it's super simple to use.You can get the fixed 5v version for 2$, here's a link :https://www.digikey.com/product-detail/en/texas-instruments/LM1085IT-5.0-NOPB/LM1085IT-5.0-NOPB-ND/363564Or you can get the adjustable version for which you can set the output voltage with 2 resistors (costs the same, 2$) :https://www.digikey.com/product-detail/en/texas-instruments/LM1085IT-ADJ-NOPB/LM1085IT-ADJ-NOPB-ND/363565You'll need a heatsink for it, because the regulator will dissipate the difference between input voltage and output voltage as heat on the regulator.So from the heatsinks section, these will do :https://www.digikey.com/short/q3rmb10.4$ :https://www.digikey.com/product-detail/en/assmann-wsw-components/V2006B/A10761-ND/34761550.85$:https://www.digikey.com/product-detail/en/advanced-thermal-solutions-inc/ATS-PCB1049/ATS2091-ND/50304691.5$ :https://www.digikey.com/product-detail/en/aavid-thermalloy/529802B02500G/HS350-ND/1216357All kinds of them in lots of sizes. Don't buy the smallest, look at the technical specs and buy something with a bit of surface.You'll need a bridge rectifier to convert your AC voltage from the transformer to DC voltage. You could use 4 diodes to create one, but a bridge rectifier is sexy. You want 2A max, but the price difference between models is so cheap you may as well go with something bigger :0.78$ 600v 6A :https://www.digikey.com/product-detail/en/smc-diode-solutions/GBU606TB/1655-1852-ND/72448800.93$ 600v 10A :https://www.digikey.com/product-detail/en/smc-diode-solutions/GBU1006TB/1655-1843-ND/7244871So the linear regulator needs around 1v above 5v to output up to 2A and around 1.5v above 5v to output 3A, so your linear regulator needs to "see" at least 6.5v DCYou'll convert the ac voltage from a transformer using a bridge rectifier which will give you a peak dc voltage of 1.414 x the AC voltage from the transformer, minus 2 times the voltage drop on one diode from your bridge rectifier - that's gonna be around 1v.You're probably going to go for a 9v AC or 10v AC transformer, since these are super common and cheap, and that means your peak DC voltage will be 1.414 x 10v - 2 x 1v = ~ 12v DCHow big the transformer should be ... well, you can sort of estimate with the formula DC Current = 0.62 x AC Currentso for example, if you have a 25VA transformer, that means it can do 25VA / 10V ac = 2.5A ... and after rectification with a full bridge rectifier you'll have around 0.62 x 2.5A = 1.55A ... so you need a slightly bigger transformer.The cheapest on Digikey is 13$ for a 10v AC 30VA transformer :https://www.digikey.com/product-detail/en/triad-magnetics/F6-10/237-1699-ND/4878628This means your peak DC voltage will be around 12v DC ( 1.414 x 10v - 1.5v.. 2v )So now you only need a big capacitor after the transformer, so that the minimum voltage the linear regulator sees is always above 6.5v .. but just be safe and go with 7vYou can estimate the capacitance with the formula :C = Current / [ 2 x mains frequency x (peak dc voltage - minimum desired voltage) ]So your current is 2A, you're in US so you're 60Hz mains frequency, your peak dc voltage is 12v and minimum desired voltage is 7v : C = 2 / [ 2 x 60 x (12-7) ] = 2/600 = 1/300 = 0.003333 Farads or 3333uFSo you could go with a 3300uF 25v capacitor but 4700uF would also work.It doesn't have to be super expensive, doesn't have to be thousands of hours at 105c rated, here's some (3300uF to 4700uF 25-50v rated, sorted by price) :https://www.digikey.com/short/q3rm2dYou may want to add a 1uF 50v or higher voltage rating ceramic capacitor on the input of the linear regulator (as close as possible to the pins) as well as a regular electrolytic capacitor on the output of the linear regulator, let's say 47uF or 100uF 10v or higher voltage rating on the output (they're so small and barely any price difference you could go with 25v or 35v rated capacitors on the output)Job done.When you're done, you'll notice a lot of energy is wasted as heat on the linear regulator .. you have up to 12v in, only 5v out .. so your efficiency may be below 50%. Then you can go on your next project, replacing your linear regulator with a switching regulator. There's loads of them out there.

