Could somebody please explain full wave rectifiers using ohms law?

peteb

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How does a full wave rectifier cause electrons to go to ground? When ohms law informs us that electrons are attracted to high voltages?


I believe these are the pieces and their functions in a full wave rectifier.

the secondary of the PT has high voltage AC on it.

on the AA864 bassman it is 305 + 305 = 610 VAC.

each end of the secondary is connected to a diode bridge.

the diodes are oriented so that electrons can only travel into the secondary, never back out, reversing course across the diode bridge.

The AC on the secondary brings electrons in from one bridge in one direction, and then brings in electrons from the other bridge, in the other direction.

some how, the electrons in the high voltage secondary, appear to go to ground when they reach the center tap, but why?


610 VAC divided by root 2 gives an ideal 431 VDC available using ss diodes, avoiding rectifier tube voltage drop.

the plate is at 425 DC, and so therefore it is anticipated that electrons at 425 DC are going to be willing to travel towards and to the 431 VDC produced by the solid state rectifier.

that explains how the electron arrives at the SS state rectifier, using ohms law. But what happens after that?

the electron is sitting at 431 DC, the idea is that the AC on the secondary is going to encourage the electron to cross the diode bridge, travel through the secondary winding, and decide that ground is a good place to be.


could somebody please explain, using ohms law, why the electron at high voltage decides to go to ground?


the AC is making use of the diodes and it’s own electrical energy to pump electrons into the ground. That much is clear. it is like the ground is a sort of trap. The electron has no reason to want to go there, but once placed there, it is unable to escape. D683962A-2AB5-4D94-A0C9-169CD26ED75A.gif
 

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popcat

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The idea is that electrons flow from the more negative node to the more positive node, such as with a battery. Therefore, on the positive half of the cycle electrons flow from the more negative ground as steered by the diodes into the more positive xfmr winding. On the negative half cycle, they flow from the negative xfmr winding into ground (again, as steered by the diode configuration).

It is often easier when interpreting schematics to use conventional flow instead of actual electron flow, and think of current flowing from a positive node (power supply) into an IC or tube instead of from ground into the circuit (with a positive supply).

Because the diode bridge "steers" the electron flow, the result is a positive and negative pulsating DC voltage (at half xfmr voltage but 2x the frequency) at the +/- bridge nodes that are then filtered by a "smoothing" capacitor into +/- DC voltages. If the negative node of the bridge is tied to the xfmr (typically ground) not using a center tap, then the DC voltage will be only positive at full xfmr voltage (or vice-versa with the positive node grounded).

: Sorry, just noticed your question and the schematic are about a full-wave rectifier and not a bridge rectifier. Nevertheless, the concept is similar where the diodes only conduct on either positive half of the AC waveform from ground into the load, and then back to the xfmr.
 
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Jon Snell

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Ohm's Law is a formula used to calculate the relationship between voltage, current and resistance in an electrical circuit.
Nothing to do with the rectification of AC Voltage.
Here is a description for you;
  1. In the first positive half cycle of the AC signal, the diodes D2 and D3 become forward biased and start conducting. At the same time, the diodes D1 and D4 will be reverse biased and will not conduct. The current will flow through the load resistor via the two forward-biased diodes. The voltage seen at the output will be positive at terminal d and negative at terminal c.
  2. Now, during the negative half cycle of the AC signal, the diodes D1 and D4 will be forward biased and diodes D2 and D3 will become reverse biased. The positive voltage will appear on the anode of D4, and negative voltage will be applied to the cathode of D1. It is worth noting at this point that the current that will be flowing through the load resistor will have the same direction as it has with the positive half cycle. Therefore, no matter the polarity of the input signal, the output polarity will always be the same. We can also say that the negative half cycle of the AC signal has been inverted and is appearing as a positive voltage at the output.
Now the Half Wave rectifier;
When we use a half-wave rectifier, a significant amount of power gets wasted as only one half of each cycle passes through, and the other cycle gets blocked. Moreover, the half-wave rectifier is not efficient (40.6%), and we can not use it for applications that need a smooth and steady DC output. For a more efficient and steady DC output, a full wave rectifier is used.


We can further classify full wave rectifiers into:

  • Centre-tapped Full Wave Rectifier

Hope this helps to explain it for you.
 

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loopfinding

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the resistance of a diode is a function of the input voltage and current. it's also not linear. i suppose it is possible to apply ohm's law for an instantaneous point of resistance, but a diode isn't in a steady state with an AC input.

