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3rd Burnt up transformer A2 Jap


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  • 3rd Burnt up transformer A2 Jap

    Lane 3 in my center has now smoked 3 transformers in 4 years. This includes the original & 2 replacements. Seems to be an over-time issue, where it runs fine to start. There are no warning signs in the days/weeks leading up to the burn-up. although I did have to replace the magnetic clutch the last time. I am going to have to put yet another new one on, just to get it running. When I do, does anyone have any ideas what to check as far as voltage in/out.... or things directly connected to the transformer that may be bad, and that can take one out over time like this? I'd really appreciate any ideas. I have 24 machines & the other 23 are still running just fine on their original transformers from the 1970s. Thanks in advance.

  • #2
    Depending on the machine, the Selenium rectifier, pin light ballast, & cycle solenoid derive their 120Volts from the Primary Side of the Transformer, (a bad idea & design). You need to determine what voltages each of these are at. Then, with the machine Main Breaker OFF and Power Cord UNPLUGGED, take resistance readings, (Ohms), for the ballast & cycle solenoid. (The rectifier will have to be tested hot for both Amp draw on the Input, (AC Side), and DC Voltage under load on the Output Side. (Unplug the Main Motor during this test). DO NOT unplug the Mag Clutch under load as this could destroy the rectifier. Compare them to a good machine or brand new ones from the shelf. Also Test the resistance of the Mag Clutch while de-energized and unplugged. When one of these components fail or is semi-failed, they may take the Transformer with them. Do you have the Slow Blow fuse between terminals F & G on TS1 ?? If not, install one. Physically check the wiring to be sure it is wired in correctly, (I've seen them wired incorrectly). It connects H2 & H3 of the transformer, (I'm assuming your machine is 208 to 240 Volt). Then check the Amp Draw & Voltage on the Primary Side. Compare them to another machine. I think eXm will have more and better info but this will give you something to start with.
    Everything has to be Somewhere !!


    • #3
      A rule of thumb I always follow when replacing the transformer is to replace the rectifier at the same time.

      Aside from what Mickey mentioned, have you checked that you have the correctly rates fuses installed?
      Factory & Converted A-2 (US, Ger, Jap)
      Comscore ECT, Matrix & DuoHD
      Walker B, Sanction Standard, Original K, Flex Walker & Ikon
      Kegel C.A.T.S


      • #4
        In order to help you find why your transformers are failing, we need some more information as to how they are failing. There are several ways a transformer can fail. Based on your post this may not be easy to find with basic voltage and current tests.

        The first thing we need to know is if it's failing in the primary side or the secondary side. We'll get to how to check this in a minute. If its in the secondary side then we need to know what loads you have attached to it, i.e. MSR, booster relay, ball lift relay, counter, ball indicator lights, etc. If its in the primary side then we need to see what components may lead to it's failure. We can do this by looking at what coil failed, i.e. H1-H2 or H3-H4.

        The first thing I would check to protect the new transformer is the wiring to TS1-F and TS1-G. While connecting a wire to either one of these will still work, if you connect a wire to TS1-F that should go to TS1-G, or visa versa, then you bypass the fuse and no matter what fuse is in it, no protection will be provided.

        The next thing to check is your capacitor. While popular sentiment is that the capacitor is there to protect the switch, it is actually there to help stabilize the magnetic clutch circuitry. Although the rectifier provides DC to the clutch, it provides rectified DC and not pure DC like a battery would. This means the voltage goes from 0 to 90 volts and back to 0. It does this 120 times per second. As the magnetic field in the clutch collapses each time the voltage goes from 90 to 0, it induces a large voltage into the circuit which includes the transformer and rectifier. This increases the current through the other components and can cause them to fail prematurely. The capacitor absorbs this reactive load and stores it in the capacitor to be released back into the clutch during the next 0 to 90 volt cycle thus reducing the current through these devices. You would relate this to the power factor of a motor and this is why power companies penalize companies for having low power factors. It requires the power company to use larger equipment and conductors to handle the larger currents that bounce back and forth, which they don't get paid for by the way. So if you have capacitance checking on your meter, disconnect one side of the capacitor from the circuit and test it or replace it with a known good one.

        Take your transformer that just failed, or several if you still have them, and perform the following test. With the meter on ohms and transformer disconnected from machine, test:

        1) From each lead on primary and secondary, test to transformer frame. These should all be infinite ohms. 0 ohms indicates a short.

