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Thread: hooking up an output transformer, correct phasing

  1. #1
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    hooking up an output transformer, correct phasing

    Hi All,
    I found the diagrams with the dot notation for my output transformer. As I understand it, if I hook up an input to the transformer, then the dots tell me how to hook up the output leads so that I know what the phase of the output will be.
    My problem is that the Felder Deluxe (and basically all Fender) schematics don't say how the leads are hooked up, only the color of the wiring. The Deluxe has negative feedback, so Im guessing that the output leads are switched with respect to the dot notation. E.g. on the Fender Deluxe AB763, there are two output tubes. The topmost output tube is connected to the top lead ( in the schematic) to the OT, and the lower output tube to the lower pri lead.
    The Hammond 1760H has dots on the Blue (input) and black (output) leads. So for this amp, I want to hook up the blue lead to the lower 6V6 and the Brown lead to the upper one in the schematic. Then, for the sec side of the OT, the black lead will be the one that is sent to the feedback, and the green (8Rohm) lead goes to the common side of the speaker.
    Is this correct?
    Thanks
    MP

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    I leave these leads unconnected and then when the build is done just clip them in with alligator clips. If it sounds the horn just switch the alligator clips, turn on the amp again and confirm that there is no more horn.

    The black lead on secondary side is typically the common aka ground. It goes to ground on the speaker jack, then a ground wire is run back to the PI power supply node ground. Speaker jacks should be isolated from chassis and should only make ground connection with the wire running to PI ground, not by physical contact with the chassis.

    Green lead, 8 ohm, is where the NFB is connected.
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  3. #3
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    Thanks!

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    here is a chart for the 763 DeLuxe, we put some ref designators on there (V4, V5)

    note that Hammond has screwed up the color codes
    (they are reversed from the orig, dang Canadians, they probably drive on the wrong side of the road)

    fd.gif
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  5. #5
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    I like to use arrows to keep track of phase. Note that when feedback gets back to the phase inverter, it is in-phase with the input. The same thing happens in the gain stage - concertina type amp where the feedback goes to the cathode of the gain stage.
    Attached Thumbnails Attached Thumbnails ot_phase_1.gif  
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  6. #6
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    There are a few curve-balls that may come your way.
    Sometimes the wire colours get mixed up (seems more common with single ended OT's).
    Sometimes reversed phase will not 'honk' but give weird issues like parasitics, etc.

    So you can always check to be sure the NFB loop is in fact reducing the gain (compared to NFB disconnected). If connecting the NFB wire increases the volume, your OT phasing is reversed.
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    Quote Originally Posted by g1 View Post
    There are a few curve-balls that may come your way.
    Sometimes the wire colours get mixed up (seems more common with single ended OT's).
    Sometimes reversed phase will not 'honk' but give weird issues like parasitics, etc.

    So you can always check to be sure the NFB loop is in fact reducing the gain (compared to NFB disconnected). If connecting the NFB wire increases the volume, your OT phasing is reversed.
    Thanks! Since this is by first complete build, Im more worried about damaging something by not having it hooked up right. I naively thought this part would be easier! Transformer has color coded wires, and the schematic tells which lead goes where, voila, we are done! I come across blog posts once in a while, from a courageous all-pro, who has fixed or modified hundreds of amps, who said they hooked up something backwards and ended up spending days fixing what broke after the smoke cleared.

    So, if I put a signal generator (which I don't yet have ) across the center tap and what I think is the in-phase input, (topmost 6V6 in the Deluxe Ab763 schematic) then measure the output on a scope (which I also don't have yet), shouldn't I be able to tell what the phasing is with respect to the wire color codes?

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    Quote Originally Posted by mikepukmel View Post
    So, if I put a signal generator (which I don't yet have ) across the center tap and what I think is the in-phase input, (topmost 6V6 in the Deluxe Ab763 schematic) then measure the output on a scope (which I also don't have yet), shouldn't I be able to tell what the phasing is with respect to the wire color codes?
    You're over-thinking this. If the color codes are off so much that primary and secondary windings are mixed up, then a quick resistance check will confirm which wires go where. This will prevent 'smoke' from happening. Of course it's always a good idea to power up a first build (any build!) slowly, without tubes, then incrementally adding tubes from PSU towards the power amp.

    If the issue is primary phase, then the listen test will suffice. Leave the NFB connection off the return point at the recovery? PI? tube. listen with low level test signal, music source, whatever. Jump the NFB into the circuit. If the overall loudness went down the NFB is in the proper phase. If the loudness went up (or you get a big honkin' noise) then the phase is incorrect, and the two primary wires must be reversed (blue/brown). So leave enough primary wire to make the switch before tidy-ing up the build.
    Last edited by eschertron; 08-09-2017 at 03:38 PM.
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    If it still won't get loud enough, it's probably broken.

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    This is good stuff because I have never understood this subject. I just powered up a new Ampeg build and got 'the howl'. I knew to reverse the primaries, but didn't really know why.
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    It's weird, because it WAS working fine.....

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    Quote Originally Posted by loudthud View Post
    I like to use arrows to keep track of phase. Note that when feedback gets back to the phase inverter, it is in-phase with the input. The same thing happens in the gain stage - concertina type amp where the feedback goes to the cathode of the gain stage.
    Thanks! I think I had an ahaaa moment. The input to the phase intverter is out of phase with the top plate, and in phase with the bottom plate. But, the tap taken from the secondary of the output transformer is fed into the GRID if the bottom half of the phase inverter, so when it gets added to the signal flow, it gets added out of phase. (as all triodes, the grid and plate are out of phase). Just hit me. Thanks for the diagram.