author: Ian.M
author_rank: Super Contributor
timestamp: November 26, 2017, 12:52:19 am
content: Quote from: bitman on November 25, 2017, 11:42:08 pmA TIP31G has a Tjmax of 150°C. Its S.O.A. chart indicates its good for its full 3A Ic, (before derating) up to about 13V Vce. Assuming you can keep its junction under 87°C it will be good for 1.5A. You can make up the rest in the regulator, or parallel two TIP31Gs with 0.22R current sharing resistors in series with each individual emitter. Its Rth_JCis 3.125 °C/W, so, assuming a big-ass heatsink that you can keep under 55°C, it can handle 10.4W, which at 1.5A is fractionally under 7V drop between the average input voltage and the output voltage. That should be just enough for regulator headroom and ripple, though using two TIP31Gs would be a better choice, as the power is split between them and the current reduced so the maximum permissible voltage drop will be much higher and you can use a smaller heatsink. Note that all the boosted regulator circuits need a minimum of 0.7V extra headroom (minimum voltage between in and out) over the bare regulator, because of the drop across the resistor driving the external pass transistor. Also note, if they are going on the same heatsink, you'll need an insulated mounting kit for the regulator, because its tab isn't at the same potential as the TIP31G tab which is connected to the collector, so the whole heatsink will be sitting at Vunreg. Putting thermal pads under the transistor to have an isolated heatsink (e.g. because you want it exposed out of the case back) would increase the thermal resistance too much - you'd need a copper heat spreader to use a larger area thermal pad to make it work and still need a monster heatsink, and at that point it would be cheaper to use either two TIP31Gs or a beefier transistor.Its minimum HFEis 10 @Ic=3A, and 25 @Ic=1A. We need a reasonable extrapolation for Ic=1.5A. Figure 8. DC Current Gain in itsdatasheetshows the typical HFEdropping from 65 @Ic=1A to 45 @Ic=1.5A and 25 @Ic=3A. 65/25 is 2.6 and 25/10 is 2.5, near enough the same ratio. 45/2.6 is 17.3, so we'll design to a minimum HFEof 17 @Ic=1.5ATherefore, to use it in a Sziklai pair, for 1.5A load current you need a PNP transistor to drive its base, that can handle at least 1.5A/17 which is just under 90mA (+ handle the dissipation for the full voltage drop). If you parallel two TIP31Gs at Ic=1A each, you need 2*1A/25, which is 80mA so the base drive requirement actually drops! Its actual gain is probably quite a bit higher (see Fig 8 again), so for a one-off, if you are prepared to select on test, you could get away with a wimpier PNP driver transistor than sound design principles would require.What small to medium PNP transistors do you have handy?

author: David Hess
author_rank: Super Contributor
timestamp: November 26, 2017, 03:41:14 am
content: Higher current and power fixed 5 volt linear regulators like the LM323 used to be common and inexpensive but now the 3 amp LM350, 5 amp LM338, or their improved replacements are a better option.Multiple integrated regulators can be configured to share the current load which has power dissipation and stability advantages however it requires an operational amplifier to enforce current sharing.I like adding an external PNP power pass transistor or an NPN power pass transistor with a negative regulator. If you do it right as shown below from the National Semiconductor 1980 Voltage Regulator Handbook, the integrated regulator's built in current limit can protect the power pass transistor.Update: Had the wrong scan attached.

author: bitman
author_rank: Unknown
timestamp: November 26, 2017, 03:48:22 am
content: Quote from: Ian.M on November 26, 2017, 12:52:19 amNone useful. I got a handful of 2N3905s and similar transistors. I have some TIP transistors on order of which some are PNPs - but it will be quite a while before they arrive (if I'm lucky before Christmas). I've spent most my time using a lab supply for 5v and 200mA tops for a very long time so I never got "stuff" for higher amps/voltages. A few old grab bags had some higher caps but no transistors that I can use. Well and the fact that I haven't identified but a bit more than 50 % of what's in those bags it may very well be I have some but in a different package. From my best efforts I don't think I have any. It really strikes me how many components have NO markings what so ever on them?Thanks again - reading the replies here is making me understand some important relationships I weren't aware of. Such as the size of the caps required. Btw. I also don't have resistors less than 1ohm and having spent quite a bit of today looking at tons of diagrams it seems I need some 0.x resistors. Note, that most of the stuff I have is "cheap" ie. highly inaccurate components (10-20% off). From what I read that will also be a problem when it comes to designing simple power supplies, so I probably need to get ALL components needed instead of just getting a few things I am/was missing.It's interesting to see the "tricks" you can play to deal with heat and efficiency.Ohhh - and I stumbled across this https://www.digikey.com/en/product-highlight/t/texas-instruments/webench-design-center ... a bit too much information for a sort like me, but it sure does provide some food for thought.

