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Thread: Pre Planning For Modifying A Fender Super Reverb Amp

  1. #1
    Supporting Member TomCarlos's Avatar
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    Pre Planning For Modifying A Fender Super Reverb Amp

    MEF Members....

    As I continue to study up, test, noodle, etc, I have a project that I've wanted to dive into for some time now. Perhaps those that saw my recent post on the Peavey Vintage Amp know that I had an interest in that one- primarily the chassis, transformers, etc. I was hoping to use that and modify it to get a simple single channel amp. In lieu of that one, I am back to something that might be simpler from a design perspective. Building it will be another issue and thread.

    My goal is to have a single channel tube amp, approx 40 - 50 watts, using two 6L6 tubes. I want a spring reverb but no need for a tremolo circuit (I have a Boss Tremolo pedal that works nicely). It would be nice to have a Mid range tone control. I'd like to have a compact cab with a single 12", 8 ohm speaker. I have an output transformer that will handle this power. I think I can get there by modifying a Fender Super Reverb (AB673) circuit, building a new turret board, and using something like the D-LAB electronics conversion of a Marshall Solid State combo to a Fender Princeton Tube amp as a model. That cab and chassis would be a good candidate for what I have in mind. There may be other donor chassis and cab candidates. The chassis works because of the pre-drill holes for the pots and other front panel controls. I'd need to punch holes for the tubes and accommodate the PT.

    • I am posting the original Fender Super Reverb (AB763) below for reference.
    • Assuming I would remove parts from the Normal channel and the amp would still work, I covered those on the modified schematic.
    • I also covered the parts related to the Tremolo (that I will not use).
    • I plan on swapping the Output Transformer in favor of an 8 ohm tap (that will match up to two 6L6 tubes).
    • I also plan to swap the GZ34 rectifier tube in favor of four diodes (full wave bridge rectifier).



    I am hoping that what remains in the modified circuit is correct and will work. I am concerned about connection points between the Normal and Vibrato channels and that I have not eliminated anything I would need to make this work; especially feeding the reverb circuit and returning that to the 7025 that will feed the Phase Inverter.

    For the Power Transformer, it looks like the Classic Tone 40-18029 will work. That is one which is recommended as a replacement for the Fender 125P5D. The other option is the Hammond 290D2X. That transformer generates 710v (355 CT) as well.

    I realize that going from a GZ34 to Diode Rectifier will raise the DC voltage. So I'm wondering if I should be looking at a PT with slightly less output - maybe 325 CT instead.

    So for now, this is the plan - purely in "draft" mode. The nice thing about this is I can follow the roadmap already built and tested. I simply take the time to dive into the details and study the circuit as I go.

    Comments are always welcomed- even if you see a big flaw in the concept!! Now would be a good time to get a reality check on this idea!

    Tom
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    Senior Member Pedro Vecino's Avatar
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    I see it fine but if you want to copy exactly the vibrato channel you need to incorporate the 50K resistor (the value of the vibrato intensity potentiometer) over the junction of the 0.1 capacitor and the 220K resistor.
    The other element that needs to emulate is the existence of the 220K resistor (at the end of the normal channel).

    Also the 820R cathode resistor (marked as A) needs to be double value (1K5) since it was originally for two cathodes.

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    Last edited by Pedro Vecino; 03-10-2020 at 12:00 PM.

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    You can’t use a full wave bridge rectifier with such a PT, as the HT will be 1.4 x 710, ~1kV
    A two phase rectifier is the type to use for such an application, HT would be ~ 710 x 0.7

    See http://www.valvewizard.co.uk/bridge.html

    Why not use the PV PT?
    EDIT OK the PV is a 4x6L6 amp, so maybe you don’t want the unnecessary weight?
    I suggest the BF bassman / bandmaster replacement type PT, given they will be slightly lower winding voltage to better accommodate silicon rectification.
    eg https://www.hammfg.com/part/290EX?referer=1063

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    Last edited by pdf64; 03-10-2020 at 12:35 PM.

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    Supporting Member Daver's Avatar
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    Also, the junction of your 220K balancing resistors on the first filter cap stack needs to be connected to the junction of those filter caps.

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    Bent Member Chuck H's Avatar
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    Second what Pedro is saying. Often missed when people build stripped down versions of amps is the affect peripheral stages and mixing circuits have on the gain structures. It can be very important to keep the amp operating intuitively and even to keep symmetry in the wave form relative to gain structures. In my efforts to learn something about the design of guitar amps I've experimented with many 'variations' of known circuits and often found that the cut and paste method doesn't work as well as I'd hoped and the original design was more tweaked and engineered than is supposed.

