- Apr 20, 2008
- 10,065
- 984
- 126
As in title, how does a step-up transformer produce more voltage out of thin air? It confuses the crap out of me. Everything except the transistor In analog electronics makes sense...
How does a step-up transformer work?
- Thread starter TurtleCrusher
- Start date
Modelworks
Lifer
- Feb 22, 2007
- 16,240
- 7
- 76
transformers are really simple to understand if you think about watts.
Watts = Current x volts.
Primary of a transformer consumes 120volts x 1A = 120 watts
Secondary can then output 120 watts in ANY mix of volts and amps as long as it totals 120watts.
So if I want 500 volts on the output , 120 watts / 500 volts = .24 amps
Primary 120V@1A = Secondary 500V@.24A
Often you will see transformers listed as a ratio , like 3:1 meaning for every 3 volts supplied your output will be 1volt . The power isn't lost , it is just never taken from the primary winding in the first place.
For example a common trasnformer is a 120VAC input and 12.6VAC output, those are listed at 10:1 . For every 1 volt on the output 10 volts are needed on the input.
These are common on UPS supplies because if you turn that transformer around you now have a 1:10 , so every 1 volt on the input = 10 volts on the output. Supply it with 12VAC and you get 120VAC output. The increase in voltage comes from using up the extra current on the input.
Watts = Current x volts.
Primary of a transformer consumes 120volts x 1A = 120 watts
Secondary can then output 120 watts in ANY mix of volts and amps as long as it totals 120watts.
So if I want 500 volts on the output , 120 watts / 500 volts = .24 amps
Primary 120V@1A = Secondary 500V@.24A
Often you will see transformers listed as a ratio , like 3:1 meaning for every 3 volts supplied your output will be 1volt . The power isn't lost , it is just never taken from the primary winding in the first place.
For example a common trasnformer is a 120VAC input and 12.6VAC output, those are listed at 10:1 . For every 1 volt on the output 10 volts are needed on the input.
These are common on UPS supplies because if you turn that transformer around you now have a 1:10 , so every 1 volt on the input = 10 volts on the output. Supply it with 12VAC and you get 120VAC output. The increase in voltage comes from using up the extra current on the input.
Last edited:
- Apr 20, 2008
- 10,065
- 984
- 126
That is the explanation I needed. Thanks. My military training didn't explain it like that at all. Only that voltage went up, and that I was either full wave or half wave. They made no mention that voltage was increased at the expense of current, which explains why it's a huge inductor.
Now if I could only understand how a transistor actually works. Is there any simple way to get it that the military isn't teaching me? Essentially (common emitter) base voltage opens it up and .6v is dropped from base to emitter, and the current flows through the collector... And I'm lost...
I've gone through ac, dc, analog and digital circuits. I'm on things like radar and communcation systems now and I still don't have a complete understanding of how these components *really* work.
Now if I could only understand how a transistor actually works. Is there any simple way to get it that the military isn't teaching me? Essentially (common emitter) base voltage opens it up and .6v is dropped from base to emitter, and the current flows through the collector... And I'm lost...
I've gone through ac, dc, analog and digital circuits. I'm on things like radar and communcation systems now and I still don't have a complete understanding of how these components *really* work.
PowerEngineer
Diamond Member
- Oct 22, 2001
- 3,583
- 756
- 136
The easiest way to understand the current change between windings on a transformer is to think about the difference in the number of turns in each winding. The magnetic field induced in the transformer's core is tied to the current in the first winding times its number of turns, and this has to be matched by the current in the second winding times its number of turns (i.e. I1 x N1 = I2 x N2 or I2 = [N1/N2] x I1 ). And the (changing) magnetic field that all the turns in both winding experience (at least ideally) means that V1/N1 = V2/N2 or V2 = [N2/N1] x V1.
So, if V1 = 120 volts, I1 = 2 amps and N1/N2 = 10, then V2 = [1/10] x 120 = 12 volts and I2 = [10] x 2 = 20 amps.
