[scootsit]

I recently completed a Millet Mosfet Max and now, I've got the bug. I'm building a Millett Hybrid on some Radio Shack stripboard. I started this thread to document and post some pictures. Since the Millett PCB is no longer available, I thought I'd also post my board lay-outs from DIY Creator. I found a source for some free Intersil HA3-5002 chips, which Pete Millett discussed as one suitable replacement for the (now very, very pricey) Buf634. I need to revise the boards for that. I was wondering, why does the HA3-5002 have a V1+ and a V2+? I can't seem to tell from the original boards how power was supplied.

 

[tomb]

Hopefully, I can resolve your questions through some e-mail (as I suggested in response to your PM).  Unfortunately, the schematic is more a symbolic representation of the revMH Millett, rather than a strict documentation of the PCB.  However, there are a number of files that I can share that will shed more light on it.

 

[scootsit]

I appreciate it. I posted this a few hours ago before contacting you directly. I just figured I'd go straight to the source (or at least the person that moderates the site). ANYWAY, I appreciate it. I've read through the data sheets and some of the earlier schematics, I'm just struggling with that one issue, regarding the two voltage inputs. I'm starting to think that one is a current regulator for the output. Anyway, I could circumvent the entire problem, by just using a different buffer. As always, I appreciate your help.

 

[nikongod]

Buf634p are only $9 at newark. Yea, we all remember when buf634p were $6. But then we all remember when gas was $0.95/gal and the dollar had value in the international market too. In a few years $9 will be cheap for a buf634.
 
The To220 version is slightly more expensive, but allows for heatsinking which is totally worth it IMO. 
 
The different voltage inputs on the intersill buffer are there to allow the designer to provide power to the "input" and "output" section of the buffer separately. This allows for the use of external protection circuitry which is nice on a chip with no output protection onboard. See page 6 of the datasheet.
 
http://www.farnell.com/datasheets/32546.pdf

 

[scootsit]

The real appeal of the Intersil is that my school has a good relationship with them (read: free).

 

[scootsit]

But, I guess for the output protection, I might as well go with the TI chips.

 

[tomb]

Told you nikongod liked those TO-220 BUF634's with heat sinks.

 
FYI, it's not really the same as putting a DIP-8 heat sink on top of a chip.

 

[scootsit]

I figured as much, but I figured it would be better than nothing!

 

[scootsit]

Okay, I'm almost done. I was looking at AMB's design of putting the LED in the socket, like in the Millett Max. I assume the resistor values have to be different. I was thinking 3K for the LED resistors, given that they are in parallel with the tube heaters, I figure a higher value would keep more current going to the heater. Is 3K enough or should I go higher?

 

[tomb]

Quote:

Okay, I'm almost done. I was looking at AMB's design of putting the LED in the socket, like in the Millett Max. I assume the resistor values have to be different. I was thinking 3K for the LED resistors, given that they are in parallel with the tube heaters, I figure a higher value would keep more current going to the heater. Is 3K enough or should I go higher?

Why would the resistor values have to be different?  1K was used - but be sure it's at least a RN60 (1/2W, non-military rating).  The heaters will draw whatever they need in current - as long as the walwart/power supply can serve it up.  Each LED will pull about 12 ma, tops.  That's simply the voltage divided by the resistor (12/1000 = 0.012A).  Technically, you should add the resistance of the LED to the resistor value to calculate this, but it's close enough for sizing.  Brightness is neglibly reduced at 12ma and you won't have to worry about the LED burning out if it happens to get close to 20ma, the usual LED limit.  One would normally size it for 10ma, but hey - a 1K resistor is so common it's hard to resist (uh-oh, a pun!).  There are already several of them in the Millett circuit.  Power is 0.144W through the resistor ((0.012**2)*1000).  We have a long-held practice around here of 2X safety factor on the power-rating of a resistor, so that would become 0.288W, which is more than a RN55 (1/4W, non-military rating).
 
 

 

[scootsit]

Why did I think that? Because I don't REALLY know what I'm doing and I more or less guess/try to figure it out as I go. Thanks for the info.

 

[Breck]

I thought I'd pipe in, mostly since I love lighting my projects, and love seeing LED-lit projects.  If you're curious how to select a bias resistor for the LEDs, here's the method I use:
 
A)  Select your LED.
B)  Locate a datasheet for your LED.
C)  Choose an LED current
            i. The datasheet will usually list an optimal one, but nobody said you can't go lower.  
            ii. Do not exceed the maximum, or you'll see bright, pretty colors followed quickly by a dead LED.
            iii. I usually find somewhere in the 10-20 mA range to be quite acceptable for a 5mm LED.
D)  Locate the I-V relationship curve (graph) on the datasheet.  
            i. Look up the current you chose on the curve and find the corresponding voltage.
E)  Now you know how many volts will drop across your LED at the current you chose (the voltage you just looked up).  Subtract this voltage from your supply voltage (I'm assuming 6.3 V since this is on your heaters), and this will give you the voltage that will be dropped across your bias resistor.
F)  Compute the resistor value (Ohm's law:  R = V/I.  "V" is the voltage dropped across the resistor from part E, and "I" is the current you chose in part C).
 
If you want more than one LED and are concerned about power, you can wire multiple LEDs in series, so long as the voltage across them all (the voltage from part D multiplied by the number of LEDs) does not exceed your rail voltage.  Then to find your bias resistor you just subtract the total voltage dropped (again, the V from part D times the number of LEDs in series) from the supply voltage and use Ohm's law.  I'm guessing it wouldn't be too hard to get 3 LEDs in series running off a 6.3V heater supply, depending on the color.  That would use 1/3 the power as if you put them all in parallel (biased to the same current).
 
Can't wait to see your tubes glow!

 

[scootsit]

Breck, that was a simple, concise explanation, that will be invaluable in this project. Thank you. If I'm not mistaken, the heater voltage in the Millett is 12V.
 
Thanks again!

 

[tomb]

Quote:

Breck, that was a simple, concise explanation, that will be invaluable in this project. Thank you. If I'm not mistaken, the heater voltage in the Millett is 12V.
 
Thanks again!

Yep.
 
 

 

[scootsit]

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.
 
I'm also probably going to add an E12. I'm confused as to how the E12 can function from 12-30V, yet every component is rated to 12V. I assume every component needs to be sourced for the appropriate voltage, right? (In this case, 24V)
 
Edit: Nevermind, I didn't realize that D1 was replaced when used for single rail applications.