    And... In order for the bleed resistors in the power supply main filter totem to balance voltage across the 70uf caps you'll need to connect each one across each cap.

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    "Take two placebos, works twice as well." Enzo

    "Now get off my lawn with your silicooties and boom-chucka speakers and computers masquerading as amplifiers" Justin Thomas

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    Thank you for the replies!!

    Pedro: I added the 50K resistor over the junction of the 0.1 capacitor and the 220K resistor - noted as (1) on my schematic. I changed the cathode resistor on A to 1.5K - noted as (2) on my schematic. To emulate is the existence of the 220K resistor at the end of the normal channel, I added the resistor, the .047uf cap, and the 100K that would normally feed the 7025 tube that was removed - noted as (3) on my schematic. I hope I did that correctly.

    Daver and Chuck: Good catch on the Filter caps being tied to the center of the 200K resistors in the power supply. That change is noted as (4) on my schematic.

    Pdf64: Thanks for the comments on the power transformer. If I could not find a suitable transformer, I could use half the secondary of a HV and the CT. As for the power supply, if I can find a chassis that will easily accommodate a Rectifier Tube, I will probably do that. But if pressed for space, I will go with the Full Wave Bridge Rectifier design.

    For the Full Wave calculations, without knowing how much sag will be on the transformer, I am estimating at this point. But it does not need to be perfect science. The Hammond 290D2X is rated at 355v on the Red Leads (outer most of the secondary). The Hammond 290EX (the Bassman replacement) is rated at 330v.

    Using the info from the Valve Wizard, here is what I see and using the Hammond 290D2X as the example (and copying text from the site):

    “Typically, when the loading is light the transformer voltage will be 5 to 10% higher than the 'advertised' value.”

    355v @ 5% higher = 373v
    355v @ 10% higher = 390v

    When the transformer is fully loaded to its 200mA rating the AC voltage will fall to the nominal value (355v).

    The DC voltage will therefore fall to: 1.3 × 355Vrms = 461Vdc.

    That gets me close enough!!
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    I could use half the secondary of a HV and the CT.
    That might exceed the winding's current rating. Unfortunately there's no way to parallel the 2 sections. Your amp will draw between 200 mADC and 250 mADC at full power. And with bridge rectification transformer AC current will be at least 60% higher than the DC value.

    As for the power supply, if I can find a chassis that will easily accommodate a Rectifier Tube, I will probably do that.
    That would be my recommendation. I tried SS rectifiers in BF SRs, '59 BMs, JTM45s and AC30s. Always preferred the GZ34.

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    Last edited by Helmholtz; 03-10-2020 at 08:00 PM.
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    Quote Originally Posted by TomCarlos View Post
    ...If I could not find a suitable transformer, I could use half the secondary of a HV and the CT. As for the power supply, if I can find a chassis that will easily accommodate a Rectifier Tube, I will probably do that. But if pressed for space, I will go with the Full Wave Bridge Rectifier design.
    !
    You really need to grasp how fundamentally different the full wave bridge rectifier is to the two phase rectifier, and how a PT intended for a tube amp will be designed specifically for one or the other.
    Because the rectifier type determines the ratio of not just Vac to Vdc, but also the ratio of Iac to Idc, and also whether the Iac rating is to be for continuous current flow (FWB) or a 50% duty cycle (Two Phase).
    FWB; Vdc = 1.4 x Vac, Iac = 1.4 x Idc
    Two Phase; Vdc = 0.7Vac, Iac = Idc

    The same PT can't be used interchangeably with either rectifier type without massively affecting the HT supply capability.

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    Acc. to standard literature Iac = (1.56.... 1.92) Idc for bridge and Iac = (1.11...1.35) Idc for two-phase rectification depending on conduction angle.

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    Supporting Member TomCarlos's Avatar
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    Good point PDF... thank you for that info - Helmholtz too!

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    Supporting Member TomCarlos's Avatar
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    Just watched an excellent video from the Uncle Doug channel on YouTube - Power Transformers: Basic Design and Function. In the video, he looks at a Fender Deluxe Reverb and calculates the requirements for a transformer. The assumption is that one would still be using the GZ34 rectifier tube. But he does a nice job on going through the schematic, adding up Filament current, Plate currents, etc.

