Using Transformer Secondary to Buck/Boost primary - Page 1
Forum section: electronics / beginners

author: K1JOS
author_rank: Regular Contributor
timestamp: February 19, 2015, 01:42:41 am
content: Can someone help explain how the Turns/Voltage formula should be used when wiring a secondary to buck the primary side. This is desirable for many of us who operate old ham radio transmitters that originally were designed for 110 or 115v and run hot at 120-125vac. In these power supplies the big heat issue comes from the one or two tube 5vac rectifiers supplying the HV and Bias. Its popular to replace the old 5v rectifiers with some diodes. Some suggested to saver space and instead of an external buck to use the two unoccupied 5vac secondary's to buck the primary down from 125vac to the original 115vac.But in my discussion with other hams there is confusion how this works using an internal secondary.If I take an ideal power transformer with 100 turns on primary, and two secondaries - one with 800 turns and the other with 5 turns. Based on the Turns formula Es/Ep = Ns/Np, if I apply 100 vac to the primary then the secondaries will yield Es = 100vac * 800/100 = 800vac and the other secondary (the rectifier filament secondary) will yield 5vac.If I now wire the 5 turn secondary IN_PHASE with the primary - the new primary voltage will be boosted to 105 vac and the Turns formula would yield Es = 105 vac * 800/105 turns ratio = 800vac. No change in HV (assuming 4000 ohm load to run at 200mA)Now If the 5 turn secondary were wired OUT-OF-PHASE, like a battery, it would buck or drop the primary voltage from 100 vac to 95 vac and the turns formula would yield what? Some of us think it depends if the 5 turns added by the filament secondary are counted merely as turns regardless of phase which would lead to Es = 95 vac *800/105 turns ratio = 724 vac. On the other hand others believe that the 5 tuns out of phase leads to out of phase magnetic flux so the secondary turns should be subtracted from the primary turns in the calculation. If so the formula yields: Es = 95 vac * 800/95 turns ratio = 800vac !!! NO change???I tried this in Multisim student edition but it gave me completely different results, with the 5 turn secondary wired in -phase the HV went to 842 vac while out-of-phase it went down to 762 vac.Please can some EE give us HELP !!!

author: langwadt
author_rank: Super Contributor
timestamp: February 19, 2015, 02:03:58 am
content: Quote from: K1JOS on February 19, 2015, 01:42:41 amit is only the turns ratio that changes not the primary voltage, if I understand what you are doing100V * (100+5)/800 = 762V100V * (100-5)/800 = 842V

author: KDC
author_rank: Contributor
timestamp: February 19, 2015, 02:16:33 am
content: Your primary voltages are fixed by your input voltage, I think that's where your calculations differ from Multisim.Wiring them as your describing would result in just changing the volts per turn ratio, and gives you the numbers that multisim is spitting out for your secondary (100/105 * 800 and 100/95 * 800)What you need to do is change the actual input voltage, which is where the external bucking transformer setup comes in.

author: IanB
author_rank: Super Contributor
timestamp: February 19, 2015, 02:49:19 am
content: Short answer: you want to put your unused secondary or secondaries in series with the primary in such phase that the voltage on the used secondaries drops. So if your HV output drops from 800 V to 760 V you have got it right.Long answer:If you want to increase the supply voltage from 110 V to 120 V, you need to add 10 V more windings to the primary side. These windings need to be addedin the same directionas the main primary. It turns out that in the same direction equates to opposing voltages, so the unused secondaries need to be wired out of phase with the main primary to buck the incoming voltage (since the same direction of turns yields opposing voltages, due to back EMF).Example:Suppose you have 100 turns on the primary with 100 V coming in (suppose you are in Japan), and suppose you have 600 HT turns giving 600 V on the secondary. Your turns ratio is 1:6. Now suppose you also have a 10 V secondary; it will have 10 turns.Next, suppose your supply voltage increases to 110 V. To keep everything the same, you need 10 more turns on the primary. This will give you a turns ratio of 110:600 or 1.1:6. To achieve this you can take your unused 10 V secondary with its 10 turns and put these 10 turns in series with the 100 turns of the primary. This is going to give you the 110 turns that you need.Beware:The mains voltage isolation between the 5 V secondaries and the HT windings won't necessarily be as good as with the proper primary winding. So there is a possible voltage breakdown risk from doing this.

author: K1JOS
author_rank: Regular Contributor
timestamp: February 19, 2015, 03:06:15 am
content: Quote from: IanB on February 19, 2015, 02:49:19 amIs "series" correct? For buck or boost the examples I am seen is to place one side of the buck/boost to line/mains Neutral and use the voltage between the other end of the buck/boost and neutral? If I place the buck/boost in series with primary this only increases the number of windings on the primary side.

author: Simon123
author_rank: Regular Contributor
timestamp: February 19, 2015, 03:42:11 am
content: Just do this and add another transformer, which can be small.http://sub.allaboutcircuits.com/images/02148.png

author: K1JOS
author_rank: Regular Contributor
timestamp: February 19, 2015, 04:59:28 am
content: QuoteYes that does explain everything and reconciles my real-world findings to the Multisim Spice model. This then is not a typical buck/boost of primary voltage but rather the inductive effect of adding or subtracting turns to the primary depending on the phase of the secondary in relationship to the primary.