As ModelWorks said, the product of volts x amps for both windings should be roughly the same as transformers generally have low losses and therefore what real power goes in (on one winding) must come out (on the other winding).
So, if V1 = 120 volts, I1 = 2 amps and N1/N2 = 10, then V2 = [1/10] x 120 = 12 volts and I2 = [10] x 2 = 20 amps.
As ModelWorks said, the product of volts x amps for both windings should be roughly the same as transformers generally have low losses and therefore what real power goes in (on one winding) must come out (on the other winding).
First of all, it's worth thinking about how a transformer works.
A transformer consists of 2 coils of wire and (almost always) some sort of 'core' which links them.
When an electric current passes through a wire, it generates a magnetic field, the strength and direction depend on the current. When the wire is rolled into a spring-shaped (helical) coil, the shape of the magnetic field becomes very convenient, like a bar magnet, with the strongest field at the center of the coil.
If you apply an alternating current to the coil, the magnetic field will also alternate (constantly changing in strength and direction). Now, here's the trick. A changing magnetic field will create a voltage in a nearby wire.
So, if you have a coil of wire, connected to alternating voltage, and producing an alternating magnetic field, the magnetic field will create an equal and opposite alternating voltage in the wire. The result is that your input voltage gets cancelled out by the voltage 'reflected' back from the magnetic field. (See note 1).
So, if you apply 100 V AC to a coil with 100 turns, the 'equal and opposite' reflection will reflect 1 V into each turn of wire. The result is that this 'reflected' voltage cancels out the supply voltage, only a very small current will flow in the coil (just enough to create the magnetic field).
Now, imagine what happens if you have 2 coils of wire very close to each other - let's assume that they have 100 turns each, and 1 is connected to 100 V AC, and the other is left unconnected. Because both are exposed to the same fluctuating magnetic field, both experience the reflected voltage. So, you now have 100 V appearing "out of nowhere" on the second coil.
What happens if you put 1000 turns on the 2nd coil? Well, you have a very long wire, which picks up 1000 reflections. The input strength is 1 V per turn - which gives you 1000 volts on the second coil.
Now, what happens if you draw some current from the 2nd coil? Let's say, you draw 0.1 A @ 1000 V. The current flow will generate a magnetic field in the coil. This magnetic field is in the opposite direction to the mag. field produced by the first coil, and acts to cancel it out. Because the magnetic field has been weakened, the first coil doesn't experience the complete 'equal-and-opposite' reflection, and therefore, there is an uncancelled voltage that will cause current to flow through the primary coil. So, while the magnetic field throttles the current in the primary, once you start drawing energy out via another coil, the throttling effect declines and the primary current increases.
Note 1: While theoretically a perfrect transformer would produce an 'equal and opposite' reflection, and draw no current. This is unachievable in practice, the best that can be achieved is a 'magnetizing current' is drawn.
How big the magnetizing current is depends on a number of factors, most notably, input voltage, and frequency. Higher frequencies will reduce the magnetizing current, and higher voltages will increase it (the reflected voltage is proportional to the rate of change of current- higher frequencies increase the rate of change of current, even if the actual rms current is the same)
If you just use 'air' as the core, then the magnetizing current needed is enormous, and such a transformer is completely useless at 50 - 60 Hz, because the magnetizing current is far larger than your coils will be able to take without melting down. However, at 1 MHz or 10 MHz, they can be quite useful, and are occasionally used in things like radio transmitters.
Iron has an interesting property called magnetic permeability. This, in effect, concentrates and magnifies the magnetic field within the material. By placing coils around an iron core, the greatly magnified magnetic field dramatically reduces the amount of magnetizing current, making the transformer work nicely at 50-60Hz.
Where transformer weight and bulk are important (e.g. aviation and occasionally marine), higher frequencies (e.g. 400 Hz) are used, as less iron is needed (or alternative expensive, but lighter ceramic materials) can be used instead.
A transformer consists of 2 coils of wire and (almost always) some sort of 'core' which links them.
