Help With Buck Converter

[Laurie Forbes]

My wind machine (13' with a 1 1/2HP motor conversion) produces about 130 VAC max. Up till now I have seen simply feeding the raw 3PH AC into three halogen light bulbs but would now like to consider using the output for battery charging. The issue then is how to convert the varying AC output into 12 or 24VDC with reasonable conversion efficiency.

Conversion to DC and use of a buck converter seems like a good way of doing it but I am having trouble finding a suitable circuit and, a layman's description as to how a buck conversion circuit works. More specifically, I can't figure out how the apparent trade off for conversion of "high" DC voltage and "low" current to low voltage and high current is actually accomplished. It's easy to understand how a transformer does it for AC but how the current is multiplied in a buck converter escapes me. I expect the circuit inductor is key but I'm not sure why.

Any help would be appreciated and, a pointer to a suitable circuit.

[Dave Turpin]

The inductor is important but is not the fundamental idea behind a buck invertor.

In its most basic case, a buck convertor is simply a switch and an inductor in a circuit. (Or simpler yet, simply a switch, since all wires must be modeled with inductance and capacitance)

The idea is, when the switch is shut, and the circuit is steady-state, the voltage across the switch is circuit voltage. But, when you open the switch, you get inductive kick due to the "intertia" of the moving electrons. In this case, you will probably see a small spark as you open the switch. In electronic terms, however, this spark is actually a rapid increase in voltage across the switch until the voltage is high enough to ionize the gas between the contacts on the switch, and the electrons flow, bringing the circuit back to steady state.

For a real buck circuit, you do this at a high frequency using a silicon-controlled rectifier, aka thyristor, which is microprocessor-controlled to produce the higher voltage that you want. An actual buck circuit is therefore a little too complicated to DIY.

(this explanation is simplified, and actually ignores the "buck" phase of the circuit, where voltage is ported to ground for a small amount of time to control the rise in voltage)

For your case, you would probably be better served with other circuitry than this. You can either transform the AC voltage down before rectifying, or use half-phase rectifiers to produce a lower DC voltage at the output.

[Laurie Forbes]

Dave, maybe I misunderstand the terminology but it sounds to me you are referring to a "boost" converter (ie. increases voltage). As I understand it, a "buck" converter decreases voltage and a "buck-boost" converter would do either with a varying input to produce a constant output voltage.

The trouble with using a transformer (I think) is that the frequency varies as well as the voltage.

Is "half-phase" rectifier the same as "half-wave"?

[Rob Beckers]

Laurie, I'll give explaining a try. Let's start with a buck converter in its simplest form (click the graphic for a larger version):

1016

There's a DC voltage (and power) source on the left side, Vg, a toggle switch, and a filter built from an inductor L and capacitor C. The resistor R represents the load on the output (ie. your batteries for example).

Let's say we have 100 Volt on the input for Vg, and the toggle switch spends half its time in position 1, half its time in position 2. We're switching rapidly, say 1000 times per second.

The voltage right after the switch, at Vs, will therefore show Vg half the time, and it's shorted to ground with zero Volt the other half of the time. That makes the average voltage at Vs (measured over, say, half a second) to be half of the input voltage, or 50 Volt.

What should be clear at this point is that by changing the ratio of the time the switch spends at position 1 vs. position 2, you will be changing the average voltage at Vs. If it spends all its time in position 1 the average voltage will simply be Vg, if it spends all its time in position 2 the average voltage will be 0 Volt. In fact, you can write this up in a simple equation for the average voltage:

Vs = D * Vg

Where D is the ratio of the time spent in position 1 vs. overall time (the duty-cycle; eg. 1/3 of the time in position 1 and 2/3 of the time in position 2 will get you an output voltage of 1/3 * Vg, or 33 Volt in our example).

