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PSU - dropping resistors?
PSU - dropping resistors?
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Old 27th February 2008, 04:02 AM   #1
AndrewF is offline AndrewF  Australia
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Default PSU - dropping resistors?

Hi all
If this question has been asked before, my apologies. I searched but am unsure of the correct terminology.

Am building my first tube amp
basically a Byrith 4-30/Elektor version.

I am restricted to using what I can salvage (poor student )
so have a power transformer putting out 385VAC resulting in around
540V DC.

This amp requires 440-450V, 410-420V and 160-185V
(depending which version, but I assume around these values is acceptable)

My questions -
What is the best way to drop 540V to the required levels?

If using RC filtering, is there a method to calculate the resistor values? Most webpages I have looked at seem to gloss over this

I have simulated the circuit in psudII but am unsure what value the load resistor should be. Is there a way to estimate this?

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Old 27th February 2008, 04:58 AM   #2
jnb is offline jnb  Australia
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Look at how much current a stage will draw at rest. Add each of the stages' currents for each separate rail. The necessary voltage drop divided by this current equals the resistor value. The voltage multiplied by the current equals the power dissipation.

How much current are we talking about here?
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Old 27th February 2008, 07:03 AM   #3
AndrewF is offline AndrewF  Australia
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Thanks for your reply

how to calculate the current of each stage at rest - stage voltage/anode resistor?

The Byrith article says it is a 30W amp, I am building a stereo version with a single PSU. In his article he wrote the EL34 tubes operate at 40mA each.

the article is here -http://www.lundahl.se/claus_b.html
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Old 27th February 2008, 08:40 AM   #4
soundbrigade is offline soundbrigade  Sweden
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One way to try to optimize the PSU, and this is a way I often go, is to use the PSUD II from Duncanamps. However I need to know the various currents, but a rude guess can help.

You have the complete article here with PSU:
Claus Byrith 30W
Due to economical cutbacks, the light at the end of the tunnel has to be switched off.
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Old 27th February 2008, 08:58 AM   #5
jnb is offline jnb  Australia
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Stage current would be (rail voltage - anode voltage) / anode resistor value.

Look at the 5-10 and 5-20 for examples of series dropping resistors. Knowing the rail voltages and these resistor values you can calculate the current intended to run through them.

If you just add an extra dropping resistor early so even the ouput stage is at the correct voltage, it stands to reason that the others would be correct with the publshed values of resistor.

Make sure your filter cap is large enough for the output stage and if this stage is run in class B at all, you may get a slight reduction in performance by doing it this way, YMMV.
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