When an electric current passes through a wire, it generates a magnetic field, the strength and direction depend on the current. When the wire is rolled into a spring-shaped (helical) coil, the shape of the magnetic field becomes very convenient, like a bar magnet, with the strongest field at the center of the coil.
If you apply an alternating current to the coil, the magnetic field will also alternate (constantly changing in strength and direction). Now, here's the trick. A changing magnetic field will create a voltage in a nearby wire.
So, if you have a coil of wire, connected to alternating voltage, and producing an alternating magnetic field, the magnetic field will create an equal and opposite alternating voltage in the wire. The result is that your input voltage gets cancelled out by the voltage 'reflected' back from the magnetic field. (See note 1).
So, if you apply 100 V AC to a coil with 100 turns, the 'equal and opposite' reflection will reflect 1 V into each turn of wire. The result is that this 'reflected' voltage cancels out the supply voltage, only a very small current will flow in the coil (just enough to create the magnetic field).
Now, imagine what happens if you have 2 coils of wire very close to each other - let's assume that they have 100 turns each, and 1 is connected to 100 V AC, and the other is left unconnected. Because both are exposed to the same fluctuating magnetic field, both experience the reflected voltage. So, you now have 100 V appearing "out of nowhere" on the second coil.
What happens if you put 1000 turns on the 2nd coil? Well, you have a very long wire, which picks up 1000 reflections. The input strength is 1 V per turn - which gives you 1000 volts on the second coil.
Now, what happens if you draw some current from the 2nd coil? Let's say, you draw 0.1 A @ 1000 V. The current flow will generate a magnetic field in the coil. This magnetic field is in the opposite direction to the mag. field produced by the first coil, and acts to cancel it out. Because the magnetic field has been weakened, the first coil doesn't experience the complete 'equal-and-opposite' reflection, and therefore, there is an uncancelled voltage that will cause current to flow through the primary coil. So, while the magnetic field throttles the current in the primary, once you start drawing energy out via another coil, the throttling effect declines and the primary current increases.
Note 1: While theoretically a perfrect transformer would produce an 'equal and opposite' reflection, and draw no current. This is unachievable in practice, the best that can be achieved is a 'magnetizing current' is drawn.
How big the magnetizing current is depends on a number of factors, most notably, input voltage, and frequency. Higher frequencies will reduce the magnetizing current, and higher voltages will increase it (the reflected voltage is proportional to the rate of change of current- higher frequencies increase the rate of change of current, even if the actual rms current is the same)
If you just use 'air' as the core, then the magnetizing current needed is enormous, and such a transformer is completely useless at 50 - 60 Hz, because the magnetizing current is far larger than your coils will be able to take without melting down. However, at 1 MHz or 10 MHz, they can be quite useful, and are occasionally used in things like radio transmitters.
Iron has an interesting property called magnetic permeability. This, in effect, concentrates and magnifies the magnetic field within the material. By placing coils around an iron core, the greatly magnified magnetic field dramatically reduces the amount of magnetizing current, making the transformer work nicely at 50-60Hz.
Where transformer weight and bulk are important (e.g. aviation and occasionally marine), higher frequencies (e.g. 400 Hz) are used, as less iron is needed (or alternative expensive, but lighter ceramic materials) can be used instead.
Modelworks
Lifer
- Feb 22, 2007
- 16,240
- 7
- 76
I was in the military for electronics as well and I think part of the reason some of the training they give is lacking is because the instructors want to get you to the point of what your daily job will be and they skip the details. I didn't really learn how things work until I took a college course on electronics. The teacher started really low level with diodes and explaining how the elements used to make a diode work and why those elements like germanium were used and not some other element.
Transistors are basically diodes with a third wire. There are two different elements in a diode and that is what blocks the flow one way and allows it another way. The third wire (base) that is added to the diode is added at the junction between the two elements. When you apply voltage to that third wire you add some electricity to the gap making the flow from collector to emitter much easier than without it.
