CT Transformer Voltage Question

[dbishopbliss]

I finished putting together my linestage breadboard last night and I got very unexpected voltages from my power supply.  My transformer has the specifications indicated in the diagram (500VCT, 40mA). 

I implemented a center tapped full-wave rectifier using UF4007 diodes.  When I measured the DC after the diodes and before CRC filter, the value was around 300VDC.  This is much lower than I was expecting.  Without getting into a lot of technical detail, can someone explain why this value isn't closer to 500V?

 

[JC]

Is the CRC filter hooked up at this point?  Where are you connecting your meter probes?

More importantly, perhaps:  What is the open-circuit AC Voltage from the secondary?

 

[Doc B]

You should have 250VAC present on each leg, relative to the CT. A full wave rectifier will give very roughly 1.4 x 250, about 350VDC with a nominal load.  Your preamp circuit may be pulling more current than the transformer is really designed for and that is pulling the voltage down.

 

[dbishopbliss]

I am getting 250VAC on each leg.  I thought I would have somewhere are 500V after rectification before the CRC filter.  So, the 500VCT is what confused me - I was thinking you added each side together.  I was hoping to have around 500V before the CRC filter then drop the voltage down where I want it after that.  The actual value may be close to 350VDC I will have to check again.  I just remember that it wasn't the 500VDC I wanted.

Based upon the formula Doc posted, I should look for a 350VAC transformer to get me in the ballpark.  This is why you breadboard and connect everything using terminal strips.

 

[JC]

You know, in reading over your original post again, I came to realize that I hadn't actually addressed your question.

Using  a two-diode full wave rectifier with a center-tapped transformer, you should figure on getting something in the neighborhood of the peak Voltage of one half of the full secondary winding.  As Doc points out, this should be perhaps just north of 350 Vdc.

Why only half?  In this version of the full-wave rectifier, you might think of it as combining two half-wave rectifiers; one diode rectifies the top half of the winding. During the time it is "off" ( the half of the AC wave when it is not forward-biased), the second diode rectifies the bottom half of the winding.  The resultant rectified Voltage, then, is a combination of the two as they trade off.  The two amplitudes don't add, though, because they are never "on" at the same time.

This type of rectifier circuit was particularly useful with dual-diode rectifier tubes.  It has the additional advantage of supplying essentially all the secondary current available from the transformer.

If you need something nearer the full Voltage of the secondary, you can always go back to using a full-wave bridge; simply tape off the center-tap, and ignore it.  The trade-off (seems like there is always a trade-off!) is that you will no longer be able to count on the full secondary current.

 

[dbishopbliss]

This makes sense now. 

Is there a formula for determining the secondary current if I implement a full-wave bridge? 

I will have two tubes drawing around 10 to 15mA each.  Do you think the transformer will be able to handle it, or would I be better off adjusting my operating points?

 

[Beefy]

I once stumbled across the following PDF from Hammond. It explains the voltage and current conversions you can expect from various transformer and rectifier schemes:

http://www.hammondmfg.com/pdf/5c007.pdf

So for your situation, you have a full wave capacitor input loaded, so voltage is 0.71 x 500 = 355VDC (this is the same as 1.4 x 250 for each leg that Doc did above), and DC current should equal the AC current.

 

[JC]

You know, to be honest with you, I know of no reliable formula.  I generally rely on what the transformer manufacturer says its product can do with different rectifier and filter arrangements.

One rule of thumb I have heard that seems conservative to me is as follows: For a full-wave (two-diode with CT transformer) rectifier, feeding a CRC, you will need a transformer capable of about 20% more than the current you intend to draw.  This allows for overhead for charging current, which can be considerably higher than load current.  For a full-wave bridge (four diode), the transformer should be sized to supply about 80% more current than you expect the DC load to draw.

It gets even more complicated when you factor in different filtering arrangements, temperature rise in the transformer, etc.

Using this rule of thumb, if you want to use a full-wave bridge on the full 500 Vac winding, with a CRC filter, to supply 20 mA of current to your load, you would want the transformer to be able to supply 36 mA from its secondary.  If you need 30 mA DC for your load, the transformer would need to supply 54 mA.

 

[Paul Joppa]

Lots of good stuff here in this thread. I'll throw in my two cents.

First, it's confusing that the terms "full wave CT" and "full wave bridge" are so similar, when the performance is so different.

Second, the peak voltage from a rectifier is only achieved when there is no load on the power supply. In most cases, if the supply is running anywhere near the normal current, you will get 10% to 30% more than the rms voltage - not 41% which is the peak. This is due to the losses in the resistance of the transformer.

Third, ratings are confusing as well. (Confusing as hell?) Traditional high voltage tube power transformers are usually rated by the DC current you can draw from a fullwave CT power supply into a capacitor input filter. Unless otherwise noted, that is. :^) Filament transformers, and most general-purpose transformers, are rated for the rms current in the secondary winding, which is usually 1.5 to 2 times as much as the DC current you can draw from a fullwave bridge, or 1 to 1.5 times what you can draw form a fullwave CT arrangement into a capacitor input filter. Choke input filters generate less voltage and more current. Also, the rated voltage of a transformer is the voltage when the transformer is delivering the rated power; at low current the actual voltage is higher.

To get any closer, you really must do the analysis. There are books and curves plotted from the old days, but I highly recommend the simulation tool called PSUD, which you can download from the Duncan Amps website. It takes some time and effort to learn to use it, but it's time well spent.