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inductorless SMPS

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Thunderchild

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Diagram is attached, feel free to offer "real" world considerations. It works a treat in multisim

only thing I can't figure is why below 7 ohms load it all goes dead.....

Putting an inductor in btween the output condenser and transistor does not really do that much that I can see

now how would i rearange it to work with an N channel mosfet (VNP20N07) ? and stay under a frequency of 50 KHz

its for a solar charge controller so big ripple is not an issue and I'm sure it can be refined for proper power supply usage
 

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ronsimpson

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In the real world Q1 will get very hot.
Q1 will not work with 8 ohm load because the base current is too low with 1k resistor.
If you have no inductor there is no need for D1.
You could replace Q1 with a P-FET.
********************
I think you should use a PWM-IC and an inductor.
 

Hero999

Banned
This circuit is a classic. :D

I've seen variations on this theme before but none of them work.

The only thing limiting the charging of the capacitor is the transistor which will heat up. If the transistor, power supply and capacitor were perfect then the capacitor would charge instantly when the transistor switches on, then discharge exponentially via the resistor so there would be no voltage regulation.

It's possible to use a charge pump to do this but it's only efficient at very low currents.
 

crutschow

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You don't understand how an SMPS works. An PWM SMPS that converts one voltage level to another, needs an inductor to store the energy when the switch is on, which is then transferred to the load when the switch is off. The inductor is not just to reduce ripple on the output (although it does that also).

Your circuit just connects the source directly to the load with some duty-cycle and does no voltage conversion, in other words the switch serves no purpose other then to dissipate heat.
 

Thunderchild

New Member
well I did have it working with an inductor as well, thats why there is a diode there. I expect that in real life the BJT/Mosfet will have resistance which will limit the current. I actually found that with an inductor there was more ripple, although its all theoretical simulation so I expect it probably is too good to be true. So how would I go about putting in an appropriate inductor ? how would I calculate it ? in this case i do have some idea of the current output and that this will be pretty much constant if not considerably less.
 

Hero999

Banned
That's how I always figure it out.
 

dknguyen

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An inductor is not there to limit the current (like your theory about the transistor resistance). If you recall, if you leave an ideal inductor conducting DC for long enough it will become a short-circuit. THe inductor is there to slow down the rise of current so that it is controllable. The resistance of your transistor will not only limit current, but it will allow the current to rise to peak level almost instantly which does not help much when the resistance is ultra low. The inductor does the opposite, it does not limit the peak, but slow down the rate which is what is important. because...

The inductor's voltage drop starts out equal to the input voltage which makes a zero volt output. BUt this voltage drop decreases over time which makes the output voltage rise...SLOWLY. This means you can stop the rise when the output voltage gets to your desired level to get your lower, regulated voltage. Resistors don't do this either. Their voltage drop is instantaneous and depends on current (which sucks when your load current is changing). So not only would this make your output voltage jump to a particular value instantly so you could not regulate it, this voltage value would also depend on the current draw of the load (kicking regulation a second time while it's on the ground).
 
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Thunderchild

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for that matter how do I calculate the output capacitor ? I found that there is a tendency to oscillate with higher values. and with some LC combinationes the output eventually becomes VCC although I can't figure out how, perhaps I need to do more work on the "control logic" to take into account all possible states.
 

Ubergeek63

Well-Known Member
ΔI/ΔT=V/L

ΔT being 1/f ?

V being difference between input and output ?

ΔI being the current output required ?
no, delta as in the change in current and change in time. a constant voltage across an inductor will cause the current in the inductor to ramp up.

if you put 1V across a 1H inductor the current changes 1A in 1S
 

crutschow

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The ΔT is either the "on" time or the "off" time of the switch depending upon which you are calculating.

The V is the voltage across the inductor. It is roughly Vin - Vout when the switch is on (neglecting switch resistance) where the inductor current is increasing, and 0.7 - Vout when the switch is off (current flowing through the diode) and the inductor current is slowing down. You use these two values to calculate the inductor ripple current (it's the main criteria for deterimining the inductance you need)

The output capacitor determines the ripple voltage. It is approximately Vrip = (ΔI x t) ÷ C where ΔI is the ripple in the inductor current and t is 1/2 the switch period.