So, without the extra inductor and capacitor we do have a buck converter as far as the average output voltage is concerned, but it's an ugly one, since the output is chopped-up DC. If we put a battery on its output, without the inductor and capacitor, we would alternate between charging for a very short time, with a very high current, and shorting out the battery for a very short time, again with a very high current. The result would be short-lived, and produce much smoke... :confused:

That's where the inductor and capacitor come in; they form a low-pass filter that smooths out the chopped DC into real DC with barely a ripple. Inductors are components that do not like to change current quickly: You can put a high voltage on them, and the current will slowly rise (and keep rising until you remove that voltage source). Once a current is going you can take away the voltage source, and the current wants to keep going, by instantly creating a very high voltage (and a spark if needed to keep the circuit closed). Capacitors are the opposite, they don't like to change voltage quickly: Put a voltage source on an empty capacitor and it will charge up very quickly, with a very high current. Try to change the voltage on a charged capacitor and you'll get a very high current as it changes its voltage and charge.

That works to our advantage for a buck converter because when the switch is in position 1 it will (initially) start a current going in the inductor, slowly, charging up the capacitor. When the switch goes to position 2 that current will continue to flow, in the same direction as before (remember that an inductor doesn't like to change current quickly), the inductor will continue to charge that capacitor. Averaged out over time it works out that the capacitor gets charged to the average voltage Vs, but it will be DC on the output.

To get back to our example; with a duty-cycle of 50% (or D = 0.5) and Vg of 100 Volt, we will get 50 Volt on the output. Say R = 10 Ohm, so the output current will be 5 Amp. That means the inductor will need to supply 5 Amp to keep that capacitor from neither charging nor discharging, and the output voltage constant. The inductors is only connected to the source Vg for half the time, so only half the time Vg is supplying 5 Amp, it's zero the other half of the time. On average Vg is therefore supplying 2.5 Amp. In the diagram above there are no losses, and that's the beauty of buck: It converters voltages/currents very efficiently.

To move on to a more realistic buck converter take a look at this diagram:

1015

The switch has been replaced by a MOSFET (stands for Metal-Oxide-Silicon Field Effect Transistor, just one of many types of electronic switches) and a diode. You could use a second real switching semiconductor instead of the diode, but there is no need, a simple diode will do. When MOSFET Q1 is conducting (switched 'on') the diode is reversed-biased and blocks current flow (it is 'off'). When Q1 is not conducting (switched 'off') the diode will conduct (it is 'on') because the inductor will make sure to keep its current flowing and forward-bias the diode in doing so.

The circuit now has losses: Every time the MOSFET conducts there is its internal resistance turning current into heat. The diode will drop around 0.6 - 0.7 Volt (for regular silicon diodes) every time it conducts. The inductor will have a series resistance that produces heat. Still a circuit like this can be 95% efficient (or better, depending on component choice, and switching frequency). MOSFETs and its close cousin, IGBTs, also have large currents flowing every time they are switched on or off, to charge/discharge their gate capacitance. Those add up quickly as the switching frequency goes up, and are a major source of energy loss. The other side of the coin is that inductors and capacitors can be made smaller (for equal efficiency) as the switching frequency goes up.

Interestingly that little circuit above really is just about all there's to a buck converter as far as the power components go. Real-life buck converters have little else in them. For example, the Outback MX60 has two identical buck-sections in it (in parallel). Each uses 3 MOSFETs to switch the voltage, and (as the exception to the rule above) has 3 more of the same MOSFETs instead of the diode in the above diagram. They use multiple MOSFETs to distribute the current, so cheap components can be used. You could do the same, especially nowadays, with a single MOSFET or IGBT. The only other components that aren't in the above diagram, and that are important, are a snubber-circuit over the MOSFETs made out of a small capacitor and resistor. This takes care of snubbing out the switching voltage spikes that could otherwise damage the MOSFETs. There are some electrolytic capacitors on the input to smooth out the DC input voltage, so that on-off current change due to the switching doesn't have to propagate all the way over the wires to the PV panels. There is surge protection on the input of the MX60 (a few MOVs), and two relays on the output, to disengage the electronics from the batteries. That's it. The circuit is driven at around 20kHz by a micro-controller. The bulk of the rest of the electronics is there to support the micro-controller, the power supply for it, provide driver circuits for the MOSFETs, and drive the display.

Let's see, what more is there to say...
Depending on input voltage and current either MOSFETs or IGBTs are commonly used. The latter are more beefy, and tend to have higher voltage ratings, though that line is blurring. MOSFETs can generally handle higher switching frequencies, so smaller inductors and capacitors can be used, vs. IGBTs. But that line is blurring too. For simple applications there are ICs that handle the generation of the needed switching waveforms. There are also driver ICs for the MOSFETs or IGBTs, to switch them on/off. Don't ask me for type numbers though, it's been years since I last looked into this.

