Modifying a Fender Hot Rod Deluxe amp - Wirewound vs Ceramic resistor - Page 1
Forum section: electronics / projects

author: bsas
author_rank: Regular Contributor
timestamp: June 06, 2018, 06:28:18 pm
content: Hi all,I am playing on modifying my Fender Hot Rod Deluxe amp to avoid one of the most common design flaws of this circuit: two 470ohm/5W ceramic resistors (R78 and R79):https://www.thetubestore.com/lib/thetubestore/schematics/Fender/Fender-Hot-Rod-Deluxe-Schematic.pdfBasically, those are needed for the switching system on the amp, that allows you to switch between channels, etc. The biggest problem with those (very common and well documented and known problem, but Fender looks like it doesn't care at all and never fixes it) is that those ceramic resistors get SO HOT that a lot of times they can make cold solder joints around them or even burn the PCB under them. Very dumb...Anyway, a common "fix" is basically remove those and put new ones "higher" then the PCB, so, they don't "touch" the PCB. I honestly think that solution is kind of a "hack" since you still have heat dissipation over the wire and cold solder joints can still occur. You are basically avoiding the PCB burning problemSo, I was thinking about changing those to a "470 ohm 5W Watt Aluminum Housed Metal Case Wirewound Resistors", raise them above the PCB (probably with a non conductive double sided tape) to have at least "some" heat dissipation over the metal case (heatsink) of the resistor. But, I am not sure if changing from a ceramic resistor to a wirewound one can influence anything here because of the inductance. This got far deeper on my small electronic knowledge and that is why I need the forum "oracles"Thanks!

author: DaJMasta
author_rank: Super Contributor
timestamp: June 06, 2018, 06:37:07 pm
content: The inductance of the resistors really shouldn't be a problem (you're also increasing it by lengthening the leads) since it's part of the DC power supply, not a signal level thing. Moving them higher prevents insulating the underside, so it increases dissipation slightly, and then you get some additional dissipation on the length of the leads, so I would expect a decent improvement of the temperature on the board.That said, I'd go with 10W resistors if they fit. You don't need the power rating, but they'll be physically larger, which will mean there's more area to dissipate the same amount of heat. So long as that doesn't take up all the air space around where they are, it will mean a lower average temperature for the device and everything surrounding it. You can also try attaching a heatsink of some sort (probably easier with the aluminum body resistors with mounting tabs).

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 06, 2018, 07:52:41 pm
content: Quote from: bsas on June 06, 2018, 06:28:18 pmSure, and where are you gonna put 'em?You read the datasheet, right?Metal-case resistors are useless without heatsinking.They also need to be heatsunk as well as transistors do, because they contain the same magic smoke as transistors.Rather, stick with ceramic/vitreous/wirewound (they're all wirewound at heart, you'll know if it's not because you'll be paying extra for it), get a bigger size (10 or 20W) and find space to put that in.Simply put: higher surface area = lower surface temperature for the same watts. It cannot be done with a small resistor, it has to get bigger.Tim

author: bsas
author_rank: Regular Contributor
timestamp: June 06, 2018, 09:56:56 pm
content: Quote from: T3sl4co1l on June 06, 2018, 07:52:41 pmWell, a 5W with a small heatsink around it floating in the air still sounds better for heat dissipation then a 5W ceramic directly connected to the PCB, no?And I can manage to fit a 10W like "DaJMasta" suggested, even better right? Even if I don't put it against the metal chassi for full heat dissipation.I will see if I can stick it to the metal chassi, but, I don't want to mod the metal structure (for example, putting wholes to it)...

