Center-tapped transformer question - Page 1
Forum section: electronics / general

author: NewBeginner
author_rank: Regular Contributor
timestamp: July 19, 2012, 05:10:39 pm
content: Hi,I have another beginner question. Is it possible, with a center-tapped transformer to use both taps of the secondary in the same time? I mean to have loads on both of the taps? Both taps referenced to the same point. Like in the attached image?If it is possible ... is it recommended?Thank you

author: T4P
author_rank: Super Contributor
timestamp: July 19, 2012, 05:17:44 pm
content: Yes, why not.That's like buying a ferrari to race, but only using half the horsesBut the center point is usually the 0VBut R1 is taking full voltage while R2 takes half voltage the way you wired

author: G7PSK
author_rank: Super Contributor
timestamp: July 19, 2012, 06:09:11 pm
content: No problem at all if you need full voltage and half voltage at the same time. But remember you can only load the winding's to their rated total load still.

author: Simon
author_rank: Unknown
timestamp: July 19, 2012, 06:36:45 pm
content: as that is wired you will not be able to use the 2-3 part to full power because you are also using some of the 1-2 section power which also has a loan on it. Basically R1 will have a higher voltage than R2

author: nitro2k01
author_rank: Frequent Contributor
timestamp: July 19, 2012, 08:02:08 pm
content: Of course you can use both the windings. Why else would they be there?But, like you're admitting, you're a beginner, and I think there's a slight risk you'll mess something up for other reasons. So I would encourage you to describe what you're actually trying to build so we can lodge you in the right direction. Chances are you don't have a clear mental image of what you're doing.

author: David_AVD
author_rank: Super Contributor
timestamp: July 19, 2012, 09:30:44 pm
content: No problem with using the transformer like that. It's commonly used to derive 2 different voltages from the one transformer.But, as other have alluded to, the "bottom" winding will carry the current for both loads so take that into account.

author: NewBeginner
author_rank: Regular Contributor
timestamp: July 23, 2012, 02:20:47 pm
content: Hi,Thank for the help.I am trying to build by own linear power supply (and learn about electronics in the process).So the R2 load represents some control circuitry for the power supply. This will always be connected when the power supply is on.R1 represents some powered device/circuit. R1 will be connected either in parallel with the control circuitry (in order to reduce power dissipation of the voltage regulator ) when a smaller voltage is required or if a higher voltage is required, the control circuitry will switch R1 from tap 3 to tap 2.I don't want to use a switching pre-regulator at this time.So, one more question comes to my mind: how should I do the rectification in this case?Thank you all

author: MikeNye
author_rank: Contributor
timestamp: July 23, 2012, 03:58:46 pm
content: Hi,You could use two rectifiers - one for the higher voltage tap, and one for the lower voltage tap. Something like this:http://www.falstad.com/circuit/#%24+1+5.0E-6+10.20027730826997+50+5.0+50%0Av+80+128+80+192+0+1+50.0+240.0+0.0+0.0+0.5%0A169+112+128+160+128+0+4.0+0.1+-0.02191920043572424+-6.418476861114186E-15+-6.3976601794024646E-15%0Aw+80+192+112+192+0%0Aw+80+128+112+128+0%0Aw+176+192+160+192+0%0Aw+176+160+160+160+0%0Aw+160+128+224+128+0%0Ag+224+128+224+144+0%0Ad+176+272+224+224+1+0.805904783%0Ad+224+224+272+272+1+0.805904783%0Ad+176+272+224+320+1+0.805904783%0Ad+224+320+272+272+1+0.805904783%0Aw+272+272+288+272+0%0Aw+176+192+224+192+0%0Aw+224+192+224+224+0%0Ag+224+320+224+336+0%0Aw+176+272+176+368+0%0Aw+176+368+288+368+0%0Aw+288+272+352+272+0%0Aw+288+368+352+368+0%0Ax+415+329+471+335+0+24+Load%0Aw+352+272+384+272+0%0Aw+352+368+384+368+0%0Ac+288+272+288+368+2+2.1999999999999998E-4+22.30306161728712%0Ar+384+272+384+368+0+300.0%0Aw+176+160+272+160+0%0Aw+272+160+272+112+0%0Aw+272+112+352+112+0%0Ad+352+112+400+160+1+0.805904783%0Ad+304+160+352+112+1+0.805904783%0Ag+352+208+352+224+0%0Aw+304+160+304+240+0%0Ad+352+208+400+160+1+0.805904783%0Ad+304+160+352+208+1+0.805904783%0Aw+304+240+416+240+0%0Aw+400+160+416+160+0%0Ac+416+160+416+240+2+2.1999999999999998E-4+10.49766315547102%0Aw+416+160+512+160+0%0Aw+416+240+512+240+0%0Ar+512+160+512+240+0+300.0%0Ax+544+208+600+214+0+24+Load%0Ao+39+64+0+35+40.0+0.1+0+-1%0Ao+24+64+0+35+40.0+0.1+1+-1%0AIf you're a beginner, I wouldn't recommend hooking up your own transformer to the mains supply, as this can be very dangerous. Why not go dumpster diving for an old laptop power supply, or purchase a pre-made switchmode power supply module that outputs 24V at a few amps? You can feed this into your linear supply. This would be much safer IMHO. :-)I hope this helps.Kind regards,Mike Nye

