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| Amp Central Station Amps, tubes, speakers & everything AMP related. |
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#1 (permalink) |
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Tele-Holic
Join Date: Sep 2007
Location: Halifax, Nova Scotia
Posts: 511
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Converting to Cathode Bias help
Hey Guys, experimenting with my Princeton Reverb project (don't worry, no vintage value) and read converting to Cathode Bias will give a "tweedier" tone.
Two questions, from a schematic I'm following, I need around an 800 ohm resistor (used a pot and have it set at 800 now), and a 47 mfd 50v cap. I'm too sure about the mfd designation, all the caps in my amp are marked in uf. I have a 4.7 uf 50 v sitting around (all the rest I have are much higher voltages) so I used that. Any help here appreciated! Turned the amp on and everything appears OK, nothing glowing too red, plate voltage (pin 3 to ground) reads 332 V for the 6V6s, but I tried checking the amps with the "transformer shunt method", reading milliamps from pin 3 on the powertube to pin 8 of the rectifier tube, but I can't get any reading at all, although it may be my cheap mulitmetre. Am I doing this right? At present the amp just sounds kind of harsh. I'm sure I wired everything correctly but I'm unsure about that 4.7 uf cap. Any help here most appreciated!
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http://www.myspace.com/darcyhoover |
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#2 (permalink) |
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TDPRI Member
Join Date: Jul 2006
Location: Gwelf
Posts: 25
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800 Ohm seems high. What schematic are you using? The cathode resistor on a 5E3 Fender Deluxe is 250 Ohm.
What voltage are you getting on the cathode resistor? Pin 8 to ground. What voltage are you getting on the plates? Pin 3 to Pin 8. |
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#4 (permalink) |
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Poster Extraordinaire
Join Date: Nov 2004
Location: Mint Hill, NC
Age: 63
Posts: 6,230
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hope the tech-savvy types can help you sort it out ... i think if you get it solved, you'll like the results. i've done this to fixed-bias Fenders and it's worked out well. don't worry about collectibility, it's an easily reversible mod.
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Truth is stranger than fact ... www.myspace.com/woodymitchellmusic BAND PAGES: www.myspace.com/stragglerswing (Stragglers - Western Swing) www.myspace.com/loafersgloryband (Loafers Glory - '70s country-rock) |
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#6 (permalink) |
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Tele-Holic
Join Date: Sep 2007
Location: Halifax, Nova Scotia
Posts: 511
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Thanks guys, I'm at work now and can't check, but I think I'll wait until I replace that 4.7 uf with a 47 uf 50 v cap, can probally get one tomorrow.
The numbers/schematic are from Torres's amp book.
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http://www.myspace.com/darcyhoover |
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#7 (permalink) |
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Friend of Leo's
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22 uF to 100 uF will work fine, but 100V power ratings are better than 25V or 50V, IMO - they give you a little more of a safety margin.
800 ohms is too high with that low of a plate voltage, IMO. While you'd probably be safe with a 250 ohm like a 5E3 uses, I typically get 'em in values of 400, 500, and 600. I then typically bypass one with another to try and hit the sweet spot - usually a 500/600 pair works well (~270 ohms). I love cathode bias in low watt amps. The Deluxe Reverb is probably the one exception.
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#8 (permalink) |
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Tele-Holic
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The traditional tweed setup is 250ohm cathode resistor and a 25mfd bypass cap. I'd try that as a first pass, as it is a "tried and true"configuration that should work well. After that, you can tweak if you feel the need.
The voltage rating (there is no such a thing as a power rating for a cap) should be 50V to be safe. You will probably see a voltage of approximately 20 -24V across the cathode resistor. One note: the tremolo on the PR is bias driven, and this works much less well when the output stage is cathode biased. If you don't use the tremolo much, this won't matter to you.
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---------- Tech Geek and Sensitive Artiste String bender ordinare! |
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#9 (permalink) |
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Tele-Holic
Join Date: Sep 2007
Location: Halifax, Nova Scotia
Posts: 511
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Is there any problem having a 1K pot in place of a resistor? I have 800 set now but I'll try those numbers depending on what size caps I can find in my parts bin!
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#10 (permalink) |
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Tele-Holic
Join Date: Sep 2007
Location: Halifax, Nova Scotia
Posts: 511
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OK, set the pot to approximately 250 ohms, and just happened to have a 25uf 100V sprague cap doing nothing, please forgive my ignorance, but where should I take these measurements and which are volts (DC) and which are mA?