author: Ian.M
author_rank: Super Contributor
timestamp: November 26, 2017, 04:21:01 am
content: 2N3905 is a little wimpy, but will probably be fine driving the base of a TIP31G in a Sziklai pair configuration to pass up to 1.5A as above.Q1 is the 2N3905, Q2 is the TIP31G, then you can use the combo as the power PNP in the regulator datasheets external pass transistor circuits. N.B. the simple ones don't have a current limit so don't short them or pop goes the external pass transistor(s), and when you remove the short, pop goes your 5V load as the pass transistor will now be a dead short.

author: bitman
author_rank: Unknown
timestamp: November 26, 2017, 04:36:23 am
content: Quote from: Ian.M on November 26, 2017, 04:21:01 amWhen you say wimpy are you talking about the total current? I have most of the ones in this set remaining:https://www.amazon.com/gp/product/B01LXA3O1V.How do you easiest compare components btw? It's quite long winded to look up every component's data sheet to compare properties? Are there easier ways where you can specify a list and see how they compare?

author: Ian.M
author_rank: Super Contributor
timestamp: November 26, 2017, 05:51:57 am
content: My objection was the total power dissipation. The driver Q1 sees the total voltage drop - Q2 Vbe, and may, worst case have to pass 90mA if Q2's gain is right on the lower limit. Its maximum power rating is 625mW. 625mW/90mA is fractionally under 7V, so if the input to output drop goes over about 7.5V it may burn out Q1. If so, to be 100% safe, you'd need to look for one rated more than 625mW. A good clue would be a PNP with a larger package than the usual TO-92 small plastic D shaped one. However as the typical TIP31G gain is more than double its worst case gain, odds are the 2N3906 will have plenty of margin, so if you can keep your finger on it without burning at full load current, it will do for a one-off.The Old-Skool way to quickly find transistor comparison data was to look it up in Towers International Transistor Selector - a dead-tree book that listed most current and recently obsolete transistors in alphanumeric order of part number, with key parameters in a tabular format, and when the part was fairly generic, a brief list of possible substitutes. There's a scan of an old 1975 copy at archive.org - Google for it, but newer editions are much harder to find. There were similar books from other countries - the German ECA vrt books are particularly comprehensive.Nowadays you use the parametric search tools on most big international distributers websites or tryhttps://alltransistors.com/.N.B. for many Japanese transistors in through hole packages, the part number printed on the package starting with a letter is only part of the full part number, and it will often need 2S added in front. See:https://en.wikipedia.org/wiki/JIS_semiconductor_designation.When you've looked one up, if you've got stock, write a brief summary of the type, key parameters (Vceo/Vce, Icmax, Power, HFE), and typical application on your stock card or draw label.N.B. always double check the real datasheet when you find a candidate transistor - the selector guides tend to have a few mistakes in them.

author: David Hess
author_rank: Super Contributor
timestamp: November 26, 2017, 04:47:35 pm
content: In the past, a TO-39 part (the 2N3053 was popular) would have been used as the driver and there were a lot of good ones but that is no longer the case. Today, a TO-126 or TO-225 packaged transistor could be used. It might be more economical to use a pair of TO-92 packages transistors in parallel and I have seen that done even when TO-39 parts were available.It is a good idea to implement current limiting of the driver transistor by adding a collector or emitter resistor and maybe shunt diodes to draw off the base current during overload. In the past, sometimes a low value low power carbon composition resistor was used as a fuse. Today they make "fusible" resistors which are not suppose to fail with a short.

author: MK14
author_rank: Super Contributor
timestamp: November 26, 2017, 04:56:36 pm
content: https://archive.org/details/TowersInternationalTransistorSelectorIf you scroll down a bit, it lets you download PDFs and stuff.

author: bd139
author_rank: Super Contributor
timestamp: November 26, 2017, 05:47:03 pm
content: Art of electronics chapter 9, which happens to be the free sample:https://artofelectronics.net/wp-content/uploads/2016/02/AoE3_chapter9.pdfThis covers linear and switching power supply design including thermal side of things, foldback current limiting (handle shorts without blowing up), crowbars (stop overvoltage destroying the device you are powering if something fails) etc etc.I would personally, if I was to build one, build a linear supply for bench use. Transformer + rectifier + filter cap + LM723 + MJE3055T pass transistor + SCR based crowbar. Or a single LT1084.But I'd probably buy one. It's way more expensive to build versus buy here. If you're lucky you might get a canned Farnell or Lambda supply from ebay with the above characteristics.

