Millett Hybrid
Mar 6, 2012 at 9:51 AM Post #16 of 76
 
Quote:
If using the OPA551 opamp, why doesn't the (-) input (pin 2) need to be connected to ground? Pete Millett just mentioned putting it in the feedback loop by connecting it to pin 6.
 


The op amp compares the 2 inputs and tries very hard to make them equal. 
 
When you connect the - input to the output the op amp tries to make the voltage on the output (which is hooked up to the - input) equal to the + input which gives you a buffer.
 
When you ground the - input and apply signal to the + input the op amp applies full gain to the signal on the + input and bad things usually happen. Full gain on the opa551=100dB=voltage gain ~10000.
 
http://en.wikipedia.org/wiki/Operational_amplifier
 
Mar 15, 2012 at 6:24 AM Post #17 of 76
I've got the whole thing almost done. I decided to build a sigma25. I'm trying to make sure that I used big enough heat-sinks and everything for the circuit. I've tried to figure out just how hot it will get, but I am a little unclear on the math and would really appreciate some help. So, I've got a 25.2VAC transformer that puts out 450mA. The voltage regulator is a 7824, mounted on a 1.5" heat-sink (3.5C/W). The voltage regulator and transformer are in an aluminum case from radio shack. I'm not exactly sure about draw of the Millet. If someone could point me in the right direction on the math (I really don't want to start any fires!) I would appreciate it. Thanks!
 
Mar 15, 2012 at 11:51 AM Post #18 of 76


Quote:
I've got the whole thing almost done. I decided to build a sigma25. I'm trying to make sure that I used big enough heat-sinks and everything for the circuit. I've tried to figure out just how hot it will get, but I am a little unclear on the math and would really appreciate some help. So, I've got a 25.2VAC transformer that puts out 450mA. The voltage regulator is a 7824, mounted on a 1.5" heat-sink (3.5C/W). The voltage regulator and transformer are in an aluminum case from radio shack. I'm not exactly sure about draw of the Millet. If someone could point me in the right direction on the math (I really don't want to start any fires!) I would appreciate it. Thanks!


I don't have all the time I need right now to walk you through this, but check Tangent's website.  He used to have a calculator for power supplies.  My concern is that a 7824 means you'll burn up quite a bit of voltage (and power through the heat sink).  We typically advise setting a power supply to 27VDC output if you're starting with a 24VAC walwart.  You see, when rectified, that 25.2VAC will gain quite a few volts - easily over 30VDC by the time it hits the regulator.  I think a 7824 is even lower voltage drop out than an LM317, which is what I'm most familiar with.  So, you may be burning several volts through the regulator for no reason.
 
Just a thought.
 
 
 
Mar 15, 2012 at 11:58 AM Post #19 of 76
I thought about that, but went with it anyway. I have the switch after the regulator, so I switched the amp off and then used my DMM to short across it in amperage. The current is ~0.17 to 0.19A. So even at worst case scenario, it's going to maybe hit ~250mA, which is not bad at all. AMB had some of the equations on the sigma25 site. Now that I know the draw of the amp, I should be able to figure it out a bit better.
 
Mar 15, 2012 at 3:56 PM Post #20 of 76
I typed out everything below, but its all on AMB's site :facepalm:
http://www.amb.org/audio/sigma25/
 
to calculate power:
Figure out how much DC you have straight off of the transformer. 
25.25vac*1.4=26VDC
 
The biggest current-suckers in the millet is/are the tube heaters followed by the output stages. The heaters are in series, the buffers in parallel. 
tube heaters=0.15A
Buffers NOTE! Current based on a slightly wild guess! adjust if you think its more=~40mA each*2=80mA combined=0.08A
Add em up : 0.15A+0.08A=0.23A
 
 
If you want 24V out of the reg, you have 24V-26V=2V across the regulator
 
P=V*I
2V*0.23A=0.46W
Add in a hefty safety factor and you should still be OK with a little baby heatsink.
 
Mar 15, 2012 at 7:18 PM Post #21 of 76
I measured the current within the Millet at ~0.17A, so I was figuring 200mA as a fudge factor. 25.2*1.4=35.28
Voltage regulator is regulating that to 24V. 35.28-24=11.28
11.28*.2 = 2.256W
 
How big of a heatsink does that need?
 
Mar 15, 2012 at 8:59 PM Post #22 of 76


Quote:
I measured the current within the Millet at ~0.17A, so I was figuring 200mA as a fudge factor. 25.2*1.4=35.28
Voltage regulator is regulating that to 24V. 35.28-24=11.28
11.28*.2 = 2.256W
 
How big of a heatsink does that need?


The rectifiers actually have a "silicon" voltage drop - 1.4V for a bridge, so it's more like 33.88.  Plus some recommend another 80-90% of that voltage for smoothing.  Still, that's close enough to 2W.  How big of a heatsink?  Huge.  That's relatively speaking, of course.  I believe you would do OK with a 1" wide, 1" tall finned heat sink similar to a Wakefield like we use in the MiniMAX (#647-10ABP), but I'm just working from memory at the higher current in a MiniMAX vs. the lower voltage drop - should be in the ballpark.
 
 
 
Mar 17, 2012 at 12:21 AM Post #24 of 76
I checked the heater circuit today, and each heater is only at 12.2 ohms. The whole circuit runs at about 24.4 ohms. The polyfuse adds one to two ohms, so the entire amp is only at about 25.2 ohms. Which, at 24V is about 1A. I'm thinking of adding a 100 ohm resistor to the heater circuit, bringing the total draw down a bit. Is there any reason not to? Or, am I doing something wrong? It's weird that I measure 0.2A through a 25 ohm circuit. Unless that 0.2A is the maximum that the transformer can put out.
 
According to the data sheet, the heater current should be 0.15A. The circuit is at 24V, which 24/0.15 = 160 ohms. Given that the heaters contribute 24.4 ohms, adding a resistor (160-24.4=135.6 ohms), adding a ~100 ohm resistor should bring the heater current close to the correct value, right?
 
Mar 17, 2012 at 12:41 AM Post #25 of 76
The DCR of the heater changes (a lot) depending on whether the heaters are hot or cold, with the cold DCR being several times lower than hot. The start up current of a tube amp is accordingly several times higher than its idle current. What you are measuring sounds pretty normal to me.
 
Mar 17, 2012 at 12:43 AM Post #26 of 76
That explains a LOT. I've been using .5A fuses and going through them like CRAZY. Then, once it was warm, the fuse lasted all night (until I let it cool off then turned it back on). So, moral of the story? Slow blow fuses!
 
Mar 20, 2012 at 7:55 PM Post #30 of 76
I gave it to someone to start listening, I told them I would need to recheck the tube biases about 50-100 hrs in. I'll ground the transformer box then. I suppose it has well over that inside, given that the transformer is in there. I assume you mean grounding it to the mains ground, not the regulated ground. Hmmm, I'll have to find a three-conductor wire.
 
But, it came out looking really cool. The UV LEDs in the tubes and the color changing LED inside the box, peeking out around the sides. Very 80s.
 

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