How does one "Test" an attenuator (vaccuum tube amp) - Page 1
Forum section: electronics / general

author: gooseEL34
author_rank: Regular Contributor
timestamp: July 11, 2013, 05:02:20 am
content: Ok.I can read a circuit, use a scope, and am handy with a VOM.Here is my issue.I play guitar and have many tube amplifiers (most in push-pull CLASS AB design using 6L6 or EL34 tubes)I have recently come across a THD hotplate (schematics unobtanium on the net) which is a power soak orattenuator.Basically, the audio Output transformer is routed to this device BEFORE the speaker cabinet and you can crank up the volume of a tube amp (getting that nice power section tube clipping), but bring it to bedroom (not arena) volumes.Problem is, it has been opened and I cannot tell why.Looks like it contains several power resistors sunk into heat sink compound and the shell itself is a massive heatsink.It is called a Gibson Power Stealth for those interested.I have concerns because I have always been told NEVER run a tube amp with an open load or even a mismatched load.How can I be sure this thing is functioning well (assuming that when I turn the attenuation knob, it seems to decrease the output volume).Of course I can check for redplated tubes but is there a test equipment method to making sure it is doing what it is supposed to?Thanks.TimAnd if anyone has a schematic out there let me know. It is a Gibson Power Stealth 8ohm version (which is a rebranded THD Hotplate)

author: ElectroIrradiator
author_rank: Frequent Contributor
timestamp: July 11, 2013, 05:51:16 am
content: Quote from: gooseEL34 on July 11, 2013, 05:02:20 amCertainly advice worth following in the general case.OK, your question:The schematic is probably very simple, 2 or 3 power resistors at the most.To test it, find an 8 ohm resistor, 1W is probably way overkill. Connect it across the output terminals in place of the speaker.Connect a variable power supply to the input, where you normally connect the amp.Set the voltage to 1V, and measure the current into the Hotplate (in series with one of the input terminals).Check that the 1V input voltage you expect from the variable PSU is accurate.Measure the voltage across the 8 ohm output resistor, just for fun.If the hotplate only contains power resistors, then you can now check the input resistance it presents to the amp: R = 1 V / I (input). Should be close to 8 ohm.If you tell us the input and output voltages from your testing, then I can draw a similar schematic for you.The only potential trouble spot with making one of these yourself, is if the Hotplate contains reactive components like capacitors or chokes.

author: John Coloccia
author_rank: Super Contributor
timestamp: July 11, 2013, 11:28:47 am
content: The THD is a resistive load. Just measure the resistance with a DMM. Most dangerous would be running with NO load....you can blow up the output transformer with all sorts of nasty arcing and things like that. Tube amps don't like to see higher resistances than they're expecting, but you could generally short them out if you wanted to as the tubes themselves tend to limit the current before blowing things up...at least for a while, anyway. In fact, a lot of output jacks short when you remove the plug, just in caseYou could check for red-plating once you're up and running. You may want to rebias a little colder. It depends on the amp and the exact setup. Marshalls especially, the legend goes, would sometimes get upset at seeing a purely resistive load, and tube life goes down, but you're decreasing tube life anyway because you'll be running everything louder than you used to. I generally ignore all of that and just play. When the tubes wear out, they wear out. As long as it sounds good, I'm happy.

author: gooseEL34
author_rank: Regular Contributor
timestamp: July 11, 2013, 04:15:56 pm
content: Unfortunately, I run Fender's not Marshalls, and those tend to be 4ohm OTs. I got this for a song, and I do have some vintage Traynor's that want to see 8ohms so that;s where it will mostly be used. Thanks for the help.I am trying to wrap my head around what this actually does.... Does it put a power resistor in parallel with my speaker load? And if so, wouldn't that make my amp "see" a load that is half of what it should be?ThanksTim

author: ElectroIrradiator
author_rank: Frequent Contributor
timestamp: July 11, 2013, 07:02:48 pm
content: It is probably easiest just to show an example calculation, based on Ohm's law. You actually need at least two power resistors in your black box, in order to present a fixed 8 ohm load to the amp.I'm making up some of the numbers as I go along, don't know which attenuation is actually used.But say you have a pair of EL34 in class AB, and they distort nicely at 50W peak output. You'd like to use this in your bedroom, so maybe ... 2W does the job without breaking the windows?That is a power reduction factor of 25. Power into a resistor is proportional to the voltage across the resistor squared. So P = E^2 / R. If we want to reduce the power by a factor of 25, then we need to drop the voltage by the square root of 25, which is 5, of course.So our resistor setup needs to have an input resistance of 8 ohm, with an 8 ohm speaker across the output, and drop the voltage across the speaker down by a factor of five. You need at least two power resistors to do this, one, Rp, in parallel to the output speaker connection, and one, Rs, in series with the parallel pair.Make a 'test voltage' across the input. We know we want the 8 ohm input resistance, so if we present a voltage of 8 V, the circuit will draw 8V / 8 ohm = 1 Amp from the amplifier output.The voltage across the output needs to be one fifth the input, 8V / 5 = 1.6V. Our speaker across the output will then draw 1.6V / 8 ohm = 0.2 A.As the speaker load and the parallel resistor has to carry a total current of 1A, equal to the input current from the amp, the parallel resistor must be 1.6V / (1A - 0.2A) = 2 ohm.That leaves the main power resistor, which drop the majority of the voltage down to the pair of resistors (speaker plus 2 ohm): R = (8V - 1.6V) / 1 A = 6.4 ohm.It may also be interesting to consider the power dissipated in each resistor, particularly for a device built for higher power. Total power delivered by the amp is 8 W = 1A * 8V.P(8 ohm speaker load) = 1.6 V^2 / 8 ohm = 0.32 W - 4%, equal to 1 part in 25.P (2 ohm parallel R) = 1.6V ^2 / 2 ohm = 1.28 W - 16%.P (6.4 ohm series R) = (8-1.6)V^2 / 6.4 ohm = 6.4 W - 80%.Frequently you would add a third resistor, for instance one connected in series with the speaker output terminal, as this gives you more freedom in choosing the value of the resistances used. The calculations becomes slightly more involved in this case, but are basically the same thing.My suspicion is though, that the resistances are not very critical. I'd probably use 6.3 ohm in place of 6.4, and 1.8+0.18 in series in place of the 2 ohm.Also remember that many power resistors are intended to be mounted on large heatsinks, like you have already seen.PS: You could do the test I suggest with the 4 ohm attenuator as well, just use a 4 ohm resistor across the output.

author: gooseEL34
author_rank: Regular Contributor
timestamp: August 08, 2013, 08:28:36 pm
content: Thanks!