        2) On the primary side: (write down your reading for each)
        From H1 to H2. Should match H3 to H4.
        From H2 to H3. Should be infinite.
        From H3 to H4. Should match H1 to H2.
        ​​​​​​​ From H4 to H5. You may not use these but a short in them can cause problems in the transformer.
        ​​​​​​​From H3 to H5.

        3) On the secondary side: (write down your reading for each)
        Yellow to tan.
        Tan to Blue.
        Blue to orange.
        Yellow to blue.

        4) Check between each wire on the primary to each wire on the secondary to ensue there is no short between windings. You should get infinite ohms for each of these.

        5) What is your machine voltage?

        6) Also I see you are in the US so you should have 60 Hz power but some Japanese pinsetters had 50 Hz systems. Make sure you are replacing with 60 Hz or 50/60 Hz transformers. If your using old stock for replacements, whether new or not, they could be the wrong transformers. Sometimes you can get by with using 50 Hz equipment on 60 Hz power (you can't do the reverse) but it can cause excessive heat and cause failure.

        7) Finally, what size and types of fuses are you using on the primary and secondary sides.

        This will give us a good start and hopefully tell us how the transformer failed which can help us determine where to look for troubles.

        Post your results back here.


        • #5
          Good Point on the Capacitor exM. My old machines did Not have capacitors and switched the AC Side of the rectifier. What is the capacitor rating? I would think a 50 mf with a Full Wave Bridge Rectifier would give a very smooth DC current. Did they not also go to a 2 Amp on the Primary Side and 5 Amp on the Secondary Side of the transformer ? (Like that is going to matter in a totally shorted condition or when a rectifier is tanking). Another good point on the The 50 Hz stuff exM ! The pure 50 cycle ball lift motors we had would Not run under heavy load. If 1 ball entered the lift it would barely get it out, if 2 entered in short succession, it would come to a stop and sometimes get them both out. I also found many mis-wired circuits on the Japanese machines. They would use Black, White or GREEN as the "Hot" from machine to machine, lift to lift, with no rhyme or reason.

          Agreed !! -"As the magnetic field in the clutch collapses each time the voltage goes from 90 to 0, it induces a large voltage into the circuit which includes the transformer and rectifier. This increases the current through the other components and can cause them to fail prematurely." Again I rest my case on using the transformer for 120 Volts. If they had gone to a standard 4-wire configuration with a Neutral Bar for the 120 Volts, many of these problems would go away. I realize this would not be feasible in all situations, ( ie. Grounded "B Phase"), but it should have been the primary design or at least an option.
          Everything has to be Somewhere !!


          • #6
            Actually it's a pretty small capacitor. According to all the docs it's only 0.5 micro farads. That's 1/2 of one one-millionth of a farad. This makes it much to small to be used as a filter to reduce ripple. Setting it up as it is allows it to function as an LC Tank circuit. If you think of a water tank mounted on a pivot like a rocking cradle and then put some water in the tank to represent the energy in our circuit. Then we'll call one side L for the inductor and the other side C for the capacitor. If you lift side C, all the water/energy flows into side L. Then lift side L and all the water/energy flows into side C. Again lift side C and the water/energy flows back into the inductor. This goes on and on. That's why this is an oscillator circuit. By selecting the capacitor to match the inductor this can be set up to match the frequency of the input voltage and it will operate at resonance.

            To understand resonance, think of a child on a swing. Once you give the child a push, in an ideal world with no resistance or friction, they will swing back and forth forever . However, in the real world, some energy will be dissipated due to these losses. So if you want the child to keep swinging, you need to put some energy into the system to replace those losses. When you put that energy in is the important part!

            The rate the child swings is their resonant frequency. Let's say the child takes one second to complete one cycle of a forward and backward swing. If your pushing from behind the child, then it does no good to push when the child is in the forward part of their swing. So if you push at a rate of 2 cycles per second then half of your pushes will be when the child is forward and so you waste energy. If you instead push at a frequency of one push every ten cycles then you have to put more energy in each time you push to make up for the extra energy loss due to the missed opportunities to push. However, if you match the child's frequency and push at a rate of 1 cycle/push per second, then you make the most efficient use of your energy. This is why the capacitor is necessary at the value chosen to match the inductor and set the oscillator up to operate at the same frequency as the input source.