author: NewBeginner
author_rank: Regular Contributor
timestamp: July 23, 2012, 05:26:32 pm
content: Thank you for the link to the simulation. I was thinking to the same setupbut I wasn't sure if it is ok. Are there any drawbacks for this setup?I have another question: the control circuit needs to have the same "common" as the load. Is this already achieved by the fact that the bridge rectifiers are referenced to the same transformer tap (common/ground) or the "-" output of the bridge rectifiers should be tied together to achieve this? I don't have (yet) a clear image in my mind about this.QuoteYes, it would probably be much safer. But I think I would learn less that way; and my goal is to learn.I can assure you that I will take the necessary precautions when I will work with the mains power.Anyway I will not power up the circuit until I have a clear image in my mind and I fully understand what is going on.Thank you

author: NewBeginner
author_rank: Regular Contributor
timestamp: August 12, 2012, 08:53:08 pm
content: Hi,This may be a foolish question but I would really appreciate it if somebody could clarify this for me:QuoteThank you

author: NewBeginner
author_rank: Regular Contributor
timestamp: August 14, 2012, 09:01:25 am
content: Nobody?

author: David_AVD
author_rank: Super Contributor
timestamp: August 14, 2012, 09:47:56 am
content: I suspect nobody really understood your question. (I don't)Maybe post a diagram of what you're proposing?

author: NewBeginner
author_rank: Regular Contributor
timestamp: August 15, 2012, 07:21:36 pm
content: I tried to achieve the circuit shown in the attached image (simplified).After lots of thinking I realized that my question was already answered by userHero999in the following thread:https://www.eevblog.com/forum/general-chat/liniar-adjustable-power-supply/.I try to make my own linear power supply. To reduce the power dissipation on the regulator (not shown in the image) I want to use a transformer with multiple taps and switch between the taps as required, but in the same time I want to have a common reference for the control circuit (Rcontrol in the image) and the load (Rload in the image).Thank you

author: IanB
author_rank: Super Contributor
timestamp: August 15, 2012, 07:34:06 pm
content: Your extra diode (attached to the centre tap of the transformer) appears to be unnecessary. You can remove it from the circuit and the result will be the same.

author: NewBeginner
author_rank: Regular Contributor
timestamp: August 15, 2012, 08:04:05 pm
content: Yes, you are rightThank you

author: Simon
author_rank: Unknown
timestamp: August 16, 2012, 06:20:37 am
content: what are you trying to achieve ? It looks like a split supply but is certainly not wired as one.

author: G7PSK
author_rank: Super Contributor
timestamp: August 16, 2012, 06:37:33 am
content: What you have there is a three wire split supply with half wave rectification, you could achieve the same with two diodes. You have a negative rail and a positive rail with a central common but if you throw that switch you short out half your bridge.

