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Oct12-04, 10:06 AM
There is a fixed power a.c supply connected to a transformer which is then connected to a full wave rectifier and then to a load which is parrellel to a capacitor( for smoothing).
If there is smoothing, the voltage across the load will be smoothed which will means that the average voltage across the load is increase which means that the power dissipated in the load is increased. There is also power loss in the charging up of the capacitor. But the power supply is fixed. Where do the extra power comes from????? Or is there anything wrong in my facts.
Oct12-04, 11:35 AM
Why do you say "the average voltage across the load is increased". If you could do "smoothing" without any loss, the average voltage would be exactly the same- the actual voltage a little higher at the low end, a little lower at the high end. Allowing for dissapation, the average (smoothed) voltage will be a little lower than before.
Oct18-04, 04:15 AM
But why would the voltage be lower at the high end??? If the voltage across the primary coil is constant, the voltage across in the secondary coil should also be the same. And since the capacitor and the load is in parellel, the voltage should stay the same at the higher end.....
Oct22-04, 09:25 AM
Smoothing a rectified a.c current
Without a smoothing capacitor after a rectifier, the output to the load is a series of pulses that rise & fall between zero volts and the maximum voltage of the rectifier.
The average voltage will obviously be somewhat less than the peak in this case.
By adding a capacitor after the rectifier, the pulses begin joining together.
A small capacitance will cause the tip of one pulse to make a sharp line towards the based of the next incoming pulse - since it can't store much charge to power the load between pulses.
Now the pulses are spending more time at elevated voltages, so the average voltage begins to increase.
If you add enough capacitance, the pulses turn into smoothed bumps that taper down to the next peak.
The supply is now spending much more time with it's voltage close to the peak voltage of the rectifier.
In a Hi-Fi, the big electrolytic capacitors in it's power supply are the first source of electrons for the amp. The capacitors are have very little resistance to sudden loads and they can supply massive amounts of sharp power.
So imagine your system isn't loaded.
The rectifier turns on and see the capacitor as an empty container that needs filling with electrons to bring it's voltage in line with the voltage across it.
The rectifier begins sending out pulses of charge to the capacitor and the capacitor fills with electrons, raising it's voltage. Eventually, the capacitor's voltage reaches the same value as the rectifier's output.
The capacitor's voltage is now stable and the rectifier now longer send pulses out to the capacitor - except very small ones to make up for the stray losses in the circuit.
Now imagine you drop a paper clip across the output of the circuit.
The capacitor will then discharge as fast as possible through the paper clip. Some of the power will come from the wall socket, but most of it will come from the capacitor.
If the capacitor is very big and the power supply low rated, the capacitor's voltage will now momentarily rocket down and the electrons rush off the plates and through the paper clip. The electrons are trying to stabilise the voltage differences in the circuit so that there is zero volts difference between them if you think about it. Inside the capacitor, they are sitting right on the edge of doing this. When you short it out, they run to the other pole of the circuit to try and balance it out.
Imagine that the discharge is powerful enough to melt the paper clip, and so, remove the load.
Now the rectifier begins supplying pulses of electrons to the capacitor again to recharge it.
Obviously, amplifiers aren't designed to melt after one pulse. They supply continuous pulse like loads.
Say I'm playing my electric guitar very quitely, the voltage on the capacitor in the supply will remain close to the rectifiers peak as only small loads are asked of the supply.
Now I turn up the gain and hit a chord.
A big load of current is expected from the supply, and this will be sourced from the capacitors.
Immediately after this, I carry on playing.
Since the capacitor it's self represents a heavy load when it's drained, the supply will begin recharging the capacitor before it sends out more electrons to whatever else is connected after the capacitor.
As a very extreme example of what this would be like, which never actually happens, I might hit a chord and it'll come through sparkling clean. Immediately after, whatever I play would barely make a sound.
As the capacitor recharged, the sound level would return.