            By doing this, the energy needed to re-establish the magnetic field in the inductor each cycle is stored in the capacitor when the magnetic field collapses each time. Without the capacitor then the supply would need to supply the energy for the circuit plus the energy to establish the magnetic field each time. The energy for the magnetic field then has to travel back through the supply every time the magnetic field collapses. This ping pong effect of sending power back and forth is what happens when motors have low power factors and why power companies penalize you for it. By keeping the energy for the magnetic field stored locally, this means a smaller current is needed through the supply each cycle and so reducing the load through the rectifier, transformer, light, wires, etc. This is exactly what happens when we add a capacitor to a motor in order to improve it's power factor.

            An additional benefit of this would be to limit the inductive spike delivered to the rectifier each time the field collapses and especially when the switch opened. Each cycle, this spike could easily be several hundred volts and the early rectifiers simply weren't capable of withstanding this very well. Today's silicon rectifiers are capable of handling this much better. During a switch opening, the voltage spike could easily reach a thousand volts or more and certainly enough to destroy the selenium rectifier with continuous spikes like this. While I've never seen the engineers notes on the design changes, I would suspect excessive rectifier failure was the cause for the modifications. There is a post on this site that makes an argument for the capacitor to protect the switch but the analysis is incorrect and doesn't support the conclusion. I've been going respond to that post for sometime but never returned to do so. I guess now that I've opened Pandora's box I might have to do that. When I previously said the capacitor wasn't to protect the switch, I fully expected someone to call me out on that and use that post as their argument. That's why I spent extra time on this post.


            • #7
              Now to respond to the rest of your post Mike. Yes, the Japanese pinsetter uses a 2 amp fuse for the primary and a 5 amp for the secondary. This could be due to running at 50 Hz instead of 60 Hz. The lower frequency will have lower inductive reactance and therefore lower resistance to current flow and higher currents for the same given components on a 60 Hz system. So in order to use the same equipment they would need to up the protection limits. Just speculating here but that would be my initial guess.


              • #8
                When I changed a 24 lane house over to 3 amp breakers for the secondary because the price of 3.2 fuses got too high, I converted 4 Japanese pinsetters as well as the 20 American pinsetters. They never tripped. If you use a fuse that is too small the worst you can do is replace fuses often. A fuse that is too big can get expensive.


                • #9
                  Oooowwwww..... My head hurts LOL Seriously exM, very Nice ! as always. Sooo, if I get your drift, the resonance would basically be out of sinc as the very large capacitor would require too much energy for charge/discharge ??? You also spoke of low power factors from motors, (and transformers, ballasts, etc.). This is a problem for Power Plant Generators and why we penalized them for this. A "lagging" power factor, (vars-in), does not make the generator and power grid happy and its ability to produce real power is lessened. A "leading" power factor, (vars-out), allows for better efficiency and better power production. Our line distribution centers would bring in various large capacitor banks at various times to help control this. I'm reaching back almost 35+ years for this stuff so it is a little foggy.
                  Everything has to be Somewhere !!


                  • #10
                    Very true mrB. And even with the 1 Amp slow-blow to protect the Primary side of the transformer, I've seen a rectifier take out the transformer. The original machines had all the components running on 208 - 240, including the selenium rectifier, ( they simply placed a big resister in series to lower the voltage to it). They did Not have the 1 Amp fuse. There was a change soon after which called for the re-wire, installing the fuse AND changing out the transformer. I doubt many transformers were changed out and I have found various machines Without the fuse ! which I corrected when I started at Olympic Lanes in the mid 70's. I did not change the transformers as they continue to work to this day.
                    Everything has to be Somewhere !!


                    • #11
                      Sorry Mike, didn't mean to hurt your head. You may understand whats about to follow but others may not so maybe it is worth covering.

                      The thing to understand about capacitors, inductors and resistors is that the each function differently when placed in a circuit. In an AC circuit the current will be in phase with the voltage with a purely resistive load. With, a capacitor, the current will lead the voltage by 90 degrees while in an inductor the current will lag the voltage by 90 degrees.

                      To see this in action you can take a battery and a light bulb. When you connect the bulb to the battery you will see it turn on immediately and stay on until you disconnect the battery at which time it will go out immediately. This is the current acting in phase with the voltage.