author: NewBeginner
author_rank: Regular Contributor
timestamp: August 16, 2012, 09:01:09 am
content: Hi,I do hope it is not a split supply. That is not what I am trying to achieve.I think that a split supply would require a positive voltage, a negative voltage and 0V in the middle; I am a beginner in electronics so please correct me if I am wrong.Here there is no negative voltage after rectification (at leas that is what the simulation shows- and that is what I measured with the oscilloscope, with the switch in the position shown in the image; I did not measure yet the case with the switch in the other position).QuoteI try to make my own linear power supply. To reduce the power dissipation on the regulator (not shown in the image) I want to use a transformer with multiple taps and switch between the taps as required, but in the same time I want to have a common reference for the control circuit (Rcontrol in the image) and the load (Rload in the image).QuoteYou may be right.But according to the simulation located at this link, it should work:http://www.falstad.com/circuit/#%24+1+5.0E-6+10.20027730826997+56+5.0+50%0A169+96+128+208+128+0+4.0+0.1+-0.06624508666696448+0.0956681193319852+8.326672684688674E-17%0Av+64+176+64+144+0+1+50.0+230.0+0.0+0.0+0.5%0Aw+64+144+64+128+0%0Aw+64+128+96+128+0%0Aw+64+176+64+192+0%0Aw+64+192+96+192+0%0Ad+304+128+352+80+1+0.805904783%0Ad+352+80+400+128+1+0.805904783%0Ad+304+128+352+176+1+0.805904783%0Ad+352+176+400+128+1+0.805904783%0Aw+208+128+240+128+0%0Aw+240+128+240+64+0%0Aw+240+64+352+64+0%0Aw+352+64+352+80+0%0Aw+208+192+208+304+0%0Aw+208+304+352+304+0%0Aw+352+304+352+176+0%0Aw+400+128+544+128+0%0Aw+256+128+304+128+0%0Aw+256+128+256+400+0%0Aw+256+400+544+400+0%0Aw+208+160+288+160+0%0Aw+288+160+288+272+0%0Aw+288+272+544+272+0%0Ar+544+304+544+368+0+10000.0%0Aw+544+368+544+400+0%0Aw+544+272+544+304+0%0Aw+544+400+592+400+0%0Aw+544+128+592+128+0%0Ar+592+272+592+352+0+100.0%0Aw+592+352+592+400+0%0Aw+544+272+592+272+0%0Ao+29+64+0+35+20.0+0.2+0+-1%0Ao+24+64+0+35+20.0+0.00625+1+-1%0AIt is also possible that the simulation is not working properly.I am at work now and I can't measure it, but I'll take the measurements as soon as I get home.I have redrawn the image to better illustrate the connections.Thank you

author: NewBeginner
author_rank: Regular Contributor
timestamp: August 16, 2012, 03:25:16 pm
content: @G7PSKQuoteCould you please explain in more details what do you mean by "you short out half your bridge"?It seems that the simulator was correct. Using the oscilloscope I measured the second case (with the switch closed to point B) and it is working properly; just like in the case when the switch is closed to point A.It is still full-wave rectified and not just half-wave.AsIanBsuggested in an earlier post in this thread, it even works with the diode completely removed from the center tap.If I would remove the connection from the 24V tap into the bridge rectifier then I would get a half-wave rectification only.Thank you

author: G7PSK
author_rank: Super Contributor
timestamp: August 16, 2012, 05:19:00 pm
content: You have now revised your simulation and shown the switch in a different manner. The circuit would now switch between two voltage levels but you are switching out your bridge rectifier and using the diode, this will give you half wave either use 2 diodes or switch on the input side of the bridge.

author: NewBeginner
author_rank: Regular Contributor
timestamp: August 16, 2012, 07:00:29 pm
content: @G7PSKThank you for your answer but please read the posts more carefully.QuoteThe simulation was not revised. The circuit posted in "Image 1" already switched between two voltage levels.I did not change anything in the functionality of the circuit.The two images I posted represent exactly the same circuit. The only difference is that in the second image I colored the connections so that it can be followed more easily."Image 1" and "Image 2" illustrates the case when "Rload" is connected to a higher voltage, when the switch is switched to position A.The simulation I posted illustrates the other case when "Rload" is connected to the lower voltage and the switch is switched to position B. In this case "Rload" is in parallel with "Rcontrol".I measured both cases with my oscilloscope and in both cases the voltage on both Rload and Rcontrol is fully rectified (100Hz for the 50Hz supply). So the oscilloscope measurementshows exactly the same results as the simulation.QuoteBy "throw that switch" I suppose you were referring to turn the switch from position A to B.And this is the case illustrated in the simulation I posted.The simulation was not revised. It simply illustrated the case when the switch is set to position B.Please note that this setup is not a split supply.As far as I know in a split supply with a center tapped transformer, the common (0V) would be the center tap, but that is not the case in this setup.So in this setup the Rcontrol (the control circuit) remains always connected to the low voltage side and "Rload" can be switched to lower or higher voltage as required, so that the linear regulator for Rload (not shown in the images) won't have to dissipate too much power.So I am not sure where are you saying that the half-wave rectification occur.As I said I measured the voltage on both loads and it is fully rectified.Thank youImage 1:Image 2:

author: IanB
author_rank: Super Contributor
timestamp: August 18, 2012, 06:21:33 pm
content: The circuit NewBeginner has is fine. If you analyze it, all the currents flow where they should.NewBeginner has done the right thing by building the circuit and measuring it to verify the simulation (thumbs up!)Please don't confuse the poor guy by telling him his circuit is wrong unless you can show exactly where the problem lies.

author: G7PSK
author_rank: Super Contributor
timestamp: August 18, 2012, 08:34:53 pm
content: My apoligies when I saw the first circuit I read the switch as the top being the center after the A & B were put on I saw it as acting in a diferent Manner. But it still looks to me as the switch is now switching between two separate rectifiers one the bridge the single diode so sure the simulation will show the circuit as working but it would be better to switch on the AC side and use only one rectifier.

