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Questions on Audio Note Kit One

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aslee

Member
Joined 2006
Hello,

I am planning to build an Audio Note Kit One amplifier. When I view the circuit, I have a puzzle.

The power transformer HT secondary is 425-0-425 at 250mA. The PS uses CLC filter. The first cap is around 24uf, second is 220uf. The resistance of the choke is 80R. How is the B+ of 415V come about? It seems to me that the B+ will be approaching 500V.

Please kindly educate me.

Thanks and best regards,
Francis
 

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waltube

waltube

Member
Joined 2005
The 5U4 has an high internal impedance; following the data sheet you can understand the final Vdc available for a specific dc current.
I know very well this project and it is working very fine.
In my opinion is one of the best SE for 300B available for different reason, first of all the robust drive stage.

Walter
 
45

45

Member
Joined 2008
6A3sUMMER said:
425V rms x 1.414 = 600V peak
0.25A x 200 Ohms = 50V drop, 1/2 secondary winding DCR
40V drop, rectifier
0.25A x 80 Ohms choke = 20V drop

600 - 50 - 40 - 20V = 490V.

I also do not see how the B+ is shown as only + 415V.
Where is the additional 75V drop?
Click to expand...

There is no additional voltage drop. The output voltage at C8 is about 420V. Then add 5-6V volts dropped into the OPT primary and that's 415V.

The typical DC voltage from datasheet with 450V per plate, 67 ohm total plate-supply resistance, 40 uF input cap is "just" 460V for 275 mA current. It would be some 50V more for 150 mA.

Here the total current draw is closer to 150 mA but the total series resistance is a lot more, the input cap is about 1/2 and each AC plate supply is 25V lower.
 
6

6A3sUMMER

Member
Joined 2016
45,

You proved my point. With no additional voltage drop, you can not start with 600V peak, and have 415V at the plate of the output tube. That is a total of all the drops of 185V.

With no additional voltage drops, the B+ would have to be a modified to have a much smaller input capacitance than 47uF/2.
 
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waltube

waltube

Member
Joined 2005
The official manual of kit One says that the Dc voltage at pin 8 of 5U4 must be 450 volt.
At the out of the choke there are 435 Vdc; if it has 80 ohm of Rdc and there are 200 mA ( it is real) of max dc current we got the value listed on manual. And it is in line with the datasheet.
The current for each 300B is around 80 mA. At the anode of 300B we must found around 425 Vdc so the Rdc of OT is around 125 ohm

This is , less or more, what I also found in the past.


Walter
 
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6

6A3sUMMER

Member
Joined 2016
It seems as if one of the numbers is not right:

Unless marketing is cheating, transformers are rated Under Load.
425V 0 425V should be when it is under load.

425 x 1.414 = 600V
425Vrms = 600Vpeak

50Hz becomes 100Hz when full wave rectified

Capacitive Reactance = 1 / (2 x pi x f x C)
Xc = 1 / (2 x 3.14 x 100 x 23.5uF)
Xc = 68 Ohms

200 mA x 68 Ohms = 13.6V peak to peak ripple at the top of the two stacked 47uF caps.
The average of 13.6V peak to peak is 6.8V

600V Peak - Rectifier drop - 6.8V = 425VDC

600V - 6.8V - 425V= rectifier drop

rectifier drop = 168V That would be one HOT rectifier. Not possible without meltdown or fire.

Better check the voltages on the schematic, manual, and the working amp.
 
6

6A3sUMMER

Member
Joined 2016
OK, so the ripple is 85V at the first cap, 23.5uf.
That means the center of the ripple is 42.5V

Now the question is, are these two stacked 4.7uF, not two stacked 47uF?
is there a missing decimal, or perhaps an readable decimal in post #1?
Is the 425VAC on the schematic of post #1 correct?

600 V - 42.5V = 557.5V

557.5V - 425V DC at the first cap = 132.5V drop in the rectifier.
Clearly, 425Vrms (1/2 secondary) which gives 600V peak is wrong;
or 425VDC at the first cap is wrong;
or we have an impossibly very hot rectifier (132.5V drop), yes?
 
Chris Hornbeck

Chris Hornbeck

Member
Joined 2011
I wonder if some modeling here is based on wrong assumptions. For example, 600 Volts DC *never* exists. It's only a theoretical, based on no load, no losses, perfect rectifiers, etc. But it's not real and not significant to a power supply model. Before computer modeling, power supplies were designed using graphs taken from actual measurements (analog computing!) because shortcuts just weren't reliable or usable.


A second note is that "DC Volts" in this context only has meaning if carefully defined. With a highly non-sinusoidal ripple superimposed, God and the Devil dwell together in the details.


All good fortune,
Chris
 
45

45

Member
Joined 2008
6A3sUMMER said:
OK, so the ripple is 85V at the first cap, 23.5uf.
That means the center of the ripple is 42.5V

Now the question is, are these two stacked 4.7uF, not two stacked 47uF?
is there a missing decimal, or perhaps an readable decimal in post #1?
Is the 425VAC on the schematic of post #1 correct?

600 V - 42.5V = 557.5V

557.5V - 425V DC at the first cap = 132.5V drop in the rectifier.
Clearly, 425Vrms (1/2 secondary) which gives 600V peak is wrong;
or 425VDC at the first cap is wrong;
or we have an impossibly very hot rectifier (132.5V drop), yes?
Click to expand...

Again, there is no 600V at 250 mA output.

https://frank.pocnet.net/sheets/093/5/5U4GB.pdf

From the datasheet one can already see that at the first cap it can never be better than 470-480V with 450V AC per plate. Here the OP is talking about 425 so it is less than 470-480V already in the best case.

600V with 425V AC per plate I don't think will ever happen! It only occurs if there are about 450V AC per plate, the output current is some 25-30mA max, the total series resistance is 67R (here is around 200R) and input cap is 47 uF (here is about 24).

Moreover 425V is the supply voltage measured at the 300B plate and has to take into account the choke drop and OPT drop.
 
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6

6A3sUMMER

Member
Joined 2016
I think I know where my calculations went wrong in my other posts. I forgot to account for this:
Often, the peak rectifier current in a cap input filter is 3, 4, or 5 times the DC current the load.

The maximum specified plate current of the 5U4G is 800mA (0.8A).

Suppose the 1/2 winding DCR of the B+ secondary is 50 Ohms, and the peak rectifier plate current is 4 times the DC current: 0.8A versus 0.2A).
50 Ohms x 0.8A = 40V drop due to DCR.

And that is probably the voltage drop that I missed.

I believe that transformer manufacturers test the voltage output using a resistive load.
And they rate the transformer according to that test, i.e. 200mA of restive load.
There are no transient currents.

Then we come along and design a 200mA DC supply using a rectifier and capacitor input filter.
The transient currents are much larger than the 200mA, but they only last for a small portion of the alternation.
But do not forget the power integral of that transient, it is ((I)squared x R) x duty cycle.

Resistive load: 200 mA (0.2A) 0.2A x 50 Ohms = 10 Watts heating of the 1/2 secondary wire. The whole secondary heat is 20 Watts.

Rectifier and capacitor load: (0.8A x 50 Ohms) x 0.25 duty cycle = 10 Watts; 20 Watts for the whole secondary.

In either the resistive, or the rectifier/capacitor, the heating of the secondary winding would be 20 Watts.
 
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