AC to rectified DC voltage math

[Greg_L]

I did a search, I swear! I found some info...but I'd like some answers specific to my project.

I'm building a late 60s Marshall replica. My PT has two high voltage secondary windings.
1) 152-0-152
2) 185-0-185

I'm planning to use the 185v winding.

The PT center tap will connect between the two main filter caps for a voltage "lift"?

Also...I'm using a full wave bridge rectifier configuration.

What kind of B+ can I expect and how do I calculate it?

I'm not an electrical engineer or a mathematician. Keep it basic for me please!

[Helmholtz]

Bridge rectifier is connected between 2*185V= 370V.
For unloaded DCV multiply by square root of 2 (1.414) giving 523VDC
At idle B+ will be around 500V. Exact value depends on actual mains voltage, power tube bias and PT regulation.

[Greg_L]

Bridge rectifier is connected between 2*185V= 370V.
For unloaded DCV multiply by square root of 2 (1.414) giving 523VDC
At idle B+ will be around 500V. Exact value depends on actual mains voltage, power tube bias and PT regulation.

Ok great thanks. Good to know. I saw that formula in my research and thought it can't be that simple.

[TomCarlos]

Hi Greg....

For your reference... See attached files.

Hammond_Design_Guide_For_Rectifier_Use.pdf

[Greg_L]

Hi Greg....

For your reference... See attached files.

[ATTACH]n987785[/ATTACH]

Wow nice, thank you. Cool reference.

So mine is a #6 with fries - full wave bridge capacitor input load.

[Helmholtz]

This design guide contains errors.
E.g. #6 V(avg)DC is wrong.

[Greg_L]

This design guide contains errors.
E.g. #6 V(avg)DC is wrong.

Can you elaborate?

[Helmholtz]

Can you elaborate?

The factor 0.9 would only apply without a smoothing cap.
With cap the average DCV can vary up to 1.4 times Vrms depending on load current.

[Greg_L]

The factor 0.9 would only apply without a smoothing cap.
With cap the average DCV can vary up to 1.4 times Vrms depending on load current.

I see. Is that because the cap fills in the ripple?

[Helmholtz]

I see. Is that because the cap fills in the ripple?

Yes.
Without load current there's no ripple at all and the cap is charged to the peak voltage.
With load the current discharges the cap between the voltage peaks.