---------------------------------------------o ground 0V DC
Diode is a low power shotky switch is actually transistor.
Ok m so when the switch is closed the current flows through the inductor, creating a magnetic field. The cap charges and the output voltage of 5V is reached. So the p.d at the output is 5V say at 500mA to keep things simple.
When the switch is opened, there is a back e.m.f generated by the inductutor. Which basically means that the p.d at the didoe and inductior is -5V. The diode is now reverse biased. So current flows through the inductutor to the cap keeping the output voltage a 5V.
What about the current does it still stay at 500mA? I thought that caps could not store energy?
The process is then repeated to said f when the switch is again opened and closed.
The inductor acts like a "current smoother", much like a capacitor "stores" voltage, an inductor "stores" current.
So think of it in terms of current, not voltage:
When the switch is closed, the inductor's current gradually increases. When it exceeds the current draw on the output, the excess is used to charge the capacitor, and the voltage increases. When the voltage on the cap exceeds +5v, the switch is opened.
When the switch is open, the inductor's current gradually decreases (using the diode). When it no longer exceeds the current draw on the output, the capacitor provides the additional current, which discharges the capacitor. When its voltage drops below +5v, the switch is closed.
The switching frequency is just the fastest that the switch changes state. Since the switch is either "on" or "off" (and never halfway, like a resistor), you minimize heat losses there. The inductor and capacitor are "ideal" in that the only heat loss is from their inefficiencies and leakages. Most of the power loss is at the diode, since it has a fixed Vf. Some switchers add a mosfet in parallel with the diode to reduce those losses also.
So when the switch is closed the inductor provides the current for the load, which provides the votlage for the load (via the cap) and the current also.
When the switch is closed the inductor discharges slowly via diode and once the CURRENT drops the cap takes over with current, once the cap has been exhausted the voltage drops and an error amp/fb loop turns the swithc on again.
thanks, I was focussing on V and then i rather than looking at i. It makes sense, L charges c abd provides load, then c provides load and v drops then repeats
Yes, it very bad. You need to use a fixed width font when doing ascii so each character gets the same space. Else it could make no sense. You could, for example, use word or notepad and select a fixed width font and then copy it into the email if you cannot change the font in the email program(you can do it with outlook if you go into options).
lol, no. Caps do store energy. They are like a tank of water. You can pour energy in and drain it out later if you want. As long as you pour more than you take it will always be full(on average). The capacitor there is to act like a temporary battery while the main battery is disconnected.
Think about it this way, if you charged the capacitor and stuck its leads across a resistor then current would flow. The big difference between a cap and a battery is that a battery supplies constant voltage while the voltage on a cap will decrease over time and it depends on the charge it has stored and its capacitance. (so it depends on how fast your removing that charge, i.e., current)
Resistors do no store energy. Inductors and capacitors do. This is reason you don't see a resistor in there.
For example, if your load is very low and your capacitor is very big then you don't need an inductor and diode there. The issue here is that you want to minimize the capacitor size cause you have to quickly fill it up when the power supply starts. If its to big you could draw excessive current from the PS and destroy something. But that would mean you could leave the switch off much longer too. Although it would probably be an extremely noisy method.
No. When the transistor is off, the diode is forward biased. It keeps the voltage from swinging very much negative, which it would do if the diode weren't there. The diode "clamps" the switch node from swinging negative.
So current flows
Well, it keeps pumping current into the capacitor and the load.
Caps do store energy. So do inductors.
The current through the inductor builds up when the transistor is on, and decays when it's off. Those phases correspond to adding energy to the inductor, and taking it out. If the L is big enough and the frequency is high enough, the current through the inductor is *almost* constant. In real life, there's a pretty-much triangular ripple current in the inductor, because nobody wants to buy huge inductors.
The ripple current in the inductor in turn causes a bit of ripple voltage across the capacitor, and the load.
"The diode "clamps" the switch node from swinging > negative."
?
So I'm a bit confused. So am I correct in thinking that thorught out the on/off phase, the current always flows from the inductor to the capacitor. The back e.m.f would go to the switch when turned off if it were not for the diode?
I thought that the diode would conduct as it were revese biased by the back e.m.f
Yes, the diode will short the back EMF, Also called wheeling voltage and thus a wheeling diode. Polarity is reversed on the EMF.
Which leads to some interesting facts here. Since this is true, the polarity on the other end must be (+) in which case can increase the voltage on the cap. So the current cycle when the switch is on will most likely not be on long enough to introduce the full voltage on the output side of the inductor because, when it shorts via the diode. Additional voltage will appear at the output. The regulating circuit must be in tuned to this process of course.
I may have over looked something but that is just about the gist of it.
--
"I\'d rather have a bottle in front of me than a frontal lobotomy"
http://webpages.charter.net/jamie_5
Current always flows through the inductor [1] in the direction of the load. If the diode weren't there, terrible things would happen when the transistor turns off. The voltage at the switch node would try to swing to negative infinity, the transistor would break down from overvoltage, and the transistor would probably die. Even if the transistor didn't die, it would get very hot and the thing wouldn't act like a switching regulator.