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 06, 2018, 10:16:36 pm
content: Quote from: bsas on June 06, 2018, 09:56:56 pmNoooo, that's not a heatsink, that's just the resistor body! You need to bolt the body to an external heatsink!That's why metal-case resistors kinda suck.Tim

author: bsas
author_rank: Regular Contributor
timestamp: June 06, 2018, 10:20:21 pm
content: Quote from: T3sl4co1l on June 06, 2018, 10:16:36 pmBut do they dissipate heat "worse" then a ceramic resistor if they are NOT bolt to an external heatsink?Also, what if you just get a big aluminium heatsink and put it the resistor itself?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 06, 2018, 10:45:30 pm
content: RTFDS -- they're usually rated, oh, 1W I think without a heatsink?"Worse" in that the maximum case temperature (before they release the magic smoke) is a lot lower than a ceramic resistor.Less temp = less watts per size.Quote from: bsas on June 06, 2018, 10:20:21 pmNot sure how to parse this.If you have a heatsink to bolt it to, yes, that's the way they're meant to be used. The steel chassis might be okay, but you said you'd rather not put holes in it, so that's out.Tim

author: bsas
author_rank: Regular Contributor
timestamp: June 07, 2018, 01:32:35 pm
content: OK, so, another question then:Is there any other (more efficient) way to do what those toasters... sorry, resistors... are doing? Based on the circuit it looks like they are just doing a voltage drop from +/- 48VDC to +/- 16VDC, but in a very inefficient way (pure heat). I thought about voltage regulators or maybe step down converters... but I am not sure...

author: DaJMasta
author_rank: Super Contributor
timestamp: June 07, 2018, 05:06:22 pm
content: There are definitely more efficient ways to do it, but it would require a fair bit of reengineering. Basically, the resistors are "isolating" the 48V supplies from the 16V supplies, which are regulated by the pair of zeners on the right side. It's a fairly power inefficient way of doing things, but it gets the job done.If you used a pair of regulators for the +-16V rails, you would remove the resistor and the zener for the appropriate regulator on both sides, and it would be more efficient in that it would require much less power just to be turned on - the current regulation scheme burns a lot of power even with no load on the rails. Otherwise, you could also do it with switching converters, but then you have to deal with switching noise and all sorts of other complexities, which probably isn't worth it given that you already have the +- voltage from the transformer taps.So a retrofit with regulators is possible, but there's a chance it will introduce a new problem and the current design is actually sufficient.... so I wouldn't bother unless power consumption was a big issue.

author: ConKbot
author_rank: Super Contributor
timestamp: June 07, 2018, 05:39:15 pm
content: Quote from: T3sl4co1l on June 06, 2018, 10:45:30 pmI think you're underselling the failure mechanism of the Al body resistors.With enough overload, they eject the end plugs, and all the contents. So you can end up with the wire winding spewed eveywhere, still potentially connected to the terminal.You can get ceramic WW resistors with radial termination, meant to act as a standoff from the PCB. As an example,https://www.digikey.com/product-detail/en/vishay-dale/CPR1010R00JE66/CPR1010R00JE66-ND/6005255The wide legs add stability, and a bit of heatsinking.

author: Zero999
author_rank: Super Contributor
timestamp: June 07, 2018, 06:41:01 pm
content: Quote from: DaJMasta on June 07, 2018, 05:06:22 pmI would get rid of the resistors and use the LM317 for the +16V and LM337 for the -16V supply. The 16V zener diodes can be connected in series with the regulators to drop the voltage to a safe level. Alternatively the LM317 and LM337 could be replaced with the LM7815 and LM7915, assuming the reduction in output voltage to 15V, from 16V won't make any difference.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 07, 2018, 06:55:11 pm
content: You would at least need LM317HV, and I forget if there's a 337HV, otherwise the rating is too close under overload conditions.The +/-48V isn't even used? They just throw in a voltage doubler here, just to burn it all away?The right way would be to use a 20VCT winding and FWB that, or a switching DC-DC to obtain all the secondary voltages from the main supply at high efficiencyTim

author: bsas
author_rank: Regular Contributor
timestamp: June 07, 2018, 07:01:46 pm
content: Quote from: T3sl4co1l on June 07, 2018, 06:55:11 pmDon't ask me how Fender design amps... they don't look like they evolved their electronics techniques since the 60s