Laurie, for your particular application there is a bit more complication to it: You want to do two things that this circuit doesn't. One is that the load provided to the turbine should match its MPPT profile. The second is that you need to have output current control to match the battery charging profile (and avoid overcharging them). Those add significant complication, almost certainly needing a micro-controller to take care of them (and current-sensing circuitry etc.).

In a nutshell though, this is it!
Hope this helps!

-RoB-

[Laurie Forbes]

Rob, thanks much for the description of the buck circuit. I can readily see why the output voltage is 1/2 the input (assuming a 50% duty cycle) and I think I see why the average output current is twice the average input (when the input is switched in, it supplies current to the inductor, capacitor and the load; when the input is off, the inductor and capacitor continue to supply current to the load(??)).

But, the problem is as I see it, the input is delivering current only 1/2 the time so it looks like only 1/2 of its capacity is being utilized. If then this circuit was connected to a wind mill or a solar panel capable of delivering say 100 V @ 5 A (500 W) steady state, only 250 W would be delivered on the average (100 V * 5/2) which is also = to the circuit's output if I understand it (50 V * 5 A = 250 W). This is efficient conversion but only 1/2 of the input's capacity is being utilized.

If one wanted to reduce the output voltage even further, say for charging a 12 V battery, as I understand it, a 12 V/100 V = 12% duty cycle would be required. This would mean the mill or solar panel is connected only 12% of the time so most of its potential output would be lost.

Now, if an AC transformer worked the same for DC, one could convert the entire input capacity to output as the input would remain connected 100% of the time. So then, how does one utilize a buck converter to use all the input capacity (if in fact that is possible)? Maybe a DC-DC converter would be required (convert input DC to AC, transform the AC to lower voltage/higher current, rectify & filter that AC to DC)??

Just had a though, how about putting a big C across the input? Then, when the input is disconnected, the big C could charge; when the input is connected, both the C and the input itself could supply current (more than the input could alone). That way, the input is always loaded and delivering more power.

[Rob Beckers]

Hi Laurie,

Actual buck converters do have a capacitor on their input. In part to deal with the limitations of current sources (such as PV modules, that are limited in the current they can supply), in part to isolate the source from the otherwise rather nasty switching current.

Being connected only part of the time makes no difference for the source; it's the average that counts. Especially since the switching frequency is normally very high (i.e. 20,000 times per second or more).

So, in case of a 12% duty-cycle from your example, if the source can deliver 500W @ 100V the average current would be 5A. Down to 12V that becomes about 42A. So, the inductor in circuit diagram would be carrying 42A to keep the capacitor charged and the output voltage at 12V. That 42A would also flow from the source for 12% of the time, so the average input power would be 100 * 0.12 * 42 = 500 Watt (pardon the rounding errors).

If the source was a pure voltage source (constant voltage, and capable of delivering any current) the input capacitor would not be needed. Conversion efficiency would be 100% if the components (switches, inductor, output capacitor) were ideal. A wind turbine is much closer to a voltage source, though not an ideal one, and an input capacitor is normally part of the circuit. PV modules are the opposite, they are much more like current sources and can't deliver 47A no matter for how short a time, and absolutely need an input capacitor, to make them behave much more like a voltage source as far as the buck converter is concerned. As mentioned for the MX60 in the previous post, it has a capacitor on its input (a set of them, totaling 6,600uF in fact).

Your example also illustrates the limitations of a buck converter: As the conversion ratio goes up, so does the current through the components. This quickly reaches values where components become very large, very expensive, and losses mount.

[Laurie Forbes]

Rob, I think I see what's happening with a buck converter now, especially since the input capacitor(s) are typically included. I'm still a bit hesitant however on the physics of how the current multiplication occurs. I'm thinking maybe the inductor acts in a similar capacity to coils in a transformer (acting as both the primary and secondary) but, the "primary" has the same number of turns as the "secondary" so how current current increase at the expense of voltage decrease (as per a normal stepdown transformer) still eludes me.

