Self Contained Bridge rectifiers

[NorCalTuna]

wow, they sure are convenient over using a bit of board and turrets, eyelets, what have you (oh yah, one hole and one bolt too). A bit pricey if you buy way too much rectifier, but I'll get to my point:

Is there any advantage at all with the traditional full wave, 6 diode string? There's obvious differences between tube and silicon recs, but what about the two different styles of silicon rectifiers? Is there anything wrong with simply not connecting a transformer's center tap, or can you connect it with zener diodes and a bit of resistance (as to avoid the short) to drop voltage like the old setup? Lastly, Are they rated in PIV or a safe working voltage?

I would really like to use a bridge in a build which traditionally uses a string, but have always wondered, as there are amps which use bridges as well...

thanks magic info window!

[Steve Conner]

Hi,

The bridge rectifier and full-wave center tap rectifier (what you call the 6 diode string) are two fundamentally different things.

If you wire a bridge rectifier to a transformer intended for a FWCT, and don't use the center tap, you'll get a nasty surprise: twice the output voltage that the FWCT would give. Not many bridge rectifiers are rated for 1kV+.

If you do use the center tap, then it degenerates to exactly the same circuit as the FWCT. Two diodes in the bridge act as the FWCT, and the other two do nothing.

The voltage rating of a bridge rectifier is the PIV of each individual diode. That is the same as the B+ output from the bridge circuit, allowing a safety margin of course. The FWCT is different: it needs diodes with a PIV of at least twice the B+. That's why they're often made with series strings of diodes.

[Gtr_tech]

Using a bridge on low voltage winding with a CT (on amps that have a seperate winding for op amps) is OK since they're deriving a pos and neg rail from it without a ground connection at the bridge. Grounding a bridge on a winding with a CT is bad mojo.

[jrfrond]

FWIW- The "6-diode string" full-wave rectfier commonly used in vintage Fender amps is a dinousaur leftover from the days when diodes weren't quite as hefty as they are now. I replace each 3-in-series string with a single 6A10, which is also what I use in new builds. Haven't had one short yet. I use these for the bias diode two. Sure, it's not original, but neither is anything else you can get nowadays.

As for bridge rectifier packages, they are convenient when you want to add-on or modify a supply. Yes, their output voltage is higher if you use a center-tapped B+ secondary vs. a full-wave, but if you are looking, by some chance, to BOOST B+, you would also need to make sure that the VA rating of the PT can handle it, and most stock transformers barely pull their load as it is.

[lowell]

So the PIV rating of a bridge rectifier package is the PIV of 1 diode within said package. What about the current rating? I purchased a few recently at 35A for my LM3886 amp I'm building. I was told that it'll work well within it's spec. What is that 35A referring to extactly. Finally this leads me to more questions regarding diodes.

Does PIV double when diodes are in series?
Parellel?

Does peak current double when in series?
Parallel?

Once a diode conducts (.7v on typical silicon types) will it draw maximum current? That is, if you were to put a diode across a 1 volt source, anode to positive, cathode to ground, would it burn up? I'm thinking yes but just wanna confim. I think this is what happens in most guitar pedals when someone uses the wrong polarity supply.

[lowell]

Ok I just found this after further research as I couldn't find an answer to this at first. Diodes in series multiplies the maximum spec'd PIV and diodes in parallel multiplies maximum spec'd current.

Still what about bridge rectifier packages. Is the current rating the maximum forward current? EG, in your typical tube amp of 400v HT and 2 6L6 we'd have apprx 50watts/400v=125ma. Add 100ma or so for preamp tube plates and bleeder resistors. Then double to get around 500ma and then use a 1a to be safe?

And what about diode current after reaching .7v? Is a diode basically a wire once forward conductance is reached?

[R.G.]

So the PIV rating of a bridge rectifier package is the PIV of 1 diode within said package. What about the current rating? I purchased a few recently at 35A for my LM3886 amp I'm building. I was told that it'll work well within it's spec. What is that 35A referring to extactly.

The 35A is the average DC output current for the bridge. That is, the bridge will be right at the edges of its capability if it's feeding 35 amperes of DC through it. It's well within its ratings, all right.

Does PIV double when diodes are in series?

If the diodes are truly identical yes; in particular, if their leakage currents are identical under reverse bias. If you want to ensure that you get the full PIV with diodes in series, you must force the reverse voltage to share equally by paralleling each diode with an equal (and high!) value resistor to equalize them for the DC/low frequency case, and equal value capacitors to force sharing of the high speed/transient case.

Parellel?

In parallel, the lowest resistance/lowest voltage diode hogs all the forward current, and the lowest breakdown voltage diode dies first in reverse bias.

Does peak current double when in series?

No. The same current goes through both diodes when in series.

Parallel?

