I recently bought a couple of buck converters off eBay for £1.16 each with free delivery from China. The one's I bought were these buck converters that I will be using to regulate a couple of power supplies for different jobs.

One of the buck converters from China. |

After realising that not everybody knows what a buck converter is I thought I'd give a quick description here.

The buck converter is a DC-DC step down regulator that I find preferable to linear regulators such as the LM38, adjustable +ve voltage regulator or a 7805 regulator for projects where low power use is required.

A regular linear regulator wastes excess power by dissipating any excess power as heat, especially when stepping down from 12V to 5V at 2A. If anyone wants to do the math to work out the wattage wasted as heat go ahead.

A buck converter on the other hand can be up to 95% efficient at stepping down the voltage. How it works, from Wikipedia.org

Simply put, It works by taking the input DC voltage and converts it to a high frequency AC voltage. As the AC voltage rises on each cycle once the voltage hits the required level the input supply is momentarily switched off until the input level drops back to the required level giving a choppy AC output, this switching happens up to one hundred thousand times every second (100KHz) with inductors and capacitors providing the power whilst the input is switched off. This choppy AC voltage is then converted back to DC at the required voltage.The buck converter is best understood in terms of an inductor's "reluctance" to allow a change in current. Beginning with the switch open (in the "off" position), the current in the circuit is 0. When the switch is first closed, the current will begin to increase, but the inductor doesn't want it to change from 0, so it will attempt to fight the increase by dropping a voltage. This voltage drop counteracts the voltage of the source and therefore reduces the net voltage across the load. Over time, the inductor will allow the current to increase slowly by decreasing the voltage it drops and therefore increasing the net voltage seen by the load. During this time, the inductor is storing energy in the form of a magnetic field.If the switch is opened before the inductor has fully charged (i.e., before it has allowed all of the current to pass through by reducing its own voltage drop to 0), then there will always be a voltage drop across it, so the net voltage seen by the load will always be less than the input voltage source.When the switch is opened again, the voltage source will be removed from the circuit, so the current will try to drop. Again, the inductor will try to fight against it changing, which it does by reversing the direction of its voltage and acting like a voltage source. Put another way, there is a certain current flowing through the load due to the input voltage source: in order to maintain this current when the input source is removed, the inductor will have to take the place of the voltage source and provide the same net voltage to the load. Over time, the inductor will allow the current to decrease gradually, which it does by decreasing the voltage across itself. During this time, the inductor is discharging its stored energy into the rest of the circuitIf the switch is closed again before the inductor fully discharges, the load will always see a non-zero voltage. The capacitor placed in parallel with the load helps to smooth out voltage waveform as the inductor charges and discharges in each cycle.

## No comments:

## Post a Comment