Anyhow, if anyone can point me to some design info these devices, I would appreciate it (haven't been able to google anything useful so far). I'm OK on the MOSFET or IGBT part of it but have little idea how big a capacitor (both input or output) or inductor would be required.

[Rob Beckers]

An inductor is not a transformer, but what it will do is preserve current, adjusting the voltage up or down as needed to make that happen. If you get a current going in a coil and then disconnect it, the inductor will instantly raise the voltage on its terminals, all to keep the current going (there are EM-field reasons why this happens, but that gets into Maxwell's equations, and to be honest I've all but forgotten those after learning them 30 years ago). The reciprocal is perhaps easier to grasp; a capacitor will try to preserve its voltage, and any attempt to suddenly change that (short it) will result in large currents.

So, an inductor can be used to connect two points with a different voltage, without resulting in a short. That's how it's used in a buck converter; at one moment it's connected to ground, the next it's connected to the input etc.

Realize though that the output voltage is strictly a matter of the duty-cycle of the switch (IGBT or MOSFET). Once you see that, it should also be clear that the output current is a matter of the load imposed on it, and that output current carries back via the inductor to the input, but only for the part of the time as determined by the switching duty-cycle. The input voltage doesn't change, it's load, how it sees what's on the output of the buck converter does change according to the duty cycle. With a 20% duty-cycle the input will only see that current 20% of the time, so while the output voltage is 1/5 of the input voltage, the input current is also 1/5 of the output current. That adds up to the same power flow on input and output (5 times the voltage, 1/5 the current in the input - 1/5 the voltage, 5 times the current on the output).

I don't have any good ideas on how to get to a working design quickly. There are lots of integrated circuits that will take care of the timing, reference voltages, current control etc., just search digikey for "buck converter" and you'll get a list of ICs. Those still take considerable engineering though to turn into a working circuit, especially since your application is not at all 'standard'. The OtherPower board has a few projects going for MPPT wind controllers; those are all buck converters, so maybe one of them would be a good starting point (or at least an easier one).

If you build something, please post back!

-RoB-

[Laurie Forbes]

Thanks again Rob, I think it is becoming clear. If I understand it, both the duty cycle and the load determine the output voltage while the load determines the output current (simple V/R). Then, if the load resistance changes, the duty cycle would have to be adjusted to maintain a given output voltage (if that is what's wanted) - the lower the load resistance, the higher the duty cycle and vice-versa.

If the input voltage changes (as with a wind mill), the duty cycle would also have to be adjusted to maintain a given output voltage(?). So, if the converter is being used to charge a battery from a wind mill, the controller would have to track the output voltage and adjust the duty cycle to compensate(?). Similarly, as the battery charges, the duty cycle would have to be reduced to maintain proper charging voltage.

Now to see if I can determine what values of inductance and capacitance to use with a mill that generates about 150VDC at about 8A max to charge a 12V battery.

[Rob Beckers]

A little Google'ing brought up a very nice little article that not only shows how components are sized, but even goes as far as calculating losses (and shows how to improve things):

http://satcom.tonnarelli.com/files/smps/SMPSBuckDesign_031809.pdf

Another direction that may be worth pursuing is to download the evaluation copy of Orcad (http://www.cadence.com/products/orcad/pages/default.aspx). This is the default electronics simulation software that everyone and their uncle uses, and I've used it a few years ago to simulate a buck-boost converter with an H-bridge. The evaluation copy has a limit on the number of components will do. Luckily it doesn't take much to do a buck converter (there are build-in 'components' that will create the square wave with variable duty-cycle to drive the switch). You can see traces, much like an oscilloscope, of anything you want. The downside is that it's a very steep learning curve to do something useful with Orcad. Again, the Internet can help, there are samples here and there.

-RoB-

[Laurie Forbes]

OK I'll take a look at both; thanks for all the assistance Rob.

[Lauri Koponen]

Greetings, I have been working on a similar project. The biggest problem I had was finding affordable parts that could handle higher current. I did find a circuit for a 5 amp buck converter. I am currently testing four 5 amp buck converter connected in parallel. The controller board allows me to manually adjust the duty cycle from 10 to 50 %. The max current limit of the circuit is 20 amps.

I have tested the circuit with a PMA using a 12 volt deep cell as a test load.
Are you still working on this project?