 
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andrewRneumann

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@peteb I think you have a stumbled upon the difference between a mere “potential difference” (PD) and “electromotive force” (EMF). The transformer produces an EMF in the secondary winding, which consumes energy to push electrons in the opposite direction they would flow if one just looks at PD (voltage). Do a little Google searching on the difference between and EMF and PD and I think you will find better explanations from folks who understand this distinction better than I.

BTW, you could ask the same question about a simple battery + light bulb circuit. Conventional current flows through the battery from negative to positive. Electrons flow through the battery from positive to negative. How can this be? The battery expends energy to provide an EMF. If you can begin to understand this, you are starting to understand a core foundation of electrical theory.
 

philosofriend

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The center tap of the transformer is grounded and so is the cathode of the power tube. What I am about to say would work just as well and be just as true if the center tap and cathode were just hooked by a wire and not grounded, there is nothing magic about the ground. The electrons still have to go in a complete circle.

The center tap of the transformer pushes electrons to the cathode. On the preamp tubes they go through a bias resistor first. The electrons jump off the hot cathode and fly to the plate. From there they flow through the load resistor (preamp) or the output speaker transformer. The electrons then go back through the diodes to the power transformer. The power transformer adds energy to the electrons and the energy is used up bit by bit as the electrons go through the resistance of the bias resistor, the tube, and the load (which is a resistor or a speaker transformer.)

The power transformer excites the electrons up to a voltage. At each step of the way through the bias resistor, the tube and the load, there is a voltage drop that will follow Ohm's law. When the electrons get back through the diodes they are all tired out but the transformer excites them again, pumping them back up to a negative voltage.

The tube has a resistance that you can learn about from the RCA tube manual or any serious book about tube basics. The speaker has an effective AC resistance (called impedance) that depends on the design of the transformer and the load placed its output, which is the speaker.

Another good book about this stuff is the Dover reprint of a Navy book called "Basic Electronics". It is cheap used on Amazon. Darr's book about guitar amps is good too.
 

Ten Over

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The power transformer excites the electrons up to a voltage. At each step of the way through the bias resistor, the tube and the load, there is a voltage drop that will follow Ohm's law. When the electrons get back through the diodes they are all tired out but the transformer excites them again, pumping them back up to a negative voltage.
You're just messing with the OP, right?
 

dogmeat

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since the OP was thinking in terms of Ohms law, I'll add a side note. testing diodes and transistors with an analog meter will yield a different "resistance" reading than what you get with a digital meter. thats because the biasing voltage is pretty much the same on any given solid state device, but the meters use different voltages to test resistance. the reason this matters is that charts showing test values in old repair manuals were made with analog meters. any circuit with a solid state device will have a different reading with a digital meter. I've run into this on car & motorcycle repair manuals for instance. anything older than about 1990 has a good chance of being made with an analog. I can think of one Honda manual from 1986 that had 2 charts... depending on which analog meter you used (no digital). and for instance, I've seen guys ditch good ignition controls because they didn't match the test chart (then have to find the real problem)
 

Blue Bill

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How does a full wave rectifier cause electrons to go to ground? When ohms law informs us that electrons are attracted to high voltages?


I believe these are the pieces and their functions in a full wave rectifier.

the secondary of the PT has high voltage AC on it.

on the AA864 bassman it is 305 + 305 = 610 VAC.

each end of the secondary is connected to a diode bridge.

the diodes are oriented so that electrons can only travel into the secondary, never back out, reversing course across the diode bridge.

The AC on the secondary brings electrons in from one bridge in one direction, and then brings in electrons from the other bridge, in the other direction.

some how, the electrons in the high voltage secondary, appear to go to ground when they reach the center tap, but why?


610 VAC divided by root 2 gives an ideal 431 VDC available using ss diodes, avoiding rectifier tube voltage drop.

the plate is at 425 DC, and so therefore it is anticipated that electrons at 425 DC are going to be willing to travel towards and to the 431 VDC produced by the solid state rectifier.

that explains how the electron arrives at the SS state rectifier, using ohms law. But what happens after that?

the electron is sitting at 431 DC, the idea is that the AC on the secondary is going to encourage the electron to cross the diode bridge, travel through the secondary winding, and decide that ground is a good place to be.


could somebody please explain, using ohms law, why the electron at high voltage decides to go to ground?


the AC is making use of the diodes and it’s own electrical energy to pump electrons into the ground. That much is clear. it is like the ground is a sort of trap. The electron has no reason to want to go there, but once placed there, it is unable to escape. View attachment 976715
Woobagoobajooba. Sorry, Pete, your question is all over the place. As @2L man alluded to, maybe you are a bit distracted by thinking in terms of where electrons "decide" to go, or are "attracted to", or how the "AC is making use of the diodes and it’s own electrical energy to pump electrons into the ground." Sounds kinda anthropomorphic.

