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When a boost/buck converter would output low/high current output than input?

**I_in =/ I_out**

I'm teaching myself the basics of circuits, understood KCL and KVL, but this point is confusing me when relating power converters(of all classes).

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- #1

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When a boost/buck converter would output low/high current output than input?

I'm teaching myself the basics of circuits, understood KCL and KVL, but this point is confusing me when relating power converters(of all classes).

- #2

berkeman

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No, most DC-DC switching converters are constant power converters, so Iin does not equal Iout (unless Vin = Vout).

When a boost/buck converter would output low/high current output than input?

I_in =/ I_out

I'm teaching myself the basics of circuits, understood KCL and KVL, but this point is confusing me when relating power converters(of all classes).

The input current in a Buck converter only flows when the high-side switch transistor is conducting. When it snaps off, the flywheel diode at the output keeps the current flowing. Since the buck circuit is driving into an inductor, the inductor current (Iout) ramps up while the switch is on, and ramps down when the switch is off and the flywheel diode is conducting.

Does that make sense? Use Google Images to find some current and voltage waveforms for Buck DC-DC converters...

- #3

berkeman

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Like this

http://www.analog.com/library/analogDialogue/archives/45-06/AD45-06_BB_FIG_02.jpg

http://m.eet.com/media/1050719/C0295-Figure3.gif

http://www.analog.com/library/analogDialogue/archives/45-06/AD45-06_BB_FIG_02.jpg

http://m.eet.com/media/1050719/C0295-Figure3.gif

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Acting similar to the behavior of constant voltage/current sources.

- #6

berkeman

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I'm not sure I understand the question. Normally you will use voltage feedback from the output in order to adjust the PWM circuit to maintain the output voltage at the desired value, independent of the input voltage.

Acting similar to the behavior of constant voltage/current sources.

Alternately (as is used in some LED driver circuits), you can use feedback from the output current to control the PWM circuit to maintain a constant output current at some set value, independent of the input voltage.

Does that help?

- #7

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It should be a change input(V or I) to maintain a constant (V or I) depending on the setup.

- #8

berkeman

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Like this Maxim circuit for driving LEDs with a constant current:

https://www.maximintegrated.com/en/images/appnotes/3668/3668Fig01.gif

- #9

meBigGuy

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- #10

Baluncore

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Fundamentally there are two independent circuits in a buck converter. The first circuit is from the supply, through the switch, the inductor, to the load, then back to the supply via the ground return. The second is from the ground, through the diode, inductor and load.

The polarity of the voltage across the inductor is reversed during each phase of the cycle. Since V

- #11

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Assume a constant DC power supply, that is 30W connected to a boost-converter to output higher voltage(with lower current) to the load, since Pin = Pout(assume 100% efficiency). The wire's resistance is 0.5ohms(total), the load's resistance is 5 ohms(so net resistance is 5.5), I

- #12

berkeman

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I'm not following your example at all. First you set the output voltage at 12.84V, then to 30V. What do you set your boosted output voltage to? What is your load resistance? That gives you your output power. What is the input voltage set to? That determines the input current drawn to provide the output power.

Assume a constant DC power supply, that is 30W connected to a boost-converter to output higher voltage(with lower current) to the load, since Pin = Pout(assume 100% efficiency). The wire's resistance is 0.5ohms(total), the load's resistance is 5 ohms(so net resistance is 5.5), Ithat the applied voltage from the PS would be 12.84V and the current is 2.33A. Now the input range of this boost-converter is 5-20VDC,0.5-5A(example numbers,their all made up), and I set the output voltage to 30V, in my head I know it has to be 1A current since P = 30W, but if I apply ohms law... I always get higher current than 1A which can't be true, so what am I doing wrong here?think

- #13

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That would be the input from 30W PS, since resistance is 5.5ohms, that 12.84VDC will be inputted to the boost-converter.I'm not following your example at all. First you set the output voltage at 12.84V, then to 30V.

Output would be set to 30V(from 12.84 input), load resistance is 5ohms(with 0.5ohms from the wire so I totaled the resistance to 5.5ohms).What do you set your boosted output voltage to? What is your load resistance? That gives you your output power. What is the input voltage set to? That determines the input current drawn to provide the output power.

Input voltage should be set to 12.84V, I assumed the input current would be 2.33A(not sure though).

- #14

berkeman

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Can you just draw a sketch? The input current is determined by the output power and the input voltage.

- #15

Baluncore

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You have an input power of 30W from your supply, then the output power from a 100% efficient converter will be 30W.Assume a constant DC power supply, that is 30W

If the output load is 5.5 ohm total, then Vo/Io = 5.5R and Vo*Io = 30W. Then Io = 2.3355 amp and Vo = 12.845 volt.

You are assuming too many fixed parameters.in my head I know it has to be 1A current since P = 30W, but if I apply ohms law... I always get higher current than 1A which can't be true, so what am I doing wrong here?

You cannot change the output voltage of the 30W power supply to 30 volt without increasing the load resistance from 5.5 ohm to 30 ohm. That is because that would require more than the 30 watt power available at the input.

You should ignore power as an input parameter in your computational games.

Energy is conserved in a 100% efficient converter. Power is the rate of flow of energy.

Power is the only “computational bridge” between input and output.

Follow this computational process.

Specify the input voltage and the output voltage. Vi, Vo.

Specify the output load resistance. Ro.

Compute the output current. Io = Vo / Ro.

Compute the output power. Wo = Io * Vo.