A good analogy is a water valve where the input and output are the collector and emitter and the handle is the base. If you put a really tiny handle on a huge valve then even the slightest of turns will cause a huge flow of water on the output.
To get a good understanding of electronics I think you should look at some books that focus more on the basics and why things work the way they do and not just talking about parts and what they do and that start tossing out formulas from page 1. One of the compliments I got from a professor was I was the first student in a long time that actually wanted to know why something like a diode worked like it did, not that it just worked the way it does, dropping voltage or blocking the flow of current. It sounds like you are the same way in wanting to know why the parts behave like they do and not just accepting it as a given, and that is great because it is so rare in the industry now. Most people can tell you that a diode drops voltage but they can't tell you why.
Cogman
Lifer
- Sep 19, 2000
- 10,284
- 138
- 106
For a bit more detail on how transistors work. here it goes.
First, there are two common classifications (and probably 1000x more that we never cover) BJTs and MOSFETS. Both work differently and have different characteristics.
BJTs, are as ModelWorks described, they are essentially two diodes facing each other (though... you couldn't build a bjt. The inner parts have to be touching). BJTs come in two different flavors NPN and PNP. The three letters described the ordering of the substrates used in the transistors.
P stands for P-substrate. P-substrates have an abundance of "positive" charge carriers ( http://en.wikipedia.org/wiki/Electron_hole ) these carriers are attracted to negative charges and repulsed by positive charges.
N stands for N-Substrate. N-Substrates have an abundance of electrons. Electrons are attracted to positive charges and repulsed by negative charges.
These properties are the foundation for how transistors, in general, work. With a BJT, when a small voltage is applied to the base. 1 of two things can happen. If the transistor is a NPN and the voltage is positive, all of the holes will be pushed away from the voltage source and compacted, this in turn forms a "bridge" for the electricity to flow from the collector to the emitter. If, however, the charge is negative or neutral, no bridge is formed (well, a small one still exists, this is why BJTs constantly leak more so than MOSFETS). PNPs are the same thing but opposite.
BJTs have VERY fast reaction times, which makes them pretty good for stuff like high frequency amplification. BJTs are also damn near indestructible (you can't easily shock them to death). But, because electricity is constantly leaking, they are power hogs. This is why you won't commonly (ever?) see a BJT in say a processor or some sort of digital processing environment. They are far better suited for analog signals.
MOSFETS are different and somewhat harder to explain in just text
http://en.wikipedia.org/wiki/MOSFET
Check out some of the pictures there to get a better picture of what I'm trying to describe.
Mosfets still rely on a bridge between terminals being formed. However, instead of building the bridge by pushing away charge carriers, they build the bridge by pulling them in. When a strong enough voltage is applied to a mosfet gate, a bridge is formed that links the two terminals. The mosfet gate has a VERY thin insulator between it and the substrate to keep the charge carriers from just going straight into the gate. This, combined with the fact that size of the gap, makes MOSFETs very leak free.
However, mosfets are slow (interestingly enough) compared to their BJT counter parts. Electrons or holes need to travel a good distance before electricity is allowed from one terminal to the next. This used to make them somewhat of a bad choice for stuff like high frequency analog analysis, though, now that they are as small as they are they can keep up.
The thin insulator between the gate terminal and the substrate also causes issues. Provide too high of a voltage and ZAP, your transistor is fried forever. This makes MOSFETS somewhat easy to destroy compared to BJTs. Enough static and you can complete wipe these things out.
MOSFETS, however, shrink REALLY well. BJTs as they get smaller, leak worse. MOSFETS, on the other hand, consume less power the smaller they get (hence the reason we are now making them with gate widths of 32 nm). MOSFETS have almost no leakage which makes them ideal for digital computing (you don't want those billion transistors to be constantly sucking a large amount of power).
... I don't know if these disjointed paragraphs help.. So just to simplify it, transistors work by forming bridges for the electrons to cross, By applying a voltage to the base or gate of a transistor, you create that bridge. BJTs are indestructable fast analog beasts that suck a lot of power. MOSFETS are somewhat slow, dainty things that sip power and shrink really well (as they shrink, they become faster, and more dainty).