author: IanB
author_rank: Super Contributor
timestamp: August 19, 2012, 04:28:26 am
content: Quote from: G7PSK on August 18, 2012, 08:34:53 pmWe can gain some clarity by erasing the parts of the circuit that are not in use (open circuit) and examining just what remains.Firstly, consider the switch in the lower position:We see that Rcontrol and Rload are connected in a classic half bridge full wave rectifier configuration on the center-tapped transformer. Everything is fine.Next, move the switch to the upper position.This isolates Rload from the half bridge and so it disappears from this part of the circuit. However, Rcontrol is still working fine.Lastly, consider the part of the circuit that has now been switched in.There is a classic full wave bridge rectifier supplying Rload and the center tap is not involved.Note that I have removed the redundant single diode. It is doing nothing except to add an extra voltage drop in the system.Also with an analysis like this it is important to ensure that no short circuit current paths exist. In this case none do.

author: NewBeginner
author_rank: Regular Contributor
timestamp: August 20, 2012, 07:32:41 am
content: @IanBThank you for clarifying this

author: tbsmith
author_rank: Contributor
timestamp: October 08, 2014, 03:17:17 pm
content: This is an old topic, but I have a very similar question. I'm using a 120V in to 24V out center tapped transformer, and want to switch between using the full windings (24V) and only half the windings (12V). My questions is when putting a load on just half the windings, do I have to derate the power I can get from it?Thanks,Tom

author: IanB
author_rank: Super Contributor
timestamp: October 08, 2014, 03:35:59 pm
content: Quote from: tbsmith on October 08, 2014, 03:17:17 pmYes, because at half the voltage you will be drawing twice the current and this will increase the copper losses in the winding. Since the wire will have been designed for the expected current at 24 V you should not exceed the rated secondary current.

author: tbsmith
author_rank: Contributor
timestamp: October 08, 2014, 03:48:14 pm
content: Thanks for the quick reply. Just to make sure I got it, let's say I have a transformer rated at 1000 watts, 24V out. That gives about 41 amps. If I only use half the windings, I can still draw 41 amps, but I'd be okay because I'm effectively only drawing half the power (12 x 41), or 500 watts. Is that correct?

author: IanB
author_rank: Super Contributor
timestamp: October 08, 2014, 04:06:30 pm
content: Yes, 41 A would be OK as that is the rated secondary current. If you only use half the secondary you might be able to take a little more than 41 A as long as you keep within the overall VA rating of the transformer. However, you would have to check the temperature and make sure the transformer does not get too hot.

author: tbsmith
author_rank: Contributor
timestamp: October 08, 2014, 04:23:55 pm
content: Perfect, thanks! I'm doing it this way for the same reason as the original poster. I'm building a linear power supply and don't want to drop all that voltage in the pass transistors, but I want to maintain the ability to run it at high voltage so I'll add a high current switch to select secondaries (with no load). I envision most of my high current needs will be at 12 volts.Unfortunately, I pulled the transformer from surplus equipment and don't have any specs other than the power rating. Is there a rule of thumb of what temperature not to exceed for big transformers? Is the main concern melting the wire insulation?

author: SeanB
author_rank: Super Contributor
timestamp: October 08, 2014, 04:35:08 pm
content: Main concern is not to get the inner windings so hot that the insulation either degrades or becomes conductive. You typically will not want the inner windings to go above 70C, which typically means the core must be below 40C, which will apply to old insulation. If the insulation is modern it will be rated to typically 110C, and will be fine up to about 60C at the hottest point on the core surface.Simple rule of thumb is if you can smell it when operating it is too hot, which is true for any unit that has been powered for a year or so. New ones will smell of varnish, but if it starts to get to being able to make your eyes water in close proximity it is too hot.

author: bktemp
author_rank: Super Contributor
timestamp: October 08, 2014, 04:38:14 pm
content: The power rating of a transformer is mainly based on the temperature rise. Measuring the temperature is the most accurate way to determine the power rating.If you use only half of the windings, it can take about 1.41x rated current for both windings to get the same losses.

author: tbsmith
author_rank: Contributor
timestamp: October 08, 2014, 05:18:54 pm
content: The transformer is about 8-10 years old, so would that be "modern" enough to assume it would be good for 110C? I have an IR thermometer and will bench test it, looking for a core surface not to exceed 60C. I'll bring up the load gradually since there is a fair amount of thermal mass (or is that 110C/60C already taking the thermal mass into account?).The rest of the supply can't handle more than 40 amps, so it is good to know I have some reserve (1.41). Also this supply is for home hobby use, so it will never see continuous duty.