If the power supply can only supply so many electrons per second, more of them will end up going towards the capacitor. If it can supply more, it may load the rectifier down very heavily to reload the capacitor as fast as possible.
The effect is known as a transient voltage.
The capacitor is acting like a big squishy spring in the circuit. The more capacitance you add, the bigger the instanteous effect, but the longer it takes to recover it's charge.
Audiophiles notice the effect sometimes because they usually use huge electrolytics in their power supplies. These provide the instantaneous current required for a strong bass signal, like thunder and lightning or bass drum.
But, immediately after the loud bass sound, the amplifier kind of muffles it's self as the capacitors recharge themselves.
In essence, the capacitor almost always come first in this kind of circuit. They charge first, they discharge first and then they get first priority in recharging.
The problems are not as bad as I've made them out to be in the real world.
Guitar amplifiers, for instance, are rated at 50 - 100 watts of output usually. A lot of people make the mistake of thinking that the power supply would need to be capable of 100 - 200 watts continually to make up for pulses or stray losses in the circuit.
The reality is, 100 watts creates truely ear piericing volume. And, even with the amplifier turned up that high, it is very rarely actually outputting 100 watts of power.
The waveform creating the loading on the circuit, coming from the guitar, is modulated and spends very little time at it's peak. The big gaps between peaks allow the power supply to recharge it's self ready for the next peak. The bigger the rectifier, the quicker it can recover from the impulse.
Unless you're dealing with a particular weedy power supply, or massive capacitors, it's quite hard to see the effect in everyday circuits, since the capacitor is usually well on it's way to the peak voltage in less than a second of the transient being formed.
Another way to picture the effect would be arc welding. Imagine a simple transformer, rectifier and capacitor circuit. You strike an arc and the capacitor empties through the discharge creating a strong ignition. Immediately after, the arc's instensity will fall sharply as the capacitor recharges.
This idea is actually used in a lot of welders to help them strike sticky electrodes. Sometimes the electrodes are particularly poor when it comes to igniting and welding correctly due to the tip of the rod not being heated quickly enough.
A way to get round this would be to have a capacitor with a MOSFET (switch) on it's terminals. When the arc is struck, the capacitor empties and ignites the electrode correctly. Before the capacitor can begin sucking the power out the supply to recharge it's self, the MOSFETs could isolate it for the remaining time the weld is in progress for. When the welding stops, the MOSFETs would open and the capacitor would recharge ready for the next electrode.
Of coarse, you need to apply the complex 'rubbery & less than perfect' effect of electronics to get a more realistic idea of how the circuits will behave.
Just think of the impact on the load as upsetting the balance of the circuit. The circuit needn't take in more current power to rebalance it's self, it just has to shift which particular loads the current is going to. If the impact load is removed, the most immediate load to rebalance will be the capacitor. If the load remains and the supply can't rebalance the circuit, then the output will fall off.
Guitar pickups are another good example. They usually output a volt or so that acts as the biasing voltage on the grids of the valves in the amplifier. However, the second you expect them to produce any output current, their voltage rockets down. So the bias grids need to have a very high impedance (resistance), meaning that very little current is actually needed to pass through them for the reference voltage to be created on them. Guitar amplifiers usually have an input imedance of a million ohms - almost no current actually flows from the guitar to the grids compared to the current flowing out of the amplifier. The end difference is usually in the order of 10,000 to 100,000 times.
If you visit Nation Semiconductor's web site, they have a section for designing your own switch mode power supply. While this may not be of much interest to you, there is also a section for testing the design which includes a transient load test. This lets you see how the current and voltage move in the circuit as the load creates an impact on the supply. There is a web seminar on their site somewhere given by a guy that walks you through using it. You'll need to create an account with them to watch it I think.
Also, if you look around on the DIY audio forums, or ask someone there, you should be able to find a program called Power Supply simulator. You can design simple filtered power supplies with the program and then run a test on them to see what degree of noise they produce and how quickly they'll come up to their full voltage.
I hope this was at least of some help to you!
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