author: Zero999
author_rank: Super Contributor
timestamp: June 07, 2018, 08:10:24 pm
content: Quote from: T3sl4co1l on June 07, 2018, 06:55:11 pmThat's exactly why I suggested connecting the 16V zeners in series with the input: to drop the input down to a safe level. There's no need for expensive HV parts.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 07, 2018, 09:04:14 pm
content: Quote from: Hero999 on June 07, 2018, 08:10:24 pmRight, I read right past that part.Tim

author: TimNJ
author_rank: Super Contributor
timestamp: June 08, 2018, 02:59:47 am
content: The schematic won't open, but I'm assuming you're talking about the screen grid resistors on the push-pull output stage.My opinion on this is: It's a tube amp. It's not efficient to start with. So, I wouldn't bother going down that route.A few months ago, I finished a custom PCB for Fender 5F6A ('59 Bassman). I used these resistors:https://www.digikey.com/product-detail/en/vishay-bc-components/AC05000004700JAC00/PPC5W470CT-ND/596731Pretty small for the power rating. I would just use something like that, and as others have said, you can stretch out the length of the leads for a little extra heatsinking. I think it'll be okay.

author: JS
author_rank: Frequent Contributor
timestamp: June 08, 2018, 03:37:38 am
content: Ohhhh so much trouble for 70mA, and who knows how much of that is going to the load and how much to the zenner!Efficency isn't a problem here as a 50W amp will be throwing away about 50W just on the output power stage, add the heaters, who just heat, plus the usual losses of every system and say thanks it has silicon rectifier! (oh tube rectifier sound as much better, I had both on my home brew AC15-ish, just a flick of a switch and the smoothness of that vacuum tube rect)You could also use some transistors at the bases of the existing zenners, so to the zenner only goes a minimum bias current. After you will have a loss of the BE junction which is about the same output voltage than the 7X15 option, but just a pair of BJT.JS

author: NiHaoMike
author_rank: Super Contributor
timestamp: June 08, 2018, 04:38:50 am
content: Add a series capacitor before the rectifier, then reduce the value of the resistors.

author: Bassman59
author_rank: Super Contributor
timestamp: June 08, 2018, 04:53:01 am
content: Quote from: bsas on June 07, 2018, 07:01:46 pmWhy should they have evolved? After all, their market bitches and moans when they do anything other than their vintage products! There are people who think that using a PCB instead of a wire-and-turret design produces inferior tone.(NB: I own '62 and '64 Bandmasters, '66 Bassman, '62 Deluxe, as well as a reissue '59 Bassman 4x10 and the Hot Rod Deluxe that is the subject of this thread. The '62 Bandmaster is my favorite, for both musical reasons -- its tremolo is much different from pretty much anything else -- and personal, as it was a gift from a now-dead friend.)

author: TurboTom
author_rank: Super Contributor
timestamp: June 08, 2018, 11:07:45 am
content: The resistors in the original design are dissipating 2.2W each, provided the voltages specified in the schematic are correct. I consider they are correctly rated. Put them on some standoffs so they are just further away from the PCB and you should be done.

author: David Hess
author_rank: Super Contributor
timestamp: June 08, 2018, 01:45:28 pm
content: A series regulator is not necessarily going to be much more efficient and 2.5 watts x 2 is insignificant considering even the idle power of the amplifier. If the zener current is a conservative (worst case) half of the resistor current, then with a shunt regulator:Resistor 2.2 WattsZener 0.55 WattsLoad 0.55 WattsAnd with a perfect series regulator:Regulator 1.1 WattsLoad 0.55 WattsAs the zener current drops, the shunt regulator becomes more efficient. So that is lots of effort to save at most 1.65 watts per side? Besides using higher power and better resistors, consider using two in series or parallel for each one to distribute the heat better.Be careful using zener diodes in series with regulators to drop the input voltage. The 317/337 current limit at more than 1.5 amps which will destroy a 5 watt 16 volt 1N5353B in short order and after the zener diode shorts, high input voltage will destroy the regulators and if they short which is likely, high voltage will destroy everything on the +/-16 volt supply lines. If you want to do this, add series resistors to the inputs to limit the current or implement a lower current limit in the regulator.I like NiHaoMike's idea of using a series capacitor before the rectifiers.