author: IanB
author_rank: Super Contributor
timestamp: February 19, 2015, 05:11:18 am
content: Quote from: K1JOS on February 19, 2015, 03:06:15 amYes, increasing the number of turns is exactly what needs to happen. The voltage is proportional to the number of turns. If you want to increase the primary voltage from 110 V to 120 V you need to increase the number of turns proportionally.

author: K1JOS
author_rank: Regular Contributor
timestamp: February 19, 2015, 04:52:05 pm
content: QuoteOn looking at this again. you have the secondary and primary turns inverted in your formula.100V * 800/(100+5) = 762VBut even with the corrected equation the calculated voltages based on the TURNS formula are wrong from real world experiment as well as Multisim. If the secondary is added "in phase" then you are increasing the primary from 100 to 105 turns and in Multisim and real world experiment, the HV secondary goes UP to 842v (not down to 762v) as the TURNS formula you show. Can anyone explain this as I would like to understand the principles.

author: IanB
author_rank: Super Contributor
timestamp: February 19, 2015, 05:58:17 pm
content: If we consider ideal transformers it is all about the turns ratio.Our base transformer has 100 turns on the primary, 800 turns on the HT secondary and 5 turns on the LT secondary.For an input voltage of 100 V we than have 800 V HT and 5 V LT.If we now put the LT winding in series with the primary (in the same turns direction), we then have 105 turns on the primary and 800 on the HT.Our HT voltage will therefore become 100 x (800/105) = 762 V.We can also look at this with voltages instead of turns. Suppose we put the 5 turn LT winding in a buck configuration against the primary, and for convenience let's increase the supply voltage to 105 V.Now we will have the 5 V from the LT winding opposing the 105 V supply, and thus being subtracted from it. The original 100 turn primary will see 105 - 5 = 100 V, and the HT voltage will be 800 V as before.If we decrease the supply from 105 V to 100 V, then the output voltage will become 800 x (100/105) = 762 V as before.Simulation programs should return these results also.It is important to note that thesameturns direction is theopposingphase. The transformer coils acting as inductors will always oppose the voltages applied to them.

author: K1JOS
author_rank: Regular Contributor
timestamp: February 19, 2015, 06:11:47 pm
content: QuoteThanks IanB - that explains it well. Not intuitive to me though is why if the added secondary turns onto the primary are in the same direction as the primary turns why the phases are opposite, I guess I need to read up on mutual inductance as when two inductors on a common core are put together in series ?

author: K1JOS
author_rank: Regular Contributor
timestamp: February 19, 2015, 06:26:03 pm
content: IanB - OK I think I have it... the turns can all be in the same direction on the same core but its the current flow direction. If the current flow is in same direction they are cumulatively coupled and if opposite they are differentially coupled. Functionally this is the same as either adding or subtracting "turns"thanks !!

author: IanB
author_rank: Super Contributor
timestamp: February 19, 2015, 06:27:09 pm
content: Quote from: K1JOS on February 19, 2015, 06:11:47 pmIt has to be this way otherwise transformers would burn up.When you apply 120 V to the primary of your (unloaded) transformer you are putting the mains voltage across a wire with a relatively low resistance. If that were just an ordinary straight wire a huge current would flow and the wire would melt.However, when the wire is inside the transformer it generates an opposing voltage of nearly 120 V that bucks the mains voltage. The primary winding thus sees a net voltage of 120 - 120 which is approximately zero and very little current flows.

author: K1JOS
author_rank: Regular Contributor
timestamp: February 19, 2015, 07:41:05 pm
content: Thanks IanB. What would happen if instead of wiring the secondary to the primary in series in used an separate 5 turn inductor in series with the transformer primary. Since inductors are not directional - it wouldnt matter which way the inductor were inserted in series as the current flow would always be in the same direction relative to the transformer primary. Am I correct so far? If so then the external 5 turn inductor would have a voltage drop proportional to its inductance and the transformer primary would therefore see a lower voltage. The external inductor in series could only provide some bucking to the primary voltage?

author: IanB
author_rank: Super Contributor
timestamp: February 19, 2015, 08:03:33 pm
content: If you used an external inductor in series with the primary it would be described as a ballast, or current limiting device. Since the ballast would be wound on a different core, the effect of one turn on the ballast would be different from the effect of one turn on the transformer. The effect of the ballast inductor on the operation of the main transformer would be a more complicated calculation that I am not confident to go into.

author: K1JOS
author_rank: Regular Contributor
timestamp: February 21, 2015, 03:21:05 pm
content: Thanks again IanB for you willingness to share your knowledge. I can imagine it would be quite complicated as there would be a voltage current shift in the external inductor and the primary on the transfomer would experience this shit so the secondary winding's effects must be very complex. Maybe something I can try in a spice model software just to see what happens.