Beyond that, the math explains each situation that a transistor can be put in. From an amplification mode, to switch modes. BJTs are commonly used in their amplification mode while MOSFETs are commonly used in their switch modes.
First, there are two common classifications (and probably 1000x more that we never cover) BJTs and MOSFETS. Both work differently and have different characteristics.
BJTs, are as ModelWorks described, they are essentially two diodes facing each other (though... you couldn't build a bjt. The inner parts have to be touching). BJTs come in two different flavors NPN and PNP. The three letters described the ordering of the substrates used in the transistors.
P stands for P-substrate. P-substrates have an abundance of "positive" charge carriers ( http://en.wikipedia.org/wiki/Electron_hole ) these carriers are attracted to negative charges and repulsed by positive charges.
N stands for N-Substrate. N-Substrates have an abundance of electrons. Electrons are attracted to positive charges and repulsed by negative charges.
These properties are the foundation for how transistors, in general, work. With a BJT, when a small voltage is applied to the base. 1 of two things can happen. If the transistor is a NPN and the voltage is positive, all of the holes will be pushed away from the voltage source and compacted, this in turn forms a "bridge" for the electricity to flow from the collector to the emitter. If, however, the charge is negative or neutral, no bridge is formed (well, a small one still exists, this is why BJTs constantly leak more so than MOSFETS). PNPs are the same thing but opposite.
BJTs have VERY fast reaction times, which makes them pretty good for stuff like high frequency amplification. BJTs are also damn near indestructible (you can't easily shock them to death). But, because electricity is constantly leaking, they are power hogs. This is why you won't commonly (ever?) see a BJT in say a processor or some sort of digital processing environment. They are far better suited for analog signals.
MOSFETS are different and somewhat harder to explain in just text
http://en.wikipedia.org/wiki/MOSFET
Check out some of the pictures there to get a better picture of what I'm trying to describe.
Mosfets still rely on a bridge between terminals being formed. However, instead of building the bridge by pushing away charge carriers, they build the bridge by pulling them in. When a strong enough voltage is applied to a mosfet gate, a bridge is formed that links the two terminals. The mosfet gate has a VERY thin insulator between it and the substrate to keep the charge carriers from just going straight into the gate. This, combined with the fact that size of the gap, makes MOSFETs very leak free.
However, mosfets are slow (interestingly enough) compared to their BJT counter parts. Electrons or holes need to travel a good distance before electricity is allowed from one terminal to the next. This used to make them somewhat of a bad choice for stuff like high frequency analog analysis, though, now that they are as small as they are they can keep up.
The thin insulator between the gate terminal and the substrate also causes issues. Provide too high of a voltage and ZAP, your transistor is fried forever. This makes MOSFETS somewhat easy to destroy compared to BJTs. Enough static and you can complete wipe these things out.
MOSFETS, however, shrink REALLY well. BJTs as they get smaller, leak worse. MOSFETS, on the other hand, consume less power the smaller they get (hence the reason we are now making them with gate widths of 32 nm). MOSFETS have almost no leakage which makes them ideal for digital computing (you don't want those billion transistors to be constantly sucking a large amount of power).
... I don't know if these disjointed paragraphs help.. So just to simplify it, transistors work by forming bridges for the electrons to cross, By applying a voltage to the base or gate of a transistor, you create that bridge. BJTs are indestructable fast analog beasts that suck a lot of power. MOSFETS are somewhat slow, dainty things that sip power and shrink really well (as they shrink, they become faster, and more dainty).
Beyond that, the math explains each situation that a transistor can be put in. From an amplification mode, to switch modes. BJTs are commonly used in their amplification mode while MOSFETs are commonly used in their switch modes.
polarmystery
Diamond Member
- Aug 21, 2005
- 3,888
- 8
- 81
Lots of good info here. Good job fellow engineers :thumbsup:
This is invariably true now, but as their gate sizes get really small (I think ~17nm) their leakage current too goes up tremendously through the gate insulator layer. It'll be interesting to see what the industry does when the gate size gets too small and has to develop something else. I think Intel with their "3-D" transistor is one method, but others will emerge eventually.