For a 100% efficient converter. Wi = Wo

Compute the input current. Ii = Wi / Vi.

Or just note that since Wi = Wo then Vi*Ii = Vo*Io and Ro = Vo / Io.

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@Baluncore you made me realize my flaw here, thank you, but I'm stuck on a few things(bear with me):

For the computational process what is the resistance prior to the converter and after? It should be 30ohms all around? Or 0.5ohms(for wires) before for input calculations, and 30ohms for output calculations?

Ri = 0.5ohms(wires) ##\therefore ## Ro = 29.5ohms(load) + 0.5ohms(wires).

- #17

meBigGuy

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First off, the concept of a dynamic 30W power supply is not realistic. That in its self is a complex system guaranteed to confuse. It is neither constant voltage nor constant current. Is it an instantaneous 30W supply? Or does the supply average 30W. (for example does it limit the charge cycles of the DC to DC converter then go to infinite voltage when a switch opens). Supplying "30W" to a dynamically changing load makes no practical sense.

When we speak of power-in to power-out in a dynamic converter, we are speaking average power averaged over the energy storage times of the converters energy storage components.

So, first define your converters output characteristics. Is is constant voltage? Is it constant current? Is is current limited? Is the current limit "fold-back"?

'

Maybe you want to define some sort of non-linear input to output relationship? Well, define it and write the equations. But, playing with that can get tricky when you start talking dynamic systems with response times, feedback, damping factor, etc. That's a whole new subject.

Then define your load. Is it constant, or changing? If it is changing, write the equations.

Now, if you want to apply some weird power source to all that, then again, define the power source mathematically.

Start with a constant voltage supply, define a converter, define a load, then look at what is happening. Then change 1 thing at a time and look at the effects back at the power source.

- #18

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lol, well I was in the process of doing that... and I think I've defined a lot about the circuit from the previous posts relative to the example(hopefully)?Start with a constant voltage supply, define a converter, define a load, then look at what is happening. Then change 1 thing at a time and look at the effects back at the power source.

But I'll give that approach a go, it's perfect way of analyzing the circuit(and other things).

I need a clarification on point #16 though because my calculations are based off that.

I have a

Ri (which is just the wire connecting the PS to other components) = 0.5ohms

Vi = 3.8V

Ii = 7.75A

Pi = Vi x Ii = 30W

That is the input to the boost-converter that is meant to output 30V to the 29.5ohm load:

Vo = 30V

Iout = 1A

Ro = 30ohm( 29.5ohm + 0.5ohm)

Pout = 30W

Pi= Po ##\checkmark##

Am I right?

- #19

Baluncore

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No. Your 0.5 ohm must be modelled on the output side with the load if you add it to the load. Wire resistance on the input side will lower the supply voltage to the converter. It is unimportant to the computations.Ri (which is just the wire connecting the PS to other components) = 0.5ohms

A converter is the DC equivalent of an AC transformer. It transforms the V/I ratio from one side to the other.

Do not cross the converter with anything other than the power. Treat input and output as totally separate circuits.

- #20

Baluncore

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By introducing a poorly specified series resistance, one that can jump around the circuit and across the transformer without being transformed, you are making it both unreal and more complex than it needs to be.Ri (which is just the wire connecting the PS to other components) = 0.5ohms

The output load is 30 ohms. Now forget the 0.5 ohm wire. All wires are now perfect conductors.

- #21

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That 0.5 ohm of the wire is what made me determine the input current & voltage, If I neglect it, I don't know what the input values would be from the PS...Now forget the 0.5 ohm wire. All wires are now perfect conductors.

- #22

Baluncore

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The output current must be 1 amp. The output power is then 30 watt.

For converter efficiency = 100%, the input power must be 30 watt / 100% = 30 watt.

The power supply has a regulated output voltage of Vs =3.8 volt.

So the input current must be 30 watt / 3.8 volt = 7.895 amp.

The load appears to be 30 ohms when viewed from the converter.

But when seen from the power supply, the converter transforms the load to look like a different value resistance.

That converted load resistance then appears to be 3.8 volt / 7.895 amp = 0.4813 ohms.

So the converter is transforming the load of 30 ohms to look like 0.4813 ohms to the lower voltage power supply.

Now, the idea that there might be Rs = 0.5 ohm of supply output resistance before the converter is clearly impossible.

3.8 volt / 0.5 ohm = 7.6 amp. There is clearly insufficient voltage to supply the 7.895 amp current needed by the converter.

The maximum Rs that could function would be when series resistance = apparent converter and load resistance.

In that situation the input voltage to the converter will be reduced to half of the 3.8 volt = 1.9 volt.

Because converter input voltage is halved, the converter input current would have to rise to 30 watt / 1.9 volt = 15.79 amp.

1.9 volt / 15.79 amp = 0.1203 ohm. Which is the maximum possible supply series resistance for the circuit to still function.

Overall efficiency would then have fallen to 50% simply due to resistance of the supply and cable to the converter.

The total power supplied would then need to be 60 watt, 30W to the supply series resistance and 30 watt to the load.

That is why you should use short thick cable between the power supply and the boost converter.

- #23

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The out voltage(##V_o##) is it in series with the load? Any component from point a-b are considered in series with the output current and voltage or parallel?

I find it parallel, because of the diode and the capacitor are parallel, but not sure...

- #24

meBigGuy

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That question tells me you have no concept of voltage, current, and circuits.

- #25

Baluncore

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The current flows through the load.

Series and parallel refer to the connection relationship between components.

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