author: David Hess
author_rank: Super Contributor
timestamp: November 26, 2017, 06:25:54 pm
content: With a design where the error amplifier is external, it is easy to implement remote sense to the binding posts for much better load regulation. But integrated designs are competitive when you consider the leads between the binding posts and circuit.The integrated regulator plus a discrete power transistor is by far the simplest way to make a blow out proof design though.You can learn a lot building power supplies like these but if you want to save time, refurbishing a used HP or similar could be better. I would not bother buying anything new.

author: bitman
author_rank: Unknown
timestamp: November 26, 2017, 06:29:16 pm
content: Quote from: bd139 on November 26, 2017, 05:47:03 pmNo debate here - it's time consuming and expensive. However, the goal isn't to have a power supply (well, I need it but still) it's to learn the core concepts here like power/heat issues and options. I've read plenty about liniar vs. switching supplies but that doesn't help me understand how to size the components (or know which ones are easy replacing others etc).As an example (and this isn't an ask for answers), when I read the datasheet for the lm78xx I see formulas referring to Iq Ireg, Bq and Vbeq - I have NO clue what those refers to (Well, I get "I and V" but not the designation under them). It quite frankly wasn't clear to me how big a difference even a small difference in a resistor can make in these circuits. Which to me is a good thing - I love the learning quest.I've got plenty of books and beginner projects books, but most are simple arduino type stuff ie. external power supply and plug a module here, there and code. I know code - code isn't a challenge. I don't get the physical hardwired stuff and finding good books that guide you through that is well HARD. They're either nothing but theory or so simple you don't learn anything. At least when I watch Dave's posts I can pick up stuff every time - but finding a reason/place to apply it is well, HARDIt's beyond my capabilities right now but it's still learning.

author: tecman
author_rank: Frequent Contributor
timestamp: November 26, 2017, 06:35:56 pm
content: LM323 is a cheap, simple solutionpaul

author: bitman
author_rank: Unknown
timestamp: November 26, 2017, 06:49:41 pm
content: Quote from: tecman on November 26, 2017, 06:35:56 pmAt US$40 a pop??

author: David Hess
author_rank: Super Contributor
timestamp: November 26, 2017, 06:57:15 pm
content: Quote from: bitman on November 26, 2017, 06:49:41 pmExactly, LM323s were inexpensive once but no longer although you might find one surplus for a reasonable price. That just leads back to using an LM350, LM338, or one of the various improved TO-220 packaged adjustable regulators of which there are many.One reason I like the idea of using a 7805 or 317 with an external transistor is that the current limit can be tailored to the right level.

author: bitman
author_rank: Unknown
timestamp: November 26, 2017, 08:11:23 pm
content: 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

author: bitman
author_rank: Unknown
timestamp: November 26, 2017, 09:46:41 pm
content: 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*).

author: Ian.M
author_rank: Super Contributor
timestamp: November 27, 2017, 02:01:44 am
content: 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.

author: bitman
author_rank: Unknown
timestamp: November 27, 2017, 05:12:50 am
content: Quote from: Ian.M on November 27, 2017, 02:01:44 amHmmm - 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.QuoteI'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 ....

author: bitman
author_rank: Unknown
timestamp: November 27, 2017, 07:14:42 am
content: 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/CKJW4lYVqHjWiCWo1Measuring after the resistor:https://photos.app.goo.gl/b6miXXMSuuX9Mxds1This 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.

author: MK14
author_rank: Super Contributor
timestamp: November 27, 2017, 11:34:32 am
content: 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).

author: bitman
author_rank: Unknown
timestamp: November 27, 2017, 03:20:51 pm
content: Quote from: MK14 on November 27, 2017, 11:34:32 amInteresting 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.QuoteWouldn'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 ....QuoteI 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.QuoteAhh 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 reasonThanks for the suggestion - I hadn't thought of testing the PNP at all.

author: MK14
author_rank: Super Contributor
timestamp: November 27, 2017, 03:38:49 pm
content: 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;drSomething is a bit out with your results or my interpretation of them.

author: MK14
author_rank: Super Contributor
timestamp: November 27, 2017, 03:47:37 pm
content: Summary:So you have 10V coming in to the circuit.The PNP's base is around 9.3VThe output voltage SHOULD be at around 5VThe PNP's collector should also be at around 5VThe PNP's emitter should be at 10VHence there is around5V4.3V between the input and output of the LM7805 (which is fine).

author: Ian.M
author_rank: Super Contributor
timestamp: November 27, 2017, 04:14:05 pm
content: 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.