Cogman said:
This is invariably true now, but as their gate sizes get really small (I think ~17nm) their leakage current too goes up tremendously through the gate insulator layer. It'll be interesting to see what the industry does when the gate size gets too small and has to develop something else. I think Intel with their "3-D" transistor is one method, but others will emerge eventually.
Last edited:
Cogman
Lifer
- Sep 19, 2000
- 10,284
- 138
- 106
AFAIK, the problem hasn't been so much that gate currents have been increasing, but rather they haven't been going down (ok, they have increased a little). So exponentially larger amounts of transistors have eventually ended with those gate leaks being a pretty significant problem as a while.
I was pretty surprised by how fast Intel converted over to finfets.. Didn't they just switch over this latest generation? I thought it would take them MUCH longer to do.
- Apr 20, 2008
- 10,065
- 984
- 126
Thanks a lot for the info guys. Im in the Navy going to school to be an Electronics Technician as a Reservist. Some things they teach they go through it waaaaaaay too fast.
edro
Lifer
- Apr 5, 2002
- 24,326
- 68
- 91
It's the same in college.
If you really care about it (and your future profession), you need to study/practice on your own time.
polarmystery
Diamond Member
- Aug 21, 2005
- 3,888
- 8
- 81
Yeah I believe so. I'm guessing that Intel got nervous about gate leakage being more than desirable @ 22nm so they started incorporating finfets then. I'm not quite up to speed on what they are actually doing though with them.
I couldn't have said it better. This was the reason I did so poorly in my undergrad studies but a lot better in grad school. Definitely take the time to learn your craft, and it will pay off.
Last edited:
Do they still do that in great mistakes? I was in BEQ 635 for "tech core" which they probably don't do anymore - managed to get a GDUHC during the asmo from Tech core to ET A school.
As for transistor stuff. All the above seems a bit convoluted and unfortunately still doesn't address teh whole "what does a transistor do", so I'll give it a shot.
Taking a few liberties for brevity!
Your base-emitter "current flow" "controls" the collector-emitter current flow. Thus more current flow at B-E = significantly more current flow C-E (this increase is known as "beta" or weird Capital B).
Imagine the B-E current flow as a valve on a spigot. And the C-E current flow as the water flowing through that valve. Now you can increase or decrease the flow of water by restricting the valve. This is the "linear" region. In other words - very nearly perfectly an increase in B-E current increases C-E by a factor of Beta. This gets funky towards either end and changes with temperature.
Saturated means the valve is fully on, and jamming more current through only heats things up. Off means the valve is fully off and very little/no current flows.
If you "bias" the inputs and outputs JUST right - that is around the middle of the linear region, and such that the expected input voltage (current) change will not increase beyond saturation, nor all the way off, you have just built an amplifier! Other things to consider but thats the basics.
The inversion part is interesting and it took me a while to understand. Essentially the C-E junction acts as a variable resistor for the C-E current. So when you *increase* B-E current, the C-E resistance decreases, thus more current flows, BUT the voltage drop - being that the resistance decreases will also decrease across the collector - gnd resistor, as that resistance stays constant. Hard to show with no diagram though.
thx1128
Junior Member
- Sep 26, 2011
- 3
- 0
- 0
Prophetic! It is sad the number of interviewees we turn down for hire because they are limiting themselves to what we are willing to teach them on the job.
They should know that we filled that position with a High School graduate that self-taught PC hardware/Linux/Windows and now we are smoothing off their rough spots whereas we would normally have to start at teaching a vocabulary to a "college graduate".
I spent a few years in the Navy too (Glakes in winter 1982). Are they not teaching circuit theory?