I can't even figure out what your question is asking. Ohm's law is a simple 3-variable math formula, it doesn't address decision making, nor motivation, nor attempts to "escape". To understand the operation of half-wave or full wave rectifiers, there's several good videos on YT, let me know if you can't find a couple.
 

rdjones

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"Electron flow" and "conventional flow" are being confused here.
Pick one, and stick with it for the example.

Also, a pair of diodes in a full wave rectification arrangement is not a "bridge".

Most of this stuff is basics, Electronics 101 level fundamentals.
I strongly suggest the OP go back to the beginner's level DC theory.
Then move into introductory AC theory (capacitors, inductors, transformers)

It's really difficult to engage in productive discussion without an elementary grasp of terms and concepts.
Seriously, trying to learn beginner's basics from random strangers online is fraught with innumerable challenges.
 

David C

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The primary function of rectification is to convert AC power into quasi DC power. Forget the electron flow for a moment, the important thing here is the voltage. In 120 VAC power you have alternation between positive voltage and negative voltage of around 168 for a total voltage swing of 336 volts. Rectification simply converts AC voltage into DC voltage. So you would end up with voltage of 0 to positive 168 Volts occurring 240 times a second. Because you use a transformer you get higher voltages than line power.
 

dsutton24

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peteb, why don't you enroll in a basic electronics course at a community college? A little bit of discipline and structure will teach you a lot about the things you post about. If you can get a good grounding in the fundamentals you'd be amazed at how much of this stuff becomes self apparent. A few months of disciplined study will teach you a lot more than years of TDPRI ever could.
 

peteb

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Thanks for the replies.

I believe everything I wrote in the OP is true.

I suppose I purposefully avoided the term electromagnetic force, which is how the PT forces electrons to ground.

I like to think of electricity in terms of potential difference and ohms law, first, but realize that there is more.

it is the maxwell equations that describe the entire electo magnetic principles.


one clarification I would like to make to my OP. Thinking about this. My main question is how do you convince an electron to go to ground? Isn’t ground negative, and negatives repel each other. Not quite. The ground is not really negative. Ground is neutral and made up equally of positive and negative charges. Although ground is not the electrons first choice, the electron does not really mind going to ground.


now for the question, why we need to understand rectification.


I believe the full wave SS rectifier, as used on the bassman, would actually function if the secondary winding was grounded at the end of the winding instead of in the ceneter.

it is true that one diode bridge would no longer carry current, and could be removed. The remaining diode bridge would then become a half wave rectifier. John Snell points out that a half wave rectifier is less efficient, probably half as efficient as the full wave rectifier because only half of the cycle is rectified.


I believe that if the center tap became an ‘end’ tap, the remaining leg would only rectify half the cycle, but the AC voltage the remains leg would see would be double what it previously saw using the center tap. The diodes could be beefed up however.

1/2 times 2 equals one.


oh yeah, one other thing.

looking at the schematic of the full wave rectifier, it looks like the current in the secondary has an option. It could pass by the center tap, or it could follow the center tap to ground. Actually the only available path is the path to ground. There is no option of passing by the center tap, as the orientation of the diodes dictate that the only electron flow in either leg is towards ground. One option means no choice.


and additionally, there is no reason why rectification cannot be described in terms of ohms law. The maxwell equations fully define all facets of electricity including electro magnetic forces and potential, but even they are based on ohms law.

here is how full wave rectification can be explained using ohms law.

when the plate of the rectifier goes fully positive, the high positive charge encourages the electron to cross the diode bridge to reach the plate. The electron sits on the plate for 1/60 of a second. Now the electron finds itself at the most negative voltage, the other end of the secondary is at the highest voltage, but the electron can not see that far. The electron is sitting at -305 volts and the grounded center tap is at zero volts. That is an easy choice for the electron. Go to ground and get stuck at ground.
 
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