What voltage are you getting on the cathode resistor? Pin 8 to ground. Powertube pin 8 in DC volts, right? What voltage are you getting on the plates? Pin 3 to Pin 8. Powertube pin 8 to rectifier pin 3? DC volts again? Sorry for the stupid questions guys, I had a full proof biasing method I had built (once again, following instructions but not really knowing what I was doing) thats now disabled that made it too easy. Oh yeah, the vibrato is obviously disabled.
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#11 (permalink) | |
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Tele-Afflicted
Join Date: Jun 2006
Location: Northwest Missouri
Age: 40
Posts: 1,309
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Quote:
Also, you wire it in series to a lower resistor (try 100 or 200 ohms), not in place of a resistor. Should you you crank that pot down you'd eventually get 0 ohms without the resistor, and you know that's not right. But, 0 added to a 100 ohm resistor is still 100. |
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#12 (permalink) |
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Tele-Holic
Join Date: Sep 2007
Location: Halifax, Nova Scotia
Posts: 511
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OK, makes good sense. I may just put a 250 ohm resistor in there, I think I have something quite close.
About those measurements for biasing, I removed the bias adjust kit I had installed, also removing the 1 ohm resistor from pin 8 of the powertube to ground that I installed with the kit. Is this required for reading mA as opposed to volts?
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#13 (permalink) |
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Tele-Holic
Join Date: Sep 2007
Location: Halifax, Nova Scotia
Posts: 511
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Fuse blown in my meter, so thats why I couldn't get some readings. Trip to Radio Shack tomorrow!
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#14 (permalink) | |
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Friend of Leo's
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Quote:
As someone who's had a 50V cap blow in a Champ (of all things), I think that the extra voltage rating is cheap insurance. And if all you can get are 50V caps, you can run two 47 uF's in series.
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#15 (permalink) |
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Tele-Holic
Join Date: Sep 2007
Location: Halifax, Nova Scotia
Posts: 511
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Our banned friend Guitarslinger posted this, there were a ton of replies that just confused the heck out of me but for the most part, can I just do this?
"How to bias a cathode biased amp with just a voltmeter -------------------------------------------------------------------------------- This is taken from a post I posted in another thread. It may be of help to some of you. As with fixed bias amps, cathode biased amps sound their very best when biased correctly. The following steps are a way to figure the bias without a bias probe, and only a voltmeter. If you have a bias probe, it is much simpler, as the bias probe is already reading the plate voltage in relation to the cathode, so all you need to do is multiply the plate voltage reading by the current reading to figure power dissipation. The voltmeter only method requires a few more simple math steps to get you there. 1. Measure the voltage drop on the cathode resistor to ground. Write the value down. 2. Divide this voltage by the value of the cathode resistor. This gives you the amount of current being drawn by both power tubes in milliamps. Write this value down. 3. Measure the voltage on the plates of the power tubes to ground. Write this down. 4. Now, subtract the voltage from the cathode resistor in step 1 from the voltage measured on the plates. Write this value down. Take this value, and multiply it by the current (milliamps) from step 2. This will give you the dissipated power (in watts) of both power tubes. Write this figure down. 5. Take the figure from step 4 and divide by 2. Write this figure down. This is the power dissipation (in watts) of each tube. For 6V6s, if it is over 12 watts, then you need to install a higher value cathode resistor. If it's 10.5 watts or less, you need to install a lower value cathode resistor. 6. After installing the new cathode resistor, do ALL of the steps again to see what you now have. You may have to repeat this process several times to get it dialed in, but it is worth it, and your ears will thank you. Of course if you are biasing a single power tube amp like a 5F1 Champ, omit the "divide by 2" in step 5, as you will already have the power dissipation of the single 6V6"
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#16 (permalink) |
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Tele-Meister
Join Date: Aug 2008
Location: Long Island, NY
Posts: 284
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Off the bat, you are underbiased with an 800 ohm resistor in there. That's the reason for the harsh sound. The 250 ohm will work just fine.
The reason you can't just keep a pot in there is because even at 2W, it would still be underrated. I wouldn't personally use anything under 10W. Also, keep the cap AWAY from the cathode resistor, which will get fairly hot. This heat could dry out the cap prematurely. Any value from 10-100uF will work fine.