- Apr 20, 2008
- 10,065
- 984
- 126
It was a small part that wasnt exactly taught in my training. I could have pieced it together with the information i was given. Im already done with Tech Core (ATT) and I'm at A school now. From here on out I bet its all going to be working on specific systems and not theory like I had learned.
To both of you former Navy, what were you're rates?
To both of you former Navy, what were you're rates?
non-nuke ET - discharged post tech-core (which was where I learned the remaining 20% of what I know in electronics, the first 80% I figured out myself from books, experiments, and honestly figuring it out myself) I've gone to a more traditional college and the emphasis is on digital systems anymore with a crash in analog components and ohms law.
stormkroe
Golden Member
- May 28, 2011
- 1,550
- 97
- 91
I was surprised when we went through all these solid-state components during trade school as an electrician. I had hobbied electronics from 13 onward, and was delighted to see BJT's, SCR's, etc in the curriculum. Almost as delighted as watching the oil field 'electricians' try to figure out why they had to quit calling the 'hot' wire the 'positive'. It was quite in depth with regards to the P/N junctions (explaining the wholes, relationship to cathodic protection, etc).
Good times.
Good times.
stormkroe
Golden Member
- May 28, 2011
- 1,550
- 97
- 91
Since I'm on here with other people of similar interest, let me ask this: I was thinking of an alternative to the standard, and inefficient, DC to DC converter. If you 'stutter' the DC input, as pulsating DC, a regular transformer should step the voltage properly, with the secondary voltage being slightly lower due to averaging the 'off' pulse through an L-filter. Does something like this already exist? I'm almost certain it does, I just never had a need to find out. If not, I got dibs on the idea until Apple patents it first under the new patent law.
Modelworks
Lifer
- Feb 22, 2007
- 16,240
- 7
- 76
What you just described is called a flyback converter. They have been around for about 50 years. Apple probably will still claim the patent though
http://en.wikipedia.org/wiki/Flyback_converter
#
Yes. This is the exact description of a "forward" converter. There are a number of variants of the forward design - but broadly, this is the preferred DC-DC converter design for power levels between about 100W and 1000 W - unsurprisingly, it's the workhorse for PC PSUs.
There are a variety of ways in which the DC is 'chopped'. One transistor, Two transistor, and more recently, one transistor with active transistor clamping.
The one transistor design is not very efficient, and very stressful on the transistor. The two transistor design is more practical. Best efficiency, but highest cost is with the active clamping design. 95% efficiency PSUs will typically use active clamp designs, in combination with active PFC - although the slightly increased cost of active clamp designs, means that they are pretty rare (almost unheard of) in PC PSUs.
Last edited:
William Gaatjes
Lifer
- May 11, 2008
- 21,625
- 1,292
- 126
That reminds of snubber circuit designs. I had seen once a SMPS design ( > 3kW) where the snubber circuit was a mosfet with accompanying electronic components that started to conduct to pump back the induced transient voltages back into the main rail( the input storage capacitor at 325 Volts). This gives a higher efficiency because the energy of the induced transients is transferred back to be used again instead of dissipated as I^2*R. At these levels of power, the power to dissipate can easily be > 100W in that R alone.
http://en.wikipedia.org/wiki/Snubber
EDIT:
I looked it up, this technique is called active clamping it seems.
See page 3-6 :
http://www.ti.com/lit/ml/slup108/slup108.pdf
But i found other explanations as well when it comes to IGBT drivers.
http://en.wikipedia.org/wiki/Snubber
EDIT:
I looked it up, this technique is called active clamping it seems.
See page 3-6 :
http://www.ti.com/lit/ml/slup108/slup108.pdf
But i found other explanations as well when it comes to IGBT drivers.
Last edited:
TRENDING THREADS
-
Discussion Zen 5 Speculation (EPYC Turin and Strix Point/Granite Ridge - Ryzen 9000)- Started by DisEnchantment
- Replies: 24K
-
Discussion Intel Meteor, Arrow, Lunar & Panther Lakes Discussion Threads
- Started by Tigerick
- Replies: 19K
-
-
-
Facebook
Twitter