Bottlehead Forum

Bottlehead Kits => Legacy Kit Products => Quickie => Topic started by: rif on March 31, 2013, 03:35:34 PM

Title: Simple Quickie circuit question
Post by: rif on March 31, 2013, 03:35:34 PM
I'll be starting my build this week if all goes well.  I'm also going to decipher the circuit, since it looks fairly simple and I've never really done that before.

I'm sure I'll be posting lots of questions...

Is this basically a "grounded-cathode amplifier" design?






Title: Re: Simple Quickie circuit question
Post by: Paul Joppa on March 31, 2013, 03:59:02 PM
...Is this basically a "grounded-cathode amplifier" design? ...
Yes it is!
Title: Re: Simple Quickie circuit question
Post by: rif on March 31, 2013, 04:04:55 PM
"Excellent
Title: Re: Simple Quickie circuit question
Post by: rif on April 05, 2013, 01:41:53 PM
I've been trying to read and comprehend but my brain isn't what if used to be :( 

So please don't mistake simple questions for laziness.

In the manual, there's a 4000 ohm resistor attached to the plate labeled RL and referred to as the load.  But isn't the resistor on the plate going to B+ usually called the plate resistor Ra (or Rp)?

Also, Rout (270k) - isn't this RL, the load?

Since the 1.5v from the D cell is in series with Rk, does that voltage add to the drop over Rk to set the bias?

Thanks! (And more to come depending on the answers)




Title: Re: Simple Quickie circuit question
Post by: rif on April 09, 2013, 03:42:32 AM
Or is the load 4000 (resistor) parallel with 4000 (intrinsic tube)= 2000 ohm?  Or in series equaling 8000 ohm?

The more I think, the less rigorous and more confused I get. Help.
Title: Re: Simple Quickie circuit question
Post by: Paul Joppa on April 09, 2013, 05:15:05 AM
rp is the usual term for the internal small-signal resistance of the tube, called "plate resistance". RL is the usual term for the resistance into which the output power is dissipated, called the "load resistor". Since the power supply has a low impedance, the two are effectively in parallel when seen from the output. Yes, the cathode is the filament, which has +2v at one end and +3.5v at the other end, so the effective bias is distributed along the filament.
Title: Re: Simple Quickie circuit question
Post by: rif on April 29, 2013, 10:46:23 AM
Ok, making progress but slow. I think I understand the general circuit, now trying to figure out the particular values.

Since its dropping 9v (38-29v) across the plate load resistor (4.02kohm), the operating point is 29v, 2.25mA? That's where to look on a plate curve in the tube manual?

A current of 2.25mA through the 1.07kohm cathode resistor is a 2.1v drop. Add that to the 1.5v continuous drop across the cathode filament from the battery, gives a bias of 2.1 to 3.6v depending where on the filament the signal hits and where the electron is coming from.  Help me understand that part - how can the bias be a range and not a set number?

I think I'm almost there.



Title: Re: Simple Quickie circuit question
Post by: earwaxxer on April 29, 2013, 12:11:43 PM
Damn rif - if you're right, that some good shit!
Title: Re: Simple Quickie circuit question
Post by: rif on May 09, 2013, 06:49:56 AM
Ok so in addition to the above questions, I've moved on to the heater. If I measure the filament resistances with the tube out of the socket, I get about 10 to 12 ohm for each (pin 1-5 or 5-7). In parallel, that's the equivalent of 5 to 6 ohm. D-cell provides 1.5 volts, that's a current of 0.25 A.  That seems very high, the tube is rated max 100ma.

If I set the dmm to current, I get 60ma from pin 1 to 5, which sounds reasonable. That equates to a parallel resistance of 250ohm.

Which is right or what am I doing wrong?

Thanks again!
David
Title: Re: Simple Quickie circuit question
Post by: Paul Joppa on May 09, 2013, 09:58:56 AM
Filament resistance, like that of any wire, increases rapidly with temperature. You'll get the same result measuring a light bulb, which get even hotter and so have an even more dramatic resistance change from cold to hot. (I just dug out a 60-watt bulb and measured 22 ohms, but it would be 240 ohms at full voltage and temperature.

Incidentally, this is  a good way to measure the internal temperature of a power transformer. Just measure the resistance of the primary winding (usually that's the innermost one) cold, and again after the amp has been on for a few hours. Look up the temperature coefficient of resistance for copper wire on the internet, do the math, and viola - you know the temperature!
Title: Re: Simple Quickie circuit question
Post by: rif on May 09, 2013, 11:09:07 AM
That's really cool, I love these "tricks".  Resistance varies with temperature, that makes sense. It also explains why I couldn't get a stable measurement when measuring after the tubes heated up, then removing the power. I guess battery powered filaments cool quickly.

BTW, did I get the numbers right above in post #6? I hate to bug you but not knowing is bugging me :)

-david
Title: Re: Simple Quickie circuit question
Post by: Paul Joppa on May 09, 2013, 12:37:10 PM
Re post 6 - look up a data sheet for the 3S4 - try to find the Sylvania sheet, it's the only one with triode-mode curves. You will see that the bias on the plots is referred to the negative end of the filament, and the plot will specify whether series or parallel filament operation applies. The data sheet will even say which pin should be negative. So the operating point is (29v - 2v bias = 27volts), 2.25mA.
Title: Re: Simple Quickie circuit question
Post by: earwaxxer on May 11, 2013, 04:27:21 PM
Here is a white paper on the 3s4.