    A couple things that dawned on my today. (1) Of the transformers mentioned above, the Hammond 290D2X has a 5V winding. The heater winding is 6.3v CT @ 5A. The HV is 355-0-355 @ 300ma. The 290EX does not have the 5v winding. The other candidate would be the Classictone 40-18029. It has a 710CT @ 200ma, 6.3 CT @ 5A, and 5V @ 3A.

    After doing some additional studying, I think I now get the point made by PDF and Helmholtz. Because of the continuous current flow in the Full Wave Bridge, the current requirement of the transformer would increase by a factor of 1.56 to 1.92 that of the Two Phase requirement. I will continue to look at that. It is a good thing you reminded me of this. Thanks again.

    Back to the circuit, I hope my revised circuit comes closer to something that will work.

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    Because of the continuous current flow in the Full Wave Bridge, the current requirement of the transformer would increase by a factor of 1.56 to 1.92 that of the Two Phase requirement.
    I see you understood the principle. But the winding AC current requirement for bridge rectification is always 1.414 times (= square root of 2) higher than with two-phase.
    The factor 1.56 to 1.92 is the ratio of transformer AC current to rectified DC current with bridge.

    Often the amp's DC current demand is known or can be estimated from circuit parameters. OTOH, PT manufacturers typically only specify winding AC current. The factors above can be used to convert DC current to AC current. The conversion factor increases with reservoir capacitance. With tube rectification typically the lower value applies.

    Example: 40W tube amp with GZ34 and 40µF reservoir cap. The amp will draw a HT current (DC) at full power of around 200mA. Multiplied by 1.11 (see post #9) gives max. AC current requirement as 222mA.
    With bridge rectification AC current in the used part of the HT winding would be higher by a factor 1.4 for the same DC current, in other words Iac = 1.414*1.11*Idc = 1.57*Idc = 314mArms.

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    Supporting Member TomCarlos's Avatar
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    I doubt this info is covered anywhere!! So I am adding to my notes and will be more careful in the future. Interestingly, it is dang hard to get specs on a Peavey Transformer. I guess you'd have to conduct lab tests to figure out the specs.

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    Quote Originally Posted by TomCarlos View Post
    I doubt this info is covered anywhere!!
    This might come in handy. Hammond Design Guide For Rectifier Use.pdf

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    Quote Originally Posted by Dave H View Post
    Have you had a good look through that page? There’s a fair bit of nonsense on it, but as it’s on Hammond’s site and there’s a few correct tidbits mixed in, it seems designed to lull you into a false sense of security

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    Supporting Member TomCarlos's Avatar
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    Thanks Dave!!

    I started to create a similar sheet based on my lab tests. But this is helpful - especially the examples with the choke.

    Much appreciated.

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    Quote Originally Posted by pdf64 View Post
    Have you had a good look through that page? There’s a fair bit of nonsense on it
    Care to point out the nonsense for us?

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    Quote Originally Posted by Dave H View Post
    Care to point out the nonsense for us?
    All the way through it references a V (Avg) D.C. value, without defining what it is.
    eg for the two phase cap input example, it states
    V (Peak) D.C. = 0.71 X Sec. V A.C.
    V (Avg) D.C. = 0.45 X Sec. V A.C.
    I D.C. = 1.00 X Sec. I A.C.
    What is V (Avg) D.C. and why is it different to V (Peak) D.C.?
    Bear in mind that it states at the top that losses aren't accounted for, so it can't be fudging in some notional degree of ripple loss.
    And, as per Helmholtz's post, the Iac ratios are wonky (yes I know I stated Iac=Idc for the two phase, but that just goes to show how pernicious that page is!)

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    I think V(Peak) means the no-load voltage. V (Avg) D.C. = 0.45 X Sec. V A.C. is correct acc. to my literature.

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    Quote Originally Posted by Helmholtz View Post
    I think V(Peak) means the no-load voltage...
    No load is the only condition in the document's scope.

    Quote Originally Posted by Helmholtz View Post
    ...V (Avg) D.C. = 0.45 X Sec. V A.C. is correct acc.to my literature.
    In the case of a two phase cap input, for that to be true, the cap would have to be hopelessly inadequate for the loading. And in any case, losses (from loading or whatever) are excluded from the scope of the document.

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    No load is the only condition in the document's scope.
    I don't see that. It would make no sense, especially as the diagrams show a load resistor and they specify a DC load current.

    You're right about the 0.45 voltage factor. It only holds for resistive loads (no cap), sorry. Will look up my books.

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    Quote Originally Posted by pdf64 View Post
    What is V (Avg) D.C. and why is it different to V (Peak) D.C.?