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John R. Frondelli |
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#17 (permalink) |
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Tele-Holic
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No need to measure the current directly, which would require you to remove a wire or two. Just measure the voltage across the cathode resistor, then:
I=V/R where: I = current V = voltage across the resistor R = value of the resistor This will be the combined current for both 6V6s. Divide this by two to get the current in each (approximately). Typical values for this type of circuit are: R = 250 V = 22V Therefore I = 22/250 = .088A or 88mA (milliamps are thousandths of a Amp) or 44mA per tube. This is a high amount of current for a fixed bias amp, but typical for cathode bias designs. Assuming that your power supply is at around 400V, then the power dissipated by each tube is: P = I*V, or 0.044 * (400 - 22) = 16.6 watts Don't forget to adjust for the fact that we are looking at only one tube and hence half the total current! This is high for a 6V6, but again very common in tweed amps. To reduce this, the value of R must be increased. The power dissipated by the cathode resistor is: P = I*I*R, or (0.088)(0.088) * 250 = 1.94 watts
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---------- Tech Geek and Sensitive Artiste String bender ordinare! |
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#18 (permalink) |
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Tele-Holic
Join Date: Sep 2007
Location: Halifax, Nova Scotia
Posts: 511
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Found a fuse!
I'll replace the pot with a resistor tomorrow (haven't anything anywhere near 250 ohms so I'll need to run to Radio Shack anyway) but with the 100 ohm resistor in the system, I measure 270 ohms of resistance. Voltage drop across this resistance measures 9.12 volts. Plate Voltage pin 3 to gnd measures 366 volts each tube. 9.12v/270ohms = 0.03333 mA 366 - 9 = 357 v 357v X 0.0333 mA = 11.88 so according to the above, that works. Not quite your numbers though, do these sound reasonable? Did I do that right? Any of you guys make your way to Nova Scotia, I'm buying the beer!
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#19 (permalink) |
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Tele-Holic
Join Date: Sep 2007
Location: Halifax, Nova Scotia
Posts: 511
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Adjusting to 250 ohms (I'll get a 10W 250 ohm resistor tomorrow) gives me a 22v drop and a plate voltage of 360v, so
22/250 = 0.088mA 360-22=338v 338v X 0.088 = 29.744 w BIG CHANGE! Stick with 270 ohms? Seems like quite a difference in the voltage drop for 20 ohms, maybe I did something wrong???
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#20 (permalink) |
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Tele-Afflicted
Join Date: Dec 2005
Location: SW CR IA US NA PE
Age: 28
Posts: 1,919
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Hmm, something isn't right -- adjusting for slightly less resistance should have decreased the voltage drop and increased the dissipation only slightly. I'd rip that pot outta there and wait until you have some good resistors before proceeding. I've tried using pots to futz around with bias, and after frying one, I gave up on the idea.
Buy a 270, a 330, a 470, and maybe a couple others, and you'll have enough to get the job done. They certainly don't have to be 10 watters -- 2 would be just fine. Hope this helps! You're just a couple components away from victory here. - Scott |
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#21 (permalink) |
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Friend of Leo's
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It takes the fun out of it (and really isn't super critical for many cath bias setups, IMO), but a bias probe sure makes quick work of things! You've still got to check the B+, but probes at least help you to monitor how your tubes are doing.
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#22 (permalink) |
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Tele-Holic
Join Date: Sep 2007
Location: Halifax, Nova Scotia
Posts: 511
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Found a new supplier! Got myself a few 1W and 3W resistors from 100 to 500ohms (basically one of everything he had in that range) and got a new multimetre (can finally move past that $15 Radio Shack special), but then got called into work! I'll have to wait till tonight.
Besides the above voltage measurements, can I measure current with the "transformer shunt method", reading milliamps from pin 3 on the powertube to pin 8 of the rectifier tube?
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http://www.myspace.com/darcyhoover |
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#23 (permalink) |
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Tele-Holic
Join Date: Sep 2007
Location: Halifax, Nova Scotia
Posts: 511
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OK, here's the numbers with a 3W 270 ohm resistor and the pot is out of there.
R=270ohms (EXACTLY 270 ohms with my new meter) voltage drop 22.8 volts plate voltage 360 volts so 22.8/270 = 0.084 amps 360 - 22.8 = 337.2 volts 337.2 v X 0.084 amps = 28.3 watts And I get 38 milliamps from pin 3 on the powertube to pin 8 of the rectifier tube, with my fancy-all-singing-all-dancing-brand-spanking-new-auto-scaling meter! How do those numbers sound?
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#24 (permalink) |
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Tele-Afflicted
Join Date: Apr 2006
Location: Staten Island, NYC
Posts: 1,012
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I want to give it a shot myself...
http://www.tdpri.com/forum/amp-centr...-ac30ized.html Cathode bias a Peavey Classic 50....4 EL84s in the power amp this thread should help... |
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#25 (permalink) |
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Tele-Holic
Join Date: Sep 2007
Location: Halifax, Nova Scotia
Posts: 511
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So I take it those numbers are cool?
It certainly sounds good, quite a noticeable change, more than I expected, as I noted lifting the negative feedback made a slight difference, but with cathode bias the amp is much warmer, less headroom, certainly lost some sparkle but it sounds more bluesy, if that makes sense.
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