Current Flow Through This Rectifier

[Gary_S]

Can anyone explain the proper direction the current takes in flowing through the rectifier in this Marshall 4010 i own? I've studied a lot of electronic material and understand what's goin' on as regards the rectification of the sine wave from AC to pulsating DC. But the way it's drawn on the Marshall schematic i'm finding it hard to visualise how the current is flowing through the rectifier and the transformer. Usually current will flow through one diode and then to the load and then return to the transformer through the other diode of the pair but i'm finding this hard to visualize on this schematic.

[Willabe]

It goes out through D2,D3 for the 1st of half the sign wave and then D4,D5 for the second half of the sign wave. Both half wave signals are brought together, so you now have a full wave signal, at the junction of D2,D4 and C23 filter cap's positive end. Return for both is through the negative end of C23 back to the black center tap wire as both are connected to ground. That's the signal's loop.


                     Brad      

[jazbo8]

Like Brad said, it's just a full-wave rectifier configured with two diodes in series on each phase, that's done to increase the voltage rating, in this example, with two 1N4007 in series, the peak reverse (or breakdown) voltage is doubled to 2000V, the current through them is exactly the same since the two diodes are in series (if you want to double the current capacity, you need to put them in parallel). You can do that with capacitors as well by put them in series to handle the higher voltage, e.g., instead of trying to find a say a 40uF, 600V electrolytic capacitor, you may instead string two 80uF, 350V capacitors together in its place. Hope this answer your question.

[Gary_S]

Thanks guys appreciate the info and help. All i can say to you is  

Man i wish i had the knowledge some of you guys on here have got! But it all takes time.

I keep reminding myself that there's folks that have spent their whole lives dedicated to studying this stuff and it's just not realistic to expect to snap your fingers and it all clicks into place!!

It can be quite confusing because i've been studying the NEETS modules and in the power supplies module where they discuss everything about rectification in one example they give the transformer voltage makes the plate of one diode positive (tube diode) and the current flows through from the cathode to the plate then round through the PT and through whatever load it's going to. However here in this case on my amp it looks like the current is going the opposite way from anode to cathode through the SS diodes. This always gets me! I suppose it's the old thing about conventional current as opposed to actual direction of current flow?

[PRR]

> the current flows through from the cathode to the plate

The *electrons* (in a tube rectifier) flow cathode to plate.

*Current* always flowed (by arbitrary convention) plate to cathode.

FWIW, most tube-circuit designers draw current plate-to-cathode; even though in the hollow-state case only electrons flow and they go K-to-P.

In solid-state rectifiers, as here, both positive and negative charges move, inside a crystal, where you can't see what's happening. I always draw "current" flowing in the direction of the arrow, the heck with whatever is really going on in there.

Yes, the free NEETs material is older and more about getting radios fixed than worrying about precise physics details.

[Gary_S]

Thanks PRR, always find the electrons moving through the tube cathode to plate and yet conventional current going the other way quite confusing! 

[PRR]

+ and - charges are equal-but-opposite.

In math, we use + and - to work with such problems.

Look in your checkbook. There's $1,000 of "cash current" flowing-through every month. Your pay-check you denote with a "+". Your bill payments you denote with a "-". Except you work for the Cumberland Mine. Make good money -- five dollars a day!! Then the mine charges you $3 for room and $2 for food. The books balance. In electronics, the books always balance. So it really does not matter which is "-" and which is "+", as long as you are consistent.

[tubeswell]

A tube rectifier works differently to a solid state rectifier (as pointed out by PRR).

An SS diode rectifier works by only allowing current to pass in one direction, so when the VAC voltage swing goes -ve, the current is blocked through that diode. The diodes on each side of the PT's High Tension (HT) winding allow high voltage +ve current pulses to pass, which correspond to the high voltage AC wave which is in opposite phase to each other end of the HT winding.

For a tube rectifier, the ends of the HT winding are each connected to one of the rectifier's plates.