                      Now take an inductor, a spare cycle solenoid for example, (DO NOT DO THIS AT LINE VOLTAGE, USE A 12 VOLT BATTERY OR LESS) and connect it in series with the bulb. When you connect the circuit to the battery you will notice the bulb does not light immediately despite the voltage being applied. Over time you will notice the bulb begins to glow brighter and will stay on until the voltage is removed. This is an example of the voltage leading the current. You applied the voltage but current didn't immediately flow.

                      Remove the inductor and replace it with a capacitor. A large capacitor may be needed or add additional resistance to slow it down enough to see the reaction. This time when you connect the circuit to the battery, the bulb will glow brightly almost immediately and then slowly start to go out and will eventually be off even though it's still connected to the battery. This is an example of current leading voltage. This is how the time delay in the older TDM modules work to delay solenoid activation.

                      These simple experiments show the voltage current relationships between the three components. The inductor and capacitor are 180 degrees out of phase from each other. So in a heavy inductance circuit like this it is very hard to remove the ripple by simply using a capacitor because when the inductor is charging, the capacitor is discharging. This means the capacitor has no charge to provide the ripple smoothing when called to do so. This is why we use the capacitor in the LC tank circuit configuration instead of trying to smooth the ripple.

                      The capacitor and inductor both cause reactive loads in AC circuits but in opposite magnitudes. Think of power factor balancing like making gravy. You can have all flour or you can have all milk. It's when the two are mixed in the correct proportions that the magic happens. If it's to thick you add more milk. When it's to runny you add more flour. If you have too much inductance, you add more capacitance and visas versa. The idea is to add the correct amounts of each so when you add them together, the net reactance is 0. If you have 30 ohms of inductive reactance (which is 90 degrees positive) then if you add 30 ohms of capacitive reactance (which is 90 degrees negative in direction) you end up with a net 0 ohms of reactance and so the load is purely resistance at that point.

                      You shouldn't strive for a power factor of 100% because you can easily push an inductive load to a capacitive load. This can cause harmonics in the circuit and can destroy control circuits. A motor's power factor is a result of its loading among other things. So if you balance the power factor based on a lightly loaded motor, when you increase the load on the motor the power factor will increase and this is when it can be pushed over the top. This is also why you don't use oversized motors because the light loading will cause a low power factor. Your better off correcting the power factor of the smaller motor rather than oversizing the motor.

                      Hope this doesn't hurt to bad. This subject takes weeks if not months to cover in a classroom. Trying to cover it in a few posts is a tough task. Hope this helps someone out. If someone would like to try the experiments or have trouble getting them to work, please don't hesitate to ask.

                      If you would like to dive deeper into the subject or if I didn't make something clear, don't hesitate to ask.


                      • #12
                        If it had only been so easy in the Power Plant and the Distribution Grid. Again, very nice exM ! On another note, I am thinking of putting together a simple pictorial troubleshooting guide using TS1 & TS2 both in the powered mode and un-powered, (Continuity), for folks to troubleshoot their electrical systems. I would send it to several of you folks via PM for review before posting.
                        Everything has to be Somewhere !!


                        • #13
                          My electrical boxes have been butchered by installers of the various computer scoring systems.
                          We've had accuscore with bowltronics triggering, this was teplaced by computer score which was then replaced by Qubica.
                          I woud like a simplified schematics but im sure it won't correspond with my boxes, however it could come in handy.

                          What is needed in this day and age is a static component box. Im pretty sure it was Steve Stafford or peanut that posted a very clean and tidy electrical box with just static components.


                          • #14
                            So what size compasitor should we use. Mine says 50mf.


                            • #15
                              Which capacitor are you referring to?

                              If your talking about the capacitor on the magnetic clutch then you should stick with Brunswick's recommendation unless you replaced the clutch with a non equivalent in which case it would need to be specified to match the inductance of the clutch. The larger the capacitance the longer it takes it to charge and discharge and so the lower the frequency. Since the inductor is fixed in this case the capacitor is used to tune the circuit. Figure 1 below shows Brunswick's recommended capacitor. It is a 0.5 uF (micro Farad). This should be several hundred volt rating but I don't know what Brunswick's was rated at for voltage. Perhaps someone could take a look and post back.