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    you could expand on that chart a bit>ot_phase_1.gif
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    Quote Originally Posted by eschertron View Post
    You're over-thinking this. . . ..
    Yeah that's me.
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  13. #13
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    Quote Originally Posted by cjenrick View Post
    you could expand on that chart a bit>Click image for larger version. 

Name:	OT_phase_1.GIF 
Views:	13 
Size:	15.6 KB 
ID:	44443

    I never understood feedback until I started working with solid state power supplies and opamp circuits. If you had an oscilloscope and knew how to use an external trigger to look at the phase around a tube power amp, you would know that the red arrows are wrong except the one you drew at the output. The phase can't be opposite across the 0.1uF cap on the feedback input to the LPTI except at very very low frequencies where the gain of the power amp starts to fall. If there is too much gain at that low frequency, the power amp will oscillate. The idea that negative feedback is "out of phase" with the input is incorrect in many cases.

    The LTPI is an adaptation if the "differential pair", two triodes with the cathodes connected together, independent inputs and outputs. (For the moment ignore the connection at the bottom of the tail resistor and just consider the two triodes without feedback.) As a differential amplifier, the input is considered to be the difference between the two grids and the output is the difference between the two plates. If you ground one of the inputs (connect a cap to ground on the lower triode because there is DC on the grid), a signal applied to the other input (upper triode) will produce an output on both plates, but the balance will be less than prefect. (Let that sink in for a moment.)

    If you disconnect the cap to ground on the lower triode and apply a signal opposite the phase of the upper triode, (the red arrows) the differential amplifier sees a bigger input and the output increases. The input looks bigger to the differential amplifier because the difference between the two input is larger. This is not how negative feedback works. If the signal applied to the input of the lower triode is the same phase as the original input, but slightly smaller, the differential input sees a smaller signal. This is how negative feedback works. The differential inputs compare the input signal to a portion (from a Voltage divider) of the output. This comparison reduces the gain, lowers distortion and output impedance of a power amplifier.

    Here's another way to look at it. Connect a speaker to the circuit I posted above. Now, ground the input so the input cap charges up. Now get a 1.5V battery and ground the minus side. Disconnect the input cap from ground and touch it to the plus side of the battery. The speaker should move one way (example: out) when you touch the plus side of the battery and the other way when you touch the input back to ground. Now touch the plus side of the battery to junction of the feedback resistor and the cap on the grid of the lower triode. The speaker will jump in (opposite the way it jumped in the first test) and out when the battery is disconnected. The differential action of the inputs on the LTPI invert the phase of the feedback so that it subtracts from the input.
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    keep it simple, in order to have that bottom triode produce a positive signal off the plate, we need to make the grid negative with respect to the cathode.

    no way around that.

    now it don't matter if we hold the cathode constant and wiggle the grid, or hold the grid constant and wiggle the cathode. right? that grid has to see a NET minus signal with respect to the cathode. so if we inject a positive signal from the OPT at the grid node, are we not introducing negative feedback?

    you want to use the Calculus to analyze electronics? fine. but you will lose me.
    Last edited by cjenrick; 08-11-2017 at 08:04 AM.
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  15. #15
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    You almost have it!

    Quote Originally Posted by cjenrick View Post
    keep it simple, in order to have that bottom triode produce a positive signal off the plate, we need to make the grid negative with respect to the cathode.

    no way around that.
    A positive going input on the top triode causes it's cathode to pull up on the lower triode's cathode. The feedback signal is the same phase as the input signal, but slightly lower in amplitude. Because the cathode is being pulled higher than the feedback signal, the lower triode sees it's grid going more negative with respect to it's cathode. So the lower triode has a positive going signal on it's plate.

    Made up example: If the top triode's input goes up 1V, the common cathode point might go up 0.9V and the feedback might go up 0.8V. The bottom triode thinks it's grid went -0.1V with respect to it's cathode. No calculus needed.

    Quote Originally Posted by cjenrick View Post
    now it don't matter if we hold the cathode constant and wiggle the grid, or hold the grid constant and wiggle the cathode. right? that grid has to see a NET minus signal with respect to the cathode. so if we inject a positive signal from the OPT at the grid node, are we not introducing negative feedback?
    Neither grid or cathode are held constant.
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  16. #16
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    you know i think you are right, here is a good thread that explains this complicated inverter>

    http://www.fenderforum.com/forum.htm..._number=645399

    bottom line: since the NFB is being applied to both the grid and cathode of bottom triode, the net voltage is zero. however since it is being applied to the cathode of the upper triode and not the grid, the top cathode will see the positive phase of NFB as cathode degeneration or a net voltage that equals NFB.

    now for the 64 dollar question, if the two triodes are running 180 out of phase this would mean the cathode voltages would cancel, so how is the lower triode modulated?
    Last edited by cjenrick; 08-12-2017 at 10:30 AM.
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  17. #17
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    Quote Originally Posted by cjenrick View Post
    now for the 64 dollar question, if the two triodes are running 180 out of phase this would mean the cathode voltages would cancel, so how is the lower triode modulated?
    Kind of an odd situation. (Small signal) The triode with the more positive input pulls the other's cathode higher. When the other triode's input is higher, it "wins". So you get sort of a double frequency slightly distorted looking sine wave. For larger signals big enough to cause clipping on the output the distorted wave on the cathodes looks worse. More like ice cycles hanging from a tile roof.
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