The tube rectifier's cathode is heated so that it frees electrons into a 'space charge' 'floating' just above the surface of the cathode (like a conventional tube cathode). The tube rectifier's cathode isn't grounded, but is connected to the B+ reservoir cap.

On the +ve 1/2 of each end of the HT winding's VAC cycle, the corresponding tube rectifier's plate that it is connected to, swings positive. This pulls electrons away from the cathode's 'space charge', and this causes the cathode to rise to a high +ve voltage, which in-turn sucks electrons out of the reservoir filter cap's "B+" side, causing the filter cap to become positively charged. The 'internal resistance' resulting from the interaction between between the cathode and the plate (and the space between them, and their relative sizes, and the amount of cathode heater energy, and the amount of VAC swing at the plate, and the amount of reservoir capacitance) is referred to as the plate resistance, and this causes a voltage drop between the maximum +ve voltage swing at the plate and the corresponding maximum cathode voltage. Different tube rectifiers exhibit different amounts of this 'forward' voltage drop.

To return to the current cycle, when the rectifier plate swings to the -ve 1/2 of the cycle, nothing happens at the cathode as a result of that -ve plate voltage swing, BUT at this same time, the other rectifier plate is reaching its +ve 1/2 of the VAC cycle, causing electrons to flow from the cathode to that other plate. So the cathode is experiencing continuous back-to-back +ve pulses of charge, which is sucking electrons out of the reservoir filter cap, encouraging that filter cap to remain at a high +ve voltage (albeit at a voltage that's lower than the peak VAC swing on the plates, due to the forward voltage drop exhibited by the rectifier tube).

The final part of this is that the reservoir filter cap takes longer to 'discharge' (than it takes to 'charge' to a +ve voltage). The current that occurs in the corresponding 'charging' segments of the filter cap's charging cycle are referred to as 'charging' current. This charging current is not constant, but is in little discrete (but more-or-less instantaneous) pulses that correspond to the timing of the rectifier's cathode +ve pulses. Of course, once the filter cap charge equals the cathode's charge, the filter cap's charging current stops, and resumes again when the capacitor's charge falls to a sufficiently lower voltage to encourage another 'burst' of charging current. The bigger the reservoir capacitance of this filter cap, the more stress this charging current places on the rectifier. That is why tube rectifiers are rated for a maximum recommended reservoir capacitance.

The 'discrete' bursts of charging current going into the reservoir cap don't in themselves produce a steady B+ voltage at the reservoir cap, but produce what is known as 'ripple current' at that cap. This requires further filtering (e.g.; with a resistor and another filter cap) to get a smooth B+ voltage.

[Gary_S]

An SS diode rectifier works by only allowing current to pass in one direction, so when the VAC voltage swing goes -ve, the current is blocked through that diode. The diodes on each side of the PT's High Tension (HT) winding allow high voltage +ve current pulses to pass, which correspond to the high voltage AC wave which is in opposite phase to each other end of the HT winding.

Thanks tubeswell. 

See this is where the confusion comes in; I watched Gerald Weber's DVD "Understanding Tube Amplifiers" and in the section where he examines a SS bridge rectifier he talks about the current flowing the other way! from the cathode side of the SS diode through towards the anode side then through the PT then through the other diode of the pair and out via the load. I thought it was the other way round unless you're using a diode reverse biased to use as a negative voltage for your power tube grids for biasing. 

[tubeswell]

With an SS diode rectifier on an HT winding, the anode side is connected to the HT winding and the cathode side (banded end) points toward the filter cap. Current within the diode flows from the anode to the cathode, and when the VAC (from the HT winding) swings positive the 'forward' current flows through the diode. But when the VAC swings negative and 'backward' current flows, it is blocked by the diode, so the -ve 1/2 of the VAC swing doesn't make it past the diode.

In a bias circuit, the diode is put in the other way around so that only the negative 1/2 of the VAC swing (which produces the 'backward' current) goes through the diode, and the 'foward' current is blocked.