"Back emf" is a fuzzy and hazardous term. I only use it with reference to DC motors, where it actually means something.
When the transistor is off, current flows from ground, *forward* through the diode, through the inductor, into the capacitor. In that state, the energy stored in the inductor drives the load.
Redraw it as...
SPDT Switch Inductor
--------o\\ \\o----------LLLLLLL------+---------o +ve output 5V DC o | | ------ | ------ Capacitor | | | | | |
--------+--------------------------+---------o ground 0V DC
and wiggle the switch up and down, so that the inductor is connected to Vin half the time and ground half the time.
This is a "synchronous switcher."
Oh your ascii art had hard tabs embedded, which is why some people didn't see it right. Spaces are safer, because some weird people don't use 8-space tabs.
John
[1] as long as the mess is operating in continuous mode.
I'm jumping in late, but see you are still having a little difficulty with the basic concepts. This circuit puts the inductor in one of two situations. When the switch is on, there is a voltage drop across the inductor that is the difference between the input voltage connected through the switch and the output voltage (across the capacitor and load). This fixed voltage causes the inductor current to ramp steadily up.
This comes out of the basic formula that relates voltage and current for an inductor. V=L*(di/dt) or voltage across the inductor is equal to the inductance times the rate of change of current. V is in volts, L in henries and di/dt is in amperes per second.
In the switch on condition, V is (Vin - Vout). So the current ramps up at that voltage divided by the inductance.
Lets assume that this ramp averages the load current, so that during the early part of the ramp, the capacitor voltage is sagging a little as it provides the current deficit, and near the end of the ramp, when the inductor current is more than the load current, the capacitor absorbs the excess, as its voltage rises.
Then the switch opens and blocks inductor current. So the inductor produces whatever negative voltage it takes to suck the current (the inductor was passing, right before the switch opened) up from ground, through the diode. This means that the input end of the inductor drives a diode drop more negative than ground. This requires that the inductor voltage drop be the output voltage plus the forward bias diode drop. So that new V (negative, this time, since the inductor is the energy source, not an energy absorber) produces a negative rate of change of the current through the inductor. That means the current starts where it was when the switch opened and ramps steadily down toward zero. And again, the ramp is assumed to average about equal to the load current, so at the beginning of this ramp down, the capacitor is still absorbing excess current ands its voltage is still creeping up. But the moment the inductor current ramp falls below the load current, the capacitor starts supplying the current deficit to the load, and its voltage starts to decay a little.
Before the current reaches zero, I am assuming the switch closes again, and reverses the voltage across the inductor, starting another increasing current ramp, and the cycle repeats.
So the inductor current is a triangle wave between some minimum and some maximum current whose average is equal to the load current. The output voltage has a ripple as the capacitor sucks up current at the high parts of the inductor current triangle, and supplies current at the low parts of the inductor current triangle.
The larger the inductor, the flatter the current triangle, so the less the capacitor must absorb and contribute each cycle.
The higher the switch frequency, the closer the peak current of the triangle is to the minimum current of the triangle.
The larger the capacitor the less its voltage swings for a given current ripple it must absorb.
I think I've got it clear now. Switch closed creates a positive current ramp, that intially charges cap and load, with the current increasing to an excessive ammount to bring V out up to max. Diode is "off"
When switch is closed, diode is on, current still flows in the same direction, ie from ground to inductor to charge c, only becuase the diode is "on" due to being revese biased, by "back e.m.f". However the current starts at an excess, (the max point reached during on time) so Vout is up to max. As load drains the current the excess is drawn from inductor then cap and once excess is depleted the V out drops. At this point the switch turns on again.
No. The diode is forward biased. Diodes will only conduct current when they are forward biased. As another poster has said, the term 'back emf' is confusing in this context.
during on time) so
inductor then cap and once
The current through the inductor will start to drop once the switch is opened. The left side of the inductor will be at 0.7 volts (the forward drop across the diode), the right side of the inductor will be at Vout. Thus the voltage across the induction will be Vout - 0..7 volts. This voltage will be trying to reduce the inductor current.
Another way to look at what is happening with the inductor current is to look at that is happening with the inductor's energy. An inductor stores energy in its magnetic field. The energy in an inductor is given by
energy = (1/2) * L * i ** 2 or energy = (1/2) * L * i squared
When the switch is closed, the battery is supplying energy to both the output and to the inductor. As the inductor's current is increased, the energy stored in the inductor is increasing.
When the switch is opened, the battery is no longer supplying energy. Instead the inductor starts to provide energy into the capacitor and the load. This results in the inductor current dropping. When the inductor current becomes less than the load current, the output voltage will start to drop. (Note: The output voltage was still increasing for a short time after when the switch opened while the inductor current was greater than the load current. During this time, the excess current from the inductor was continuing to charge the capacitor.)
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