How do I get more current?

[Hero999]

At an estimate your laptop uses 50W.

You need to use a resistor that dissipates a similar amount of power as your laptop.

Testing a SMPs at less than one tenth of the power you're intending to draw is a waste of time. Your switching regulator might be 84% efficient at 3.12W but at 50W it might only be 50% efficient.

 

[jpanhalt]

Here is a 5A, $25 USD solution:

**broken link removed**

Although, it may not put out enough voltage for your needs.

Have you considered "splitting" the batteries as described on the Castle site under CC BEC Multi Pack?

John

 

[Papabravo]

...
And I disagree with your opinion of demonstration circuit boards. They are usually designed to achieve the maximum performance from the parts, since a customer may be using it to determine if it's appropriate for his own application. A bad design of the demo board would likely kill a sale. It would be like a car dealer letting you test drive a car with a engine that's out of tune.

You're certainly entitled to your opinion. He asked for mine and I gave it to him. He can be thick or thin skinned as the mood moves him. If he wants his ego stroked he needs to ask different questions or look elsewhere.

He also indicated that the board design was his own and not a demo board from Sharp. His schematic is nearly identical to the one in the datasheet. Test circuits in datasheets come with a disclaimer that they may not be suitable for use in an application. There are certainly missing details on the component characteristics on that datasheet schematic. The datsheet does not contain a layout, I believe that Allan is responsible for the layout. The layout and the component choices are where the problems lie. Regardless of how well the current circuit performs, he wants to reduce the heat dissipated in the switch and he wants to increase the available current. He cannot do that with the current board design for the reasons that have been already enunciated. Sugar coat it anyway you want to and the answer still comes up the same.

 

[Papabravo]

Lets go back and look at the actual design from SHARP. Schematic on p. 20 layout follows.

**broken link removed**

Can anybody see the difference between SHARP's layout and Allans layout? Notice the wide thick traces? Can you tell how many layers the board has? Do we notice anything about the inductor in this circuit? Do we notice anything unusual about the capacitors?

 

[AllanBertelsen]

Hero999. With 12V I got to go as low as 3 ohm. I tried with 9 ohm by using a bundle of resistors in parallel. So now the load is 16W. That’s about a third to a quarter of what I need. The efficiency at this load is 80%. I need power resistors get a higher load by using resistors.

jpanhalt thank you for your suggestion. If I got to select another solution it would be replacing the picoPSU with this one **broken link removed**

Then my own SMPS would be used for lesser demanding things on my robot.

 

[AllanBertelsen]

Now I used a 20W 12V halogen spot for load.
In: 18.97V x 1.21A = 22.95W
Out: 11.8V x 1.61A = 19.00W
19W/22.95W = 83%

 

[AllanBertelsen]

And with two halogen spots for load.
In: 18.4V x 2.38A = 43.79W
Out: 11.8V x 2.75A = 32.45W
32.45/43.79 =74%
So the efficiency is declining.

 

[ericgibbs]

And with two halogen spots for load.
In: 18.4V x 2.38A = 43.79W
Out: 11.8V x 2.75A = 32.45W
32.45/43.79 =74%
So the efficiency is declining.

hi Allan,

Look thru these links, they may help with your query.
Depending upon the point on the load curve for your SMPS the n% will vary, its not flat over its specified operating range.

An efficiency of 74% isn't too bad for a simple SMPS, especially as you are going to be powering from the car battery

These links may give you some ideas.

**broken link removed**

Switched-mode power supply - Wikipedia, the free encyclopedia

SMPS Switching Power Supply Circuit Design- Tutorial, Schematics

 

[Hero999]

The efficiency of a simple buck depends on the voltage drop. As the output voltage gets lower, the duty cycle is reduced which means the peak current needs to be higher for the same output current. This means that the inducor's current rating needs to increase and the switching losses also increase. I would expect a 12V to 3.3V converter will be much less efficient than an 18V to to 12V converter if they're both using the same buck topology.