author: bsas
author_rank: Regular Contributor
timestamp: June 08, 2018, 05:54:22 pm
content: Quote from: NiHaoMike on June 08, 2018, 04:38:50 amSounds interesting...Quote from: David Hess on June 08, 2018, 01:45:28 pmI like it too, but, what is the math for this, so I can figure it out the values and tolerances of the new parts?Thanks guys, learning a lot here with this simple question!!!

author: NiHaoMike
author_rank: Super Contributor
timestamp: June 08, 2018, 06:57:05 pm
content: https://en.wikipedia.org/wiki/Electrical_reactance#Capacitive_reactanceIn summary, for your application, a 22uF nonpolar cap would be a good starting point, lowering the resistor values to roughly half the original. Actual calculations are difficult because of the nonlinear effect of the diodes and I would suggest using LTspice or similar to simulate the circuit, then play around by adding a series capacitor and reducing the resistor values, trying to get about the same current through the zeners.

author: b_force
author_rank: Super Contributor
timestamp: June 08, 2018, 07:29:16 pm
content: What about replacing it for a good step down converter?

author: bsas
author_rank: Regular Contributor
timestamp: June 08, 2018, 08:42:54 pm
content: Quote from: b_force on June 08, 2018, 07:29:16 pmExample?

author: b_force
author_rank: Super Contributor
timestamp: June 08, 2018, 08:57:56 pm
content: Quote from: bsas on June 08, 2018, 08:42:54 pmThere are many possibilities.You could take it from the 48V or take a separate power supply from the mains.A third option is using opamps that draw less current.edit: something like the DPAN02C-15 or DLC03C-15With some additional filtering.Didn't check the total current consumption yet.You could also make your own step-down converter

author: Zero999
author_rank: Super Contributor
timestamp: June 09, 2018, 08:19:38 pm
content: Quote from: bsas on June 08, 2018, 08:42:54 pmTo name a few:https://www.xppower.com/pdfs/SF_ITW.pdfhttps://www.tracopower.com/products/tmr3.pdfwww.coselpower.co.uk/Cosel Datasheets/Cosel MGW1R5.pdfThe only potential issue with these power supplies is they're switched mode, which means the output will have some ripple on it, at the switching frequency. If the audio amplifier isn't designed to cope with this ripple, there's a possibility it can be demodulated, causing a buzzing or whining noise to be heard through the speakers.

author: b_force
author_rank: Super Contributor
timestamp: June 09, 2018, 08:43:31 pm
content: Quote from: Hero999 on June 09, 2018, 08:19:38 pmWell, it's a guitar ampBut besides, there are a lot of ways to get rid of the noise with just simple (passive) low pass filters.A lot of times a little RC network will do just fine.Also most opamps have some common mode rejection.In the end a 65-100kHz switching freq is recommended.To explain it on a more practical way.I have used these kind of the shelf converters in high end broadcasting audio equipment (from very well known brands) without to much issues.So it's definitely possible.

author: Zero999
author_rank: Super Contributor
timestamp: June 09, 2018, 08:48:43 pm
content: Quote from: b_force on June 09, 2018, 08:43:31 pmIt's certainly possible to use this type of power supply in a audio amplifier application, but it's a bit more difficult than a simple zener or linear regulator. The power supply rejection of op-amps tends to decrease, with increasing frequency. I haven't analysed this design but it uses some pretty old op-amps. I wouldn't recommend one of these switched mode supplies, without some kind of filter.

author: b_force
author_rank: Super Contributor
timestamp: June 09, 2018, 10:05:48 pm
content: Quote from: Hero999 on June 09, 2018, 08:48:43 pmI hear yo, but can I tell you that a lot of very expensive priced equipment still uses these opamps?Normal remedy is use a series resistor of about 22-100 ohms and than a 10-47uF in parallel with some NP0 capsThese opamps draw a constant current and have a pretty light load, so in most cases that will just work just fine.Another good way is cascading two RC circuits to for a 2nd order (Q=0,5) circuit.If you REALLY need to get rid of everything, drop in some kind of easy low drop linear regulator.