Liniar adjustable power supply - Page 1
Forum section: electronics / general

author: NewBeginner
author_rank: Regular Contributor
timestamp: July 11, 2012, 02:41:50 pm
content: Hi everybody,In my process of learning electronics I intend to build my own linear power supply.To limit the power dissipation of the linear voltage regulator (by limiting the voltage drop over the regulator) I was thinking of using a transformer (probably a toroidal one) with multiple taps and based on the required voltage to switch between the taps using relays.I was thinking to the following specs for the power supply:Adjustable output voltage: 1.2 - 24VMax output current: 2ASo my question is: is this solution for limiting the power dissipation a good solution? I would prefer not to use switch mode pre-regulators.Thank you,NewBeginner

author: digsys
author_rank: Unknown
timestamp: July 11, 2012, 03:06:12 pm
content: It's a VERY widely used method. If you go that way, why not look at Triacs instead of Relays?Cleaner switch and very fast. Also, you need a small say 6VAC winding to run the control logic ... in case of short cctyou want some control. Another option24V 2A is NOT a lot of power to get rid of - maybe use a TO3 package, large heatsink and a thermostatic fan?Worst case, only have 1 tap mid way.

author: shebu18
author_rank: Frequent Contributor
timestamp: July 11, 2012, 05:29:33 pm
content: @digsys: That sounds interesting, could you show a circuit where you can switch, lets say between a tap of 5V and one of 12V? THis is where i would use a relay.

author: Chasm
author_rank: Regular Contributor
timestamp: July 11, 2012, 07:09:15 pm
content: To reduce the dissipated energy you need a transformer with multiple windings. You can usually buy transformers with two secondaries, say 2x18V for 30V.Switching the input voltage with comparator and relay (or thyristors) is an option.For two voltages input voltages using pnp and npn transistor pairs is another nifty alternative.Let's see if I can find the schematic.

author: NewBeginner
author_rank: Regular Contributor
timestamp: July 11, 2012, 07:30:27 pm
content: Hi,@Chasm What would be the disadvantage of using a single secondary winding with multiple taps on it?Thank you

author: Chasm
author_rank: Regular Contributor
timestamp: July 11, 2012, 08:00:25 pm
content: That they are not widely available, at least here in Germany.Both with multiple windings and taps you have to take a look at the specs of the transformers. It is very much possible that different windings of the same voltage are build for different currents.

author: NewBeginner
author_rank: Regular Contributor
timestamp: July 13, 2012, 06:12:46 pm
content: Thank you all for your help

author: efron
author_rank: Contributor
timestamp: July 13, 2012, 09:12:27 pm
content: Sometimes the fact to accept some limitations on the technical specification can save a lot of thermal problems.For example, do you really need 24Vdc output?Do you really need 2A guaranteed all along the output range (from 1.24V to 24V)?All this can make the difference.If your down limit is 1.24V, I guess you're willing to use the famous LM317 or LM350. For a 2A output you should use the metal case option (LM350).The LM350 regulator needs about 2-2.5V min between the IN and OUT pins and can support a maximum power of less than 100W if suitable heatsink is used.If i come back to my questions about the specifications of your power supply ...If you need 24Vdc output, you must have at least 27V available at the IN pin of the LM350. Under these conditions, if your output is required to the minimum 1.24V and 2A maximum current , the LM350 will dissipate the maximum power = (27V-1.24V)*2A = 51W.If used in continuous conditions, your heatsink will have to be of 0.46 °C/W !!! --> This is possible but you will be surprised of their size and price!Although if you accept to limit your output to 12Vdc, the maximum required power is decreased by a factor of 2 (25W). In this case the heatsink to be used must be of about 2.5°C/W (about 10cm tall, about 10Euros in Europe).Or maybe if you say that you will never use 2A when delivering 1.24V you could use smaller heatsink (but your power supply will not support over-specification).I've just finished my linear power supply and it took me a lot of time thinking about all this power and heat dissipation stuff. Finally, I forward you my power supply's specifications and characteristics:* Output voltage: from 1.25 to 15Vdc* Output current (guaranteed): from 0 to 1.8A (includes current limit to this value but can be easily adapted to any value)* It uses one single secondary transformer 50VA* It uses one LM350 (metal case) with heatsink about 2.5°C/WIt is working quite well (for sure not so well as the one done by Dave's and other's in this forum).I'm working now on the LCD panel interface.Did you start your power supply?

author: Zero999
author_rank: Super Contributor
timestamp: July 15, 2012, 12:06:30 am
content: You could use a centre tapped transformer with a transistor to connect to the full voltage when the output is set past a certain voltage.