    And, as per Helmholtz's post, the Iac ratios are wonky (yes I know I stated Iac=Idc for the two phase, but that just goes to show how pernicious that page is! )
    V (Avg) D.C. is the average value of the rectified waveform without filter capacitor so it shouldn't be listed in any of the examples with a capacitor.

    It is possible for Iac to equal Idc for the two phase. It depends on the conduction angle.

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    Supporting Member TomCarlos's Avatar
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    Thanks for the continued discussion on this. I am taking notes. I must admit, for me there is lots of confusion around Vpk, Vrms, Vdc, that Vpk = Vdc, etc. So this is interesting conversation. I have studied electronics and have my books. But it is goofy when guys use terms interchangeably. If nothing else, the Hammond paper is a reference sheet - one with pictures and allows for me to capture notes.

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    It is possible for Iac to equal Idc for the two phase. It depends on the conduction angle.
    As the AC currents with bridge and two-phase rectification should differ by a factor of 1.414, only one of the formulas/factors for the DC current can be correct assuming the same conduction angle.
    I think it should read IDC = 0.88 X Sec. IAC for two-phase cap input. This would lead to:

    Iac = 1.14*Idc for two-phase and
    Iac = 1.61*Idc for bridge.

    Both these conversion factors lie within the ranges I gave and would apply for large conduction angle. Meanwhile I could confirm my numbers with other literature.

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    Quote Originally Posted by Helmholtz View Post
    I think it should read IDC = 0.88 X Sec. IAC for two-phase cap input. This would lead to:

    Iac = 1.14*Idc for two-phase and
    Iac = 1.61*Idc for bridge.
    I'm sure you are correct (assuming the same conduction angle)
    I wonder why Hammond chose the value the did?
    I'm sure I've measured Iac to be greater than 2*Idc on a few occasions in the past for the bridge.

    I've just run up a sim for the two-phase. It shows that it has to be a really lossy power supply (small cap, high series R) for Iac to be equal to Idc as I said above. Sorry about that.

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    The ranges for the AC/DC current conversion factors I posted cover conduction angles between 60° and 90°. This is what the authors call the "desirable" range.
    There are more general formulas which allow the the calculation of the conversion factor from a known conduction angle.

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    Quote Originally Posted by Helmholtz View Post
    The ranges for the AC/DC current conversion factors I posted cover conduction angles between 60° and 90°. This is what the authors call the "desirable" range.
    I've edited my post. It wasn't clear that I meant the value Hammond chose.

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    Regarding voltages things get complicated as it strongly depends on load current.

    The only thing I can safely state is that DCV must lie between the 2 extremes given as V(Peak)DC and V(Avg)DC* in the Hammond chart .

    The "peak" value would apply when the amp is in standby.

    * V(Avg)DC as used in the Hammond Design Guide is the averaged output DCV with resistive load (no reservoir cap). It represents the theoretical (but rather useless) lower bound for the averaged output DCV with capacitor input load.
    It still depends on load current and PT source impedance.

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    The green trace is the load voltage.

    Click image for larger version. 

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    V (Peak) DC and V AC (RMS) are fixed (ignoring PT source impedance). V (Avg) DC will drop with increasing load current and/or lower reservoir capacitance - associated with higher ripple voltage.

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    Supporting Member TomCarlos's Avatar
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    Helmholtz:

    I want to make sure I understand which scenarios we are talking about.

    As the AC currents with bridge and two-phase rectification should differ by a factor of 1.414, only one of the formulas/factors for the DC current can be correct assuming the same conduction angle.
    I think it should read IDC = 0.88 X Sec. IAC for two-phase cap input.
    This is #7 as shown in the photo? "Full Wave, Capacitor Input Load"?

    This would lead to:

    Iac = 1.14*Idc for two-phase
    Iac = 1.61*Idc for bridge.


    Again, Iac = 1.14*Idc for two-phase refers to #7 and Iac = 1.61*Idc for bridge refers to #6?

    Thanks!!

    Update: By the way, I am reading "Guitar Amplifiers Electronics, Basic Theory" by Richard Kuehnel. There is a section on Power Supplies Transformers. He covers "some" of what is being discussed here. So I am diving into the details as best I can.
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  32. #32
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    Again, Iac = 1.14*Idc for two-phase refers to #7 and Iac = 1.61*Idc for bridge refers to #6?
    Correct. Both types are actually full wave rectifiers.

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    More useful info here:

    http://www.valvewizard.co.uk/bridge.html

    And don't forget, with a GZ34 your DC voltage will be 20V to 30V lower than with SS diodes - except at standby.

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