Yes, the control loop can be unstable without any compensation since you have an inductor and capacitor in the loop which introduces significant phase shift. There are many articles available on how to compensate a switching power supply feedback loop, for example http://www.electro-tech-online.com/custompdfs/2009/03/slup173.pdf. It's not a trivial subject. That's why it's much easier to use a switching controller chip with the compensation circuits built-in, rather than try to roll your own.
 

Thunderchild

New Member
I think some of the trouble is that the circuit is self governing and so the frequency can vary. perhaps an inductor circuit warrants a fixed frequency to prevent some of the uncertainty.
 

dknguyen

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I think some of the trouble is that the circuit is self governing and so the frequency can vary. perhaps an inductor circuit warrants a fixed frequency to prevent some of the uncertainty.
Fixed frequency does make it easier for things like boost, flyback, and SEPIC converters (actually I've never seen one that hasn't, but I'm sure it's possible). Most SMPS work with fixed frequency, but the simplest step-down (buck converters) do not. They simply switch off the current when the

output voltage = input voltage - continually falling voltage drop across the inductor

passes a hysteresic threshold and when it falls below the threshold they reconnect the current so you get a zig zag output current that is averaged about the desired output. This results in variable operating frequency as the input-to-output voltage ratio changes. THe advantage of the "hystersic converter is very very fast transient response time since it reacts within one switching cycle.

These types are actually quite rare, and only National seems to make them. One of the primary advantages of fixed frequency is that the output filter required is easily predictable. They are called hysteresic converters. National also makes a variation on the hysteresic converter that modifies the concept a bit so that it works on fixed frequency but still tries to maintain the fast transient response time.

But what is most important is the predictability of the ramp.
 
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Thunderchild

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well my main aim was to increase efficiency by allowing the frequency to lower when there is less load. I think the problems occur when I hit the LC circuits oscilating frequency and so the output swings considerably and if it is nailed bang on its starts to oscilate
 

dknguyen

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The LC filter resonsance is supposed to be far below the circuit switching frequency to prevent that, as well as to smooth out as much ripple as possible.

Obviously, that's much easier if the converter is fixed frequency and that is why fixed frequency converters are often favoured. You limit the frequency in order to limit what the filter has to do and to make the ripple predictable over the load range.
 
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Thunderchild

New Member
hm well thats a good starting pointfor determining the values of C and L. I assume that L is more critical so L can be calculated and then C worked out to set the filter frequency
 

Ubergeek63

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well my main aim was to increase efficiency by allowing the frequency to lower when there is less load. I think the problems occur when I hit the LC circuits oscilating frequency and so the output swings considerably and if it is nailed bang on its starts to oscilate
what you are looking for then is a hysteretic controler.

most straight forward in buck mode, in it's most basic form is simply a compatator with 10-20mV of hysteresis driving the power switch. The inductor, capacitor, line and load determine the operating frequency.

National has a nice little part the drives a logic level PFET.

Dan
 

Thunderchild

New Member
what you are looking for then is a hysteretic controler.

most straight forward in buck mode, in it's most basic form is simply a compatator with 10-20mV of hysteresis driving the power switch. The inductor, capacitor, line and load determine the operating frequency.

National has a nice little part the drives a logic level PFET.

Dan
thats basically what i have designed, the "logic" is universal the output power dependant on the power BJT/MOSFET and the LC carachteristics. The logic require one more small block which is a comparator that prevents the output going 0 and turning the transistor when its over the top, this can happen in oscilation mode as the voltage can swing higher than predicted and max it all out (I think)
 
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crutschow

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As I previously noted, you will still need some feedback compensation to stabilize the feedback loop due to the LC output resonance. For example if you look at this circuit for a hysteretic converter LM3475 - Hysteretic PFET Buck Controller you will see a compensation capacitor across Rfb1 in the feedback loop. You will need similar compensation for your loop to operate stably.
 
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