Why does this buck converter output such low current?

Question

I'm making my first attempt at building a buck converter, as a constant-current supply for a 700ma LED, from a 5V or 12V supply. I have it working with the parts I had on hand, at about 88% efficiency, but even at 100% duty cycle, the output current is only around 70ma for a 5V input. For a 12V input, it'll go up to about 185ma, but my diode starts to heat up then. The switch is being driven at 62.5KHz directly from the PWM pin of an Arduino. There is no feedback currently, since it's designed to power a specific load at a specific input voltage.

I simulated the converter in LTSpice before building it, though the parts aren't the exact same part numbers as in LTSpice, so I'm sure many of the important values are somewhat different.

On the breadboard, I used:

• a DMG3420U (logic-level N-channel MOSFET)
• polarized caps
• 1N4001 diode
• a 47uh inductor I had around (I know nothing else about it's specifications, is there some way to measure the important ones?)

There are a few things puzzling me:

1. My understanding was that when the duty cycle is 100%, the diode would be reverse biased, so there shouldn't be any power dissipation there. Why would mine be heating up?
2. Why is the converter only supplying 70ma? What can I do to find out what is limiting it?

The LTSpice simulation shows it supplying around 850ma with a 99% duty cycle, but with different parts.

My circuit, with a 30% duty cycle. Channel 1 is the gate pin of the MOSFET, channel 2 is the output to the LED. The ripple is fairly low, so I think that means it's running in continuous mode.

My circuit again, with a 100% duty cycle. At this point my bench supply says it's supplying 70ma, at 5.0V. If I turn up the voltage to 12V, I get around 185ma out of it.

The FET is an N-channel FET I happened to have around. Both capacitors are 10uf. The diode is a 1N4001 (not ideal, I know.) The inductor is a 47uh inductor I happened to have around. I don't know anything else about it.

Answer 1

At 100% duty cycle, you are correct in that the diode will not conduct. Also, your buck inductor will saturate out and you'll have the input voltage (minus resistive losses) applied directly to your LED with no other means of current limiting.

You're not driving the MOSFET correctly. Your pulse voltage should be from gate to source, not gate to output return. Buck converters with N-channel series MOSFETs need a high-side supply. The MOSFET will never be able to fully turn on with gate and drain close to the same potential with respect to source, since the minimum gate threshold for this part (with respect to source) is around 1.1V.

Tangent: you said you want a constant-current supply, yet in your powertrain you're not measuring the current that you wish to keep constant. I don't quite follow how this is supposed to work. I would expect that you have some sort of current sensing element (like a resistor) in series with your LED, which would be used to generate a voltage to control the duty cycle of the buck to make the current constant.

Answer 2

Where to start?

You don't show how you are driving the FET, but I am going to guess that it is being driven referenced to the source (I see Madmanguruman also mentioned this ... and more completely). The LED (which appears to be obsolete ... if this is it Digikey, says it was discontinued in 2011) has a forward drop of about 3.2V @ 0.7A, and with a PWM frequency of 63kHz means that you will want about 330uH for L1 (about 10 times what you've got now). Here is a post that addresses choosing an inductor for a buck regulator. Since this is not a voltage output, but you would still want continuous conduction, you can probably get by with higher ripple current (in this case +/- 20%), so the equation would be:

• \$L\$ ~ \$\frac{5 V_{\text{LED}}}{I_{\text{LED}} F_{\text{PWM}}}\$ = \$\frac{5 \text{(3.2 V)}}{\text{(0.7 V)} \text{(63000 kHz)}}\$ ~ \$330 \text{$\mu $H}\$

Looking at the waveforms, the top trace of the first picture doesn't look like the gate voltage. It looks like the source or cathode of D2. Note that when the switch turns off and voltage goes low, it goes below ground like you would expect the diode cathode to do. Then note that about 2uSec later it jumps up to what appears to be the voltage across C1, like you would expect if the L1 ran out of energy and became discontinuous. That's probably about the right amount of time for an L1 of only 47uH. It should also ring at that time, but there is enough loss somewhere in the circuit (L1, D1, or maybe even M1) to dampen it out.

It is not clear how you are controlling the current, but driving a LED with constant current, you should not need C1. Just keep the ripple reasonable in the inductor and get constant current in the LED. It would also be easier to compensate the loop without C1. Also, there are parts that provide PWM control for LEDs, like this one (which is the cheapest example at Digikey).

As you pointed out, the 1N4001 is not a great choice, it will be too lossy. A schottky would be better.

Answer 3

To find out what's limiting the current, turn the duty cycle full on and look for voltage drops across each component. If you have 5V on the input, that 5V has to be dropping across something as current flows from the positive terminal of the source back to its negative terminal. That something is what's limiting your current. The obvious contenders are the FET, the choke, the input supply, and the load itself. None of those components look like they should have those kinds of drops across them, based on the part numbers in your schematic, but obviously something's not adding up.

You should also consider the solderless breadboard; they can be difficult to predict.

Answer 4

I think your magnetics is very small. With power inductors, inductance is not everything. Check its series resistance and the maximum current it can support.