How much HT current should the PT of an 4xKT88 amp be designed for?

[anson]

Hi Everybody,


I'm building an amp and I'm trying to figure out what specifications I should give to the guy who is going to wind the power transformer. So far I have:

Power tube heater (four kt88s): 4x1.5A = 6A @ 6.3V

Preamp heaters (five 12ax7s): 5x300mA = 1.5A @ 6.3V

Bias supply: = 50mA @ -100V

HT: ?

I'm a bit uncertain about the HT rating. I plan to supply the four KT88s with 540 volts, but am uncertain about the current that they will draw. Each KT88 will draw about 55mA at idle, so that means a 220mA rating at idle. But how much additional current should I design for when the amp is at full power?


Many thanks!!!!
Anson

[big_teee]

Have you ran it through the DreamTone current tool?
PT Current Draw
T

[anson]

Thanks, that was an interesting link. According to the on-line calculation, when I told it that I had four kt-88s at 548V, and 5 12ax7s, it said I need 399 mA of current.

f I have four kt88s each with 55mA at idle, and assuming that they are biased at 70% of max current, that means I need (4 x 55mA)/0.7 = 314mA for them. And if I add an additional 3.1 mA for each gain stage of 5 12ax7's, that comes to 314mA +(3.1mA x 10) = 345 mA. But the online calculator says 399 mA, and I don't know where they get the extra 54mA. Would the be for a safely margin? I figured that transformer winders would add that safety margin in AFTER they are informed of when specs they need to fulfill. ANy ideas?

[loudthud]

When a class B power amp is running at full power driving a sine wave into a resistove load, it's efficiency is about 70%. So if the output power is 100W, the DC power from the power supply (volts times amps) is about 140W. When the same power amp is driving a square wave into the resistive load, the efficiency goes up to nearly 100% and the DC power is about 200W. Problem is that guitar amps are not required to supply a continous output, so the power supply does not need to be as large as indicated above. How large does it need to be? Many builders would use a 100W power supply, some even less. A 200W power supply would run cool and last a long time. A 100W supply would run fairly warm, depending on how hard the amp is pushed, and there may be occasional failures.

The winding of a transformer gets hot primarily (pun) because of the RMS current flowing through it. The RMS current in a high voltage secondary is not a simple function of the DC current flowing in the load after rectification. The best way to determine this is with a modeling program, PSUD-2 is close enough for rock and roll. You need to know how much filter capacitance will be used and the amount of ripple allowed. 100W of DC power may require 180 VA rating of the transformer winding.

Edit: These numbers are given as an example, you haven't specified how much power your amp will produce. It's kind of a chicken-egg question but you have to start somewhere.

[Tom Phillips]

Your power output goal will affect the current requirements from your power supply. Therefore, let us know how powerful you expect your "4xKT88" amp to be. This information is missing on the PT current draw site parameter selection page. 100W would be easy for 4 x KT88 but I think you could also achieve 150W or higher with a stiff 540V B+ supply.

[bob p]

^^ This. I like to use online calculators for teaching & illustrating points, but being and old-school kind of guy I never trust building my amp to an online calculator, which could have a misteak in it. I always look to the tube data sheets for the answer.

I'm looking at the Genalex data sheet there's an example for Class AB1 P-P operation at 560V into a 4k5 load. A pair of tubes is rated for 100W output. That sounds pretty close to your situation.

http://www.r-type.org/pdfs/kt88.pdf

Look at Page 3. For a pair of tubes, Ia_max = 145mA. Ig2_max = 15mA, for a total of 160mA. Double that for 4 tubes and you're looking at 320mA of supply just for the output tubes.