In my opinion 74% isn't bad for regulator converting if it was converting 12V to 3.3V but it's pretty poor for a 18V to 12V converter. To put this into perspective, a linear regulator converting 18V to 12V would give an efficiency of 66.67% which is only 7% worse than this design.

At 50W your regulator is wasting 17.76W of heat which will require a reasonable sized heatsink. If this isn't a problem for you, then fine, but it shouldn't be hard to halve the losses if you improve the layout and use a larger inductor.

50W at 12V is 4.167A.

12V is 66.67% of the input voltage, therefore the duty cycle will be about 66.67%, but in practise it'll be a bit higher than that.

For an output current of 4.167A, the peak current will be 4.167/0.6667 = 6.25A so make sure your inductor is rated for a much higher current - use an 8A inductor minimum.

Increase the width of all the power carrying PCB traces to lower their inductance and power losses due to their I²R losses. Keep all the traces as short as possible and use a ground plane PCB, if you can.

 

[AllanBertelsen]

Thank you for the links Eric. I have already been studying the Wikipidia page. The third link is really interesting. Thanks again.

Thank you Hero999. This is useful. I noticed that the lambda-values of the inductors in Sharps cash register are higher than the example in the datasheet. In the cash register there are two parallel circuits. In the low voltage part it is 180uH and in the high voltage part it is 220uH. Datasheet says 90uH. What would a higher value do?

 

[Papabravo]

A schematic with an inductor that gives only the inductance can be misleading, because the inductance of the inductor is only part of the story. The inductance is related to the ability of an inductor to store energy. More inductance means more energy storage. The big picture is more nuanced. You can increase indcutance by increasing the number of turns, but that leads to higher DC resistance. To decrease the DC resistance you can use wire with a larger diameter which reduces the number of turns for a given core size. So you can make the core bigger, or change the core material, and so on and so on.

A rule of thumb, without detailed analysis, is to use twice the output current as the peak inductor current. The original 2.8 Amp inductor would, IMHO, be good for an output current of 1.4 A Max.

The inductance and the DC resistance has an effect on the switching frequency because the reactance of an inductor goes up as the frequency increases and the time constant of an LR circuit is just L/R. For your original inductor the inductance under bias drops from 90 uH to 60uH and the resistance is .108 ohms.

L/R = 60uH/.108 = 555 microseconds

I don't know what switching frequency you're working at but it seems like a rather large time constant for an allegedly (by the manufacturer) high frequency inductor.

 

[Hero999]

A rule of thumb, without detailed analysis, is to use twice the output current as the peak inductor current. The original 2.8 Amp inductor would, IMHO, be good for an output current of 1.4 A Max.

I think that's too much of a generalisation. A 15V to 5V converter will have a peak current of 3 times the output current and a 15V to 10V converter will have a peak current of 1.5 times the output current.

 

[Papabravo]

You're right of course, but the point was that it is almost always wrong to pick an inductor whose maximum current is the same as or even close to the intended output current and then try to fudge it on top of that.

 

[neon]

power

well they got off the subject. he wants to increase power capability well unless he has found a perpetual moving machine he cannot do it.

 

[Hero999]

That's not true at all.

He doesn't actually want to increase the amount of power from the primary power source. He just wants to increase the output current from his switching regulator which is perfectly reasonable and within the laws of thermodynamics.

 

[AllanBertelsen]

I just got to read up on LR circuitry and study all the articles at SMPS Switching Power Supply Circuit Design- Tutorial, Schematics. Until I know more, would it be right to state that when the resistors and the capacitor are about the same as in the datasheet and in the Sharps cash register. Then no harm would be done choosing a higher lambda-value for the inductor. And of course a much higher current value. I will seek one at Digi-Key.

 

[Papabravo]

You should also familiarize yourself with a capacitor parameter called Equivalent Series Resistance or ESR. Generally speaking, the capacitors used in SMPS designs, should be low-ESR types.