Stupid idea?

[Hellhammer]

I guess it is as I've never com across it before, but I'm thinking this switch would bring B+ down by half. A sort of one step power scaling. Would it work? Would half the voltage be a big enough difference for this to be worth it?

[IMG:298:298]http://i94.photobucket.com/albums/l96/s ... razyPS.gif[/img]

[Firestorm]

I think you'll wind up with the same B+ either way: the peak AC voltage or 1.4 x RMS.

If you want half voltage, ground the negative side of the bridge and take B+ from the positive end of the bridge and 1/2 B+ from the center tap of the transformer.

You'd definitely be able to hear one half of B+, but it wouldn't get to berdroom levels.

[martin manning]

The FWB (CT ungrounded) will produce 1.41 x the whole secondary minus two diode drops, and the FW version (CT grounded) will produce 1.41 x half the secondary minus one diode drop. So, yes, ~2:1 voltage ratio.

[Firestorm]

Ahh. I believe Martin is right. What are the relative currents?

[Hellhammer]

In a Full wave bridge its 0.62 of the AC current spec while in a FW its equal to the AC current spec.

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

[Cliff Schecht]

Where does the 0.62 number come from? I've always thought is was closer to 0.707 (the square root of 2).

[Bob-I]

The 70's Peavey's used the CT in another way, the SB switch would select either the HV or the CT for lower B+ on the power tubes. Since the preamp was SS this only changed the power tubes.

[Hellhammer]

Where does the 0.62 number come from? I've always thought is was closer to 0.707 (the square root of 2).

Don't know the math.. It's from the Hammond pdf.

[Super_Reverb]

Not a stupid, but rather an innovative idea IMO

One way to look at difference in full wave bridge vs full wave rectifier (like a 5U4 or a pair of silicon diodes):

Full wave bridge is using both phases of the PT secondary at the same time. It is in effect placing the two halves of secondary in series.

Full wave rectifier is using one phase of PT secondary at any given time with respect to center tap.

The differences between RMS and average voltage measurements have to do with definitions in how the voltage does work when we connect a load to the secondary voltage output.

Here's a reference to a thread about RMS:

http://www.gearslutz.com/board/masterin ... -peak.html

The 0.707 factor comes from RMS definition relative to an instantaneous AC voltage at 45deg in sine wave: sin(45deg)=0.707=cos(45deg)=1/sqrt(2)

and from Dr. Math:

For sine waves, the exact formulas for converting among peak,
peak-to-peak, rms (root-mean-square) and average values are as
follows:

Vp = Vpp/2 = sqrt(2)*Vrms = pi*Vav/2

Vpp = 2*Vp = 2*sqrt(2)*Vrms = pi*Vav

Vrms = Vp/sqrt(2) = Vpp/2*sqrt(2) = pi*Vav/2*sqrt(2)

Vav = 2*Vp/pi = Vpp/pi = 2*sqrt(2)*Vrms/pi

where Vp = Peak voltage
Vpp = Peak-to-peak voltage
Vrms = rms voltage (also called "effective voltage")
Vav = average voltage
pi = the constant pi (approx. 3.14159...)
sqrt(2) = the square root of 2 (approx. 1.41421...)

These constants were determined using calculus. For more information
on where these constants came from, see the following:

Why 0.707? Teaching R.M.S. Values of AC Voltage and Current
- L. B. Cebik
http://www.cebik.com/edu4.html

----------------------------------------------------------------------------------------

So a relevant example would be you have a 250-0-250 volt PT. The secondary voltage is 250VAC RMS from each secondary phase relative to center tap. (unloaded and unrectified)

If you look at this voltage on a scope, you will see a 250*2*sqrt(2) VAC peak-peak waveform on the display.

If you rectify this voltage using a full wave rectifier, you will see 250*sqrt(2) VDC at the first capacitor staqe (unloaded) ~ 354VDC. This is because capacitors charge up to peak voltage and VDC(peak)=sqrt(2)*VAC(RMS) in this example. Remember this is unloaded and assuming no cap. lkg current.

Things get more complex when you load the PT secondary due to DC resistance of PT secondary winding, finite charge storage capacity of real capacitors and imperfect coupling of magnetic flux from PT primary to secondary.

rob

[martin manning]

Things get more complex when you load the PT secondary due to DC resistance of PT secondary winding, finite charge storage capacity of real capacitors and imperfect coupling of magnetic flux from PT primary to secondary.

...and the size of the reservoir capacitor, and the relationship between the secondary and load resistances. The Hammond application sheet has some typical assumptions built in for these parameters, and so gives typical voltage and current fractions as opposed to the theoretical RMS/p-p relationship for sine waves.

[SteveG]

It looks like you'd have to be careful about the exact type of switch you use for this application.

A Make-Before-Break will put a momentary short across the transformer secondary (Give or take a diode).

Steve

[Andy Le Blanc]

Just make sure the filter caps have the appropriate voltage rating.

If its 650 ct, the ct connection will make around 462 peak

but 650 with a full wave bridge will make around 917 peak.

The current is different too.

fullwave (with ct) gives the full ma. rating, with a bridge its x .62

Filter caps snap to ground before they pop, I wouldn't do it unless you really plan for it.

The full wave bridge makes about twice the average volts and a little more than half the current.