Unless they are truly identical, no. Diodes vary slightly in their forward conduction voltages and the resistances which cause their forward voltage to creep up under high currents. The lowest-voltage/lowest resistance diode will be conducting all the current until current forces its forward voltage high enough to start the other one conducting. They share very unequally unless you put in small series resistors to force the high current one to have a higher voltage faster. Same reasoning as small emitter resistors in parallel output transistors, to enforce sharing.

Once a diode conducts (.7v on typical silicon types) will it draw maximum current?

Diodes don't conduct at 0.7V. They have a voltage across them that is the logarithm of the current through them, or a current which is an exponential function of the voltage across them. That is, it takes about 0.45V for an average silicon diode to get up to a few microamperes of current through it. From there to about 0.7V, the current rises exponentially, until above 0.7V, the only real limit is the resistance of the leads and bulk semiconductor material. The resistances may make this significant. A 1N4148 signal diode may have a voltage across it of 1V at 100ma. A 1N4007 may be 0.6V at 100ma and 0.7 at 1A.

That is, if you were to put a diode across a 1 volt source, anode to positive, cathode to ground, would it burn up?

Let's put that another way. If you try to force a diode's forward voltage to more than about 0.7V or so, effectively the diode no longer limits current flow. The current flow will be limited by the lead resistance and bulk silicon resistance and by the internal impedance of the source. If you have a "perfect" source that can supply any amount of current, then only the lead resistances/etc. limit the current. The diode dissipates the forward voltage times the actual current, plus the current squared times the resistances. Whether it burns up or not depends on how high the actual current gets based on the lead resistances/solder joints and other imponderables.

Which is a way of saying "don't do this unless you can figure out how the current will be limited, or have a good way to get the heat out."

[lowell]

Ok so how does one measure current through the diode? Is connecting a meter in series or adding a current sensing resistor a must?

[diagrammatiks]

current through the diode is something that you'll want to get off of the spec sheet.

I'm sure RG knows more then I do about this but most bridge packages are just multi-packages of diodes, meaning they are roughly similar in terms of how they operate at comparative specs.

So Merlin says that a bridge needs diodes that are rated for 1.4 times the secondary voltage. A full wave two phase needs 2.8.

At 360-0-360 thats 504 on a bridge and 1008 on a full-wave two phase. I know that the 6 diode string has been used before but given that most output transformers put out less then 600ma...2 1n4007s per side should be overkill at 2a 2000piv per side.

The self contained bridges/rectifiers come in packages meant for center tapped and non-center tapped transformers.

I know that my orange dual terror came with one of these guys...

KBJ1506 pdf, KBJ1506 description, KBJ1506 datasheets, KBJ1506 view ::: ALLDATASHEET :::

rated at 800vdc. It's just the equivalent of however many leaded diodes per side.

[GainFreak]

I'm still not getting this: why are you going to use a CT transformer if you have diodes?
What is the "disadvantage" of having just one non CT HT winding and a bridge rectifier?

[diagrammatiks]

i just assumed that the op was working around whatever transformer he had on hand.

diodes are cheap compared to transformers.

[R.G.]

Ok so how does one measure current through the diode? Is connecting a meter in series or adding a current sensing resistor a must?

Unless you really are running DC through it, you estimate it based on the average DC current out of the filters.

What rectifiers really do is let through short pulses of lots of current that charge the filter caps only at the peaks of the AC waveforms feeding them (for sine waves like power transformers give them). A 60Hz power line causes these peaks every 8.3mS for a full wave rectifier. Conduction is less than 1mS, and usually a lot less than 1mS. So the peaks have to be 8, 10, 50 times as big as the average DC current to charge the caps in such a short time. The bigger the cap, the less it runs down between peaks, and the shorter the conduction. That makes the peak currents bigger when they do happen.

Clever guys have calculated estimates of the RMS (heating) value of these intermittent peaks to be 1.6 to 1.8 times the average DC current out for a full wave bridge. So if your full wave bridge supplies filters that give 100ma to a load, the RMS current in the diodes and transformer winding is about 160 to 180ma. If it produces 1A average, the heating current in the diodes and transformer is 1.6 to 1.8A.

Measuring the real current takes either a current monitoring sense resistor or a true RMS current meter.

[lowell]

Thanks RG. Okay so one might use a 1ohm low tolerance suitably rated wattage resistor and find the current by the voltage across said resistor?

[R.G.]

Thanks RG. Okay so one might use a 1ohm low tolerance suitably rated wattage resistor and find the current by the voltage across said resistor?

One might. However, unless what you're measuring is the DC average current, you're going to have to take an oscilloscope picture, then do the math to figure out the RMS value of that pulsating current. If it's the DC average current, measure it by putting an ordinary current meter in series after the filter cap.