author: bitman
author_rank: Unknown
timestamp: November 27, 2017, 04:16:36 pm
content: Quote from: MK14 on November 27, 2017, 03:47:37 pmI 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).

author: MK14
author_rank: Super Contributor
timestamp: November 27, 2017, 04:24:32 pm
content: Quote from: bitman on November 27, 2017, 04:16:36 pmI 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;drIt 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.

author: bitman
author_rank: Unknown
timestamp: November 27, 2017, 04:31:02 pm
content: Quote from: Ian.M on November 27, 2017, 04:14:05 pmI 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?

author: bitman
author_rank: Unknown
timestamp: November 27, 2017, 04:35:21 pm
content: Quote from: MK14 on November 27, 2017, 03:47:37 pmHmmm - 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.

author: bitman
author_rank: Unknown
timestamp: November 27, 2017, 04:41:05 pm
content: Quote from: MK14 on November 27, 2017, 04:24:32 pmI 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.

author: MK14
author_rank: Super Contributor
timestamp: November 27, 2017, 04:42:55 pm
content: Quote from: bitman on November 27, 2017, 04:35:21 pmSorry, I was mistaken. When I wrote it, I had tried to remember the circuit, but couldn't see it while typing the response.R1 3is 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).

author: MK14
author_rank: Super Contributor
timestamp: November 27, 2017, 04:48:48 pm
content: Quote from: bitman on November 27, 2017, 04:41:05 pmAssuming 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.

author: Ian.M
author_rank: Super Contributor
timestamp: November 27, 2017, 04:54:20 pm
content: Quote from: MK14 on November 27, 2017, 04:24:32 pmYep. 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 Williamsstyle. For even more complex or critical circuits you can transition intoManhatten styleconstruction, 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 seehttps://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.

author: bitman
author_rank: Unknown
timestamp: November 27, 2017, 05:06:59 pm
content: Quote from: MK14 on November 27, 2017, 04:48:48 pmSure - 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.

author: Ian.M
author_rank: Super Contributor
timestamp: November 27, 2017, 05:16:10 pm
content: 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.

author: MK14
author_rank: Super Contributor
timestamp: November 27, 2017, 05:19:44 pm
content: Quote from: bitman on November 27, 2017, 05:06:59 pmThen 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;dr1.5 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.

author: bitman
author_rank: Unknown
timestamp: November 27, 2017, 05:33:19 pm
content: Quote from: Ian.M on November 27, 2017, 05:16:10 pmI've run at much lower loads and at those I see the expected results.QuoteI 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.

author: bitman
author_rank: Unknown
timestamp: November 27, 2017, 05:43:51 pm
content: Quote from: MK14 on November 27, 2017, 05:19:44 pmSure - 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.

author: MK14
author_rank: Super Contributor
timestamp: November 27, 2017, 05:48:11 pm
content: Quote from: bitman on November 27, 2017, 05:43:51 pmGood 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!

author: Ian.M
author_rank: Super Contributor
timestamp: November 27, 2017, 06:22:04 pm
content: Quote from: bitman on November 27, 2017, 05:33:19 pmA 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.

author: bitman
author_rank: Unknown
timestamp: November 27, 2017, 08:01:39 pm
content: Quote from: Ian.M on November 27, 2017, 06:22:04 pmDUH!! 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.

author: bitman
author_rank: Unknown
timestamp: November 28, 2017, 06:02:43 am
content: Quote from: MK14 on November 27, 2017, 05:48:11 pmThanks - 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.

author: MK14
author_rank: Super Contributor
timestamp: November 28, 2017, 07:02:40 pm
content: 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 3resistor.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 WattsSo it should run fairly cool. EDIT: Less than 1.1 Watts in practice, as some of the heat will be dissipated in the 3resistor.Current passing through TIP42C = 2 Amps (from title of thread, and ignoring the relatively small current through the 3resistor)Voltage across TIP42C = input voltage - output voltage = 5 voltsIgnoring the slight heating effects of the base currentTIP42C disipation = voltage across it x current passing through it = 5 x 2 = 10 WattsThermal resistance of TIP42Chttp://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 CIt 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 wattThermal compound will add about 0.8 deg C per watt or so, so call it 1, final total = 8 deg C per WattTotal dissipation = 10 WattsRoom temp = Ambient = 25 deg CExpected final temperature = 25 + (8 x 10) = 105 deg C TIP42C junction temperature with a 5 deg C heatsink + thermal compoundHeatsinks 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. 0.1or 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/3511421To get the best life and reliability. I prefer to use an even bigger heatsink (higher rated one). But peoples opinions on this can vary.