                              If your talking about the motor capacitor then that would be different for each motor even if the motors are the same. This is because power factor is related to the motor loading. You would have to analyse each motor and come up with a plan to best deal with what you have. You need to have a power factor (pf) meter to determine the pf. If your motor lists a pf or an efficiency at design load then this can be used to determine the true power. Then by measuring your amps at load you can use the power triangle explained below to come up with your power factor. There are other ways but we won't get into those here.


                              Power is made up of three components. The apparent power is the power the supply sees and what flows through the supply lines. The true power is what actually gets work done in the system and is related to the resistive load i.e. what you measure with an ohm meter in the run winding. The reactive power is the power necessary to create the magnetic field in the winding each cycle. Inductors aren't considered to consume any power because they return the power to the supply when the magnetic field collapses but the supply still needs to provide this power to create the field in the first place in addition to the true power which is what results in higher current loads with lower pf.

                              While more complex methods can be used to analyse the system we can use a power triangle to represent the three power systems in the motor. We can then use the Pythagorean theorem to solve for the unknowns. The following example shown in figure 2 is for an imaginary motor running at 240 V and drawing 6 amps. The triangle on the left is before pf correction and has a pf of 0.6.

                              (V = volts, A = amps, W = watts, VAR = volt amps reactive. The K before each means 1,000. Sometimes M will be used for 1,000,000)

                              By taking 240 V times 6 A we get the apparent power of 1440 VA or 1.44 KVA. We can then take the 1.44 KVA times the 0.6 pf to get the true power of 860 W or 0.86 KW. We can then use Pythagorean to calculate the reactive power of 1.15 KVAR (1150 VA reactive).

                              In the triangle on the right, we have the same motor but corrected to 0.9 pf. The true power stays the same because this is what does actual work so we can transfer our true power from triangle 1 to triangle 2. We know we want a pf of 0.9 so we can enter that in triangle 2. The pf is simply true power divided by apparent power so working backwards if we take true power divided by our pf we can get the apparent power. In this case 860 W divided by 0.9 = 960 VA. Again using Pythagorean we can calculate our reactive power of 430 VARs.

                              Our old VARs minus our new VARs is the amount of capacative reactance we need to add to get to our pf of 0.9. So 1150 VARs - 430 VARS = 720 VARs of capacative reactance is needed. Normally this is all the further you need to go and you can buy capacitors based on the VARs rating but since you asked about the capacitance we'll go one more step. Below the power triangles is the formula for figuring the capacitor for our example. Here C is the capacitance, Qcap is the VARs of the capicator we calculated above (720 VARs), f is the frequency (60 Hz in N. America and 50 Hz across the pond and the land down under), and V^2 is the voltage squared. Plugging in the values for our example would give give 33 uF (micro Farads) or 0.000033 Farads.

                              IMPORTANT: When performing pf correction you also need to resize your motor overload protection.

                              The thing to note here is what happened to our KVA. We started with 240 V x 6 A = 1.44 KVA. We ended with 0.96 KVA / 240 V = 4 A. While not a lot for 1 motor, consider if you had 36 motors times 2 A = 72 A smaller service, transformer and conductors. In a new build this could be significant cost savings for equipment. But consider the existing site that decides to oversize their motors which will now be under loaded and so the pf will be lower. Under the above example going in reverse from 0.9 pf to 0.6 pf you can overload your service, transformer and conductors by simply over sizing your motors when it isn't necessary to do so.

                              A point of consideration. In an existing business, there is no benefit to achieving higher pf unless you are charged by the KVA instead of the KW or unless you have a pf charge on your bill that penalizes you for having a low pf. The system will already be installed and your not going to pull out the existing service equipment to put in smaller. There is advantage in that the motors will run cooler so they should last longer. Also there is a benefit to the environment because the power companies don't have to generate as much power and everyone working together can prevent new power plants from having to be built. But it's hard to get owners to spend money on improvements they can't see. So the first step is looking at your utility bill. See if you are billed in KVA or KW. KVA charges you for pf while KW does not. Then look to see if you have a P.F. charge. If you have either of these then there can be a cost saving that can be seen. In this case I would contact your local power company and discuss it with them. Many will do an analysis of your business for free because it benefits them. They will analyze your entire system and determine what the system is operating like and can recommend changes to get rid of these extra costs. Many older business may not have pf penalties but they are becoming more common for commercial businesses.
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