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Old 16th March 2009, 09:47 PM   #1
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Default My first power supply

Hi there,


I've made my first power supply and I am asking for your comments on this. It is for the RH84 tube amp design which requires a 300V B+ line.

It is based upon the XPWR106 ( 275-0-275 125mA ) power transformer and user a EZ81 full wave rectifier. Attached are the .psu file for DuncanAmps PSU Designer II. It is ment to drive two EL84 and one ECC81 preamp tube.

It currently says 295V on the current tap.

Is this enough to power it? What changes do i have to make? And above all, is it right to ad a constant current tap of the combined tube currents? ( 2x48mA for the EL84 and 1x10mA for the ECC81 ).

Anyways, since it is a digital design i suerly wont mind to have to get to the drawing board again, so please fire away.

Thanks!
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Old 16th March 2009, 09:52 PM   #2
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Here is the PSU file
Attached Files
File Type: zip personal2-275-0-275-125ma.zip (283 Bytes, 51 views)
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Old 17th March 2009, 06:57 AM   #3
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A perfect current source is modeled as an infinite impedance, which means
any voltage across the current source will still produce the given current.
(A voltage source is the opposite, with zero impedance, such that there
is a fixed voltage for any current).

The current source here is providing no loading to the power supply, which
is another way of saying no power is being delivered to the current source.
You need to add some resistive load to get an idea for how much your supply
will sag. I don't know offhand what the plate impedance of the EL34 would
be, but roughly if it is supposed to draw 50 mA at 300 V, that is about
300 V/.05 A = 6kOhms. Put two of these in parallel and you have about
a 3k load. The preamp tube is typically negligible.

So put a 3k resistor in place of your current source, and I think you will find
you will be considerably below 300 V.
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Old 17th March 2009, 07:15 AM   #4
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Torrence, I don't understand your reasoning. Most people who use PSUD use a constant current sink to represent the load. It draws current and that's what you want to simulate. Whether any power is dissipated in the load doesn't matter, although admittedly it's perfectly OK to use a resistor instead.
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Old 17th March 2009, 07:44 AM   #5
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Default agree ray

the current shown is a load, not a source. PSUII calculates the sag since it knows what the impedence AND the
DC resistance of the transformer and choke are. From your info, you are pretty close to the money - the next move is to build and test in real life.

Personally, I tend to slightly overbuild the power supply - its easy to get rid of a few volts but it's tricky to put in a few more if you are short!

DOn't forget to figure in the heater loading - best way to see if you are ok is to calculate the TOTAL VA for the secondaries and compare to the rated TOTAL VA output of the transformer.

Good luck!
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Old 17th March 2009, 08:30 AM   #6
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Wel... going to buy the parts then. I had to make some changes because the Edcor trafos are very expensive here in the netherlands. I have found a Hammond classic HV 260G which has the following specs:

Primaries: 110V, 120V, 220V or 240V
Secondaries: 275-0-275 200mA
5V 3A
6,3V 5A

It is 40,- euros less compared to the Edcor trafo and is able to deliver 50mA more current (wich gives me the opportunity to, at a later stage, create a PP EL84 amp or maybe other larger type of tube SE ).

It also gave me the use of the 5V output which i could use for the heaters of another rectifier tube, the GZ34.

After a bit of twiddeling in PSU designer I was able to get the same, nice clean curve again. Which means that I indeed need to buy, build & test.

How do i make an artificial load? because i do not want to test this at the end when i have finished the whole amp ( and discover that it doesnt work ;-) ).
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Old 17th March 2009, 10:18 AM   #7
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Sorry, I have only quickly glanced at PSUD in the past.
I thought this was actually an active current source in
the diagram. My bad...
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Old 17th March 2009, 10:25 AM   #8
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No problem, everybody makes mistakes from time to time.

Does anyone know what size ( ohms and Watts ) to use as a bleeding resistor parallel after C2? or how the size is calculated?
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Old 17th March 2009, 08:44 PM   #9
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Default if you are talking a bleeder for

running down the power supply after you switch off, I generally put a 220k 1W in there (the actual resistance being pretty immaterial as long as it is somewhere between 200 and 500k).
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Old 17th March 2009, 09:22 PM   #10
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Quote:
Originally posted by hilbert_mostert
No problem, everybody makes mistakes from time to time.

Does anyone know what size ( ohms and Watts ) to use as a bleeding resistor parallel after C2? or how the size is calculated?
Hi Hilbert

Just apply ohm's law. Say you want to achieve 300V/200mA with your supply. Use the equation V=R*I, with V=300, R = desired resistor and I = 0,2A. Dividing 300V by 0,2A gives 1500R (1k5). The dissipation in this resistor will be about 60W... so you need quite a resistor here. A possibility is to put some of these in series and bold them all to a heatsink. Another cheaper and easier possibility is to employ some light bulbs in series... 300V is not really handy (220V would be much better), but start by putting 2x 40W/220V bulbs in series across the 300V, and measure how much current they draw. If it is too little, use 2x 60W/220V and so on. Measure the current drawn by applying ohm's law to the voltage read across the choke (divide this voltage by Resistance of choke)

Erik
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