author: NewBeginner
author_rank: Regular Contributor
timestamp: July 16, 2012, 11:54:39 am
content: Hi,Thank you all for your help. Sorry for the delayed reply.@efronI haven't started to build my supply yet. I am still researching and trying to learn more about this and other topics. For the moment I built a standard LM7805 based 5V supply to power my microcontroller projects.I came up with this specs mainly for the following reasons:- I thought that maybe in the future I will need the 24V and the 2A for different projects (maybe to drive some stepper motors, or something like that). So having the full 1.2-24V range seemed a good idea.- because of the high power dissipation involved I thought that this may be a good way of learning new things. I set this goal hoping that in the process I will learn more about the design of power supplies.But I am always opened to new ideas, suggestions, comments or any help you can give. Anything that can help me in my learning process is more than welcomed.Also thank you for sharing your design with me.@Hero999I don't fully understand ( yet) the circuit you posted (I am still learning about transistors and zener diodes) so if you could explain it in more details I would really appreciate it. Especially the zener part.I also have a question. What is the role of the R2 resistor?Thank you

author: Chasm
author_rank: Regular Contributor
timestamp: July 16, 2012, 07:59:24 pm
content: I'll try.Step one to understand this circuit is to unrevel the rectifier part. You can find it in several linear lab power supplies.To make it short, the voltage at D1 is half the output voltage. Or to say it in another way the voltage at the collector of T1 is twice the voltage at D1.It follows directly that as long as T1 is not conducting the input voltage of voltage regulator is only half as high. This means that you only need half the power dissipation in the LM317. - Which is the goal of this circuit and using a tapped transformer as discussed before. =)So how to drive T1 in a manner that achieves this goal?T1 is a PNP tansistor, it is driven by a negative voltage. (Base emitter voltage) Since the collector voltage is twice the emitter voltage in this circuit ground is quite negative in relation to the base.So we can drive T1 by pulling the base to ground. We want to do so only when we actually need more voltage at the output. So we use the output voltage to drive a the NPN Transistor T2. The higher the output voltage the more current through the base of T2 and thus more drive level for T1.But.The goal is not to drive T1 all the time, (if possible) only when the LM317 needs more input voltage.Zener diodes only conduct current in reverse when the voltage across them is higher than their rated value.Transistors are driven by current. By picking the right Zener voltage we can set a output voltage level under which T2 will not be driven. (Substract the Zener voltage from the -adjustable- output voltage to to calculate the base current driving T2.)Was that understandable or total glibberish?

author: efron
author_rank: Contributor
timestamp: July 16, 2012, 08:30:38 pm
content: QuoteChasm, the purpose of R2 isn't clear for me neither. With this resistor, it seems that the driving transistor (the one on the bottom in your drawing) will always be conducting because of the positive voltage at the IN pin of the regulator and R2 path through its base.Why not to remove it!

author: NewBeginner
author_rank: Regular Contributor
timestamp: July 16, 2012, 09:08:31 pm
content: @ChasmIt starts to make sens now. However I am still confused a little about the zener diode. Doesn't it require a certain amount of current in order to get the full 12V voltage drop across it? What if the current through the zener diode is smaller than the required one?Related to the R2 resistor: I read that zener diodes have a certain leakage current. This may be a silly question, but is it possible that the role of R2 is to prevent the leakage current from the zener to be amplified by the NPN transistor and thus to prevent TR1 from turning on when it shouldn't? Or does it have a completely different role?Thank you

author: NewBeginner
author_rank: Regular Contributor
timestamp: July 16, 2012, 09:49:21 pm
content: This may be another silly question, but if R2 should handle the leakage current, shouldn't it connect to ground instead of the collector of TR1? Hmm, now I am confused

author: Chasm
author_rank: Regular Contributor
timestamp: July 17, 2012, 02:43:27 am
content: Z diodesYou are right. To actually get the rated voltage we need a minimum current. Grab a datasheet and take a look at the current voltage diagram of Z diodes with different voltages. You will see that-esp. for low voltage versions-the reverse voltage is quite dependent on a minimum current while higher voltage versions have a much steeper curve..If you want a stable voltage, which is usually the case, you need a minimum current through the diode. (While limiting the maximum power at the same time.)In this case however we use the z diode to subtract voltages - and close enough is close enough.Lets do a worst case consideration:Say the current is so low that the voltage drop is 1V less than rated. (Which is unrealistically high) What happens? Does our circuit still work?As a result the pnp transistor will be driven earlier. The circuit will still work but more power has to be dissipated at the LM317.Realistically this is not so much of a problem as it may seem, we need some current to drive the transistor T2. We also need some headroom in the input voltage of the LM317.R2I hope that I'm not talking out of my ass (5am, no coffee yet) but R2 should provide negative feedback for T2. Without it the gain is too dependent on part variance.

author: NewBeginner
author_rank: Regular Contributor
timestamp: July 18, 2012, 07:25:32 pm
content: About the zener diode:You are right. I looked at the datasheet for a 12V zener and indeed it has a much steeper current-voltage curve.R2:I was hopping it was used because the zener diode leakage current.I still have much to learnThank you