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What's the difference between a DC DC Switching Regulator, and a Linear Regulator. I'm buying one to make a variable power supply (is this the best way to do it?) and I need to just output an adjustable DC voltage. I need an output voltage of 1V to 50V at up to 10A. I have an input voltage of either 12V or 50V at 20A.

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    \$\begingroup\$ Which one to use depends on the load you intend to drive, amongst other things. \$\endgroup\$ – Peter Smith Apr 23 '16 at 13:14
  • \$\begingroup\$ @Jawad: Perhaps I was not clear enough. What source voltage do you intend to use, what output voltage range and current (max) do you require? The tradeoff between switchmode supplies and linear devices depends on this to a great extent. Note it is easier (in general) to adjust the output voltage of a linear regulator. \$\endgroup\$ – Peter Smith Apr 23 '16 at 13:28
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    \$\begingroup\$ You should put this in the question. Just edit it and put your specs in the body of the question. How to be nice on SE is to make clear and complete questions that help you, us, and future visitors! \$\endgroup\$ – user65586 Apr 23 '16 at 13:52
  • \$\begingroup\$ Where are my comments? \$\endgroup\$ – Jawad Walker Apr 23 '16 at 14:01
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The key difference between a 'switching DC DC regulator' and a 'Linear Regulator' is efficiency. With a linear regulator the output voltage must be less than (or equal to) the input voltage and the minimum possible power loss is \$ (V_{in} - V_{out}) \cdot I_{out}\$ This power loss causes heat and may require heat-sinking to get rid of the excess power.

A switching regulator however can in principle produce either voltages that are lower than the input or higher than then input, or equal to the input. They are also usually more efficient and will be significantly more efficient if there is a big difference between the input and the output or if the load current is large. The down side is they have more complicated control.

There are several different but the most popular ones are the Buck converter, the Boost converter and the Buck-boost.

The Buck converter can only produce output voltages that are less than the input and the basic circuit is as follows.

schematic

simulate this circuit – Schematic created using CircuitLab

The Buck converter uses an inductor to store energy when the switch is on current ramps up in the inductor and goes into the capacitor. When you open the switch this current still wants flow so circulates through the diode initially and the current decays. The current may continue to circulate in the diode until the switch is turned on again CCM (continuous conduction mode) or be allowed to decay to zero DCM (discontinuous conduction mode). The purpose of the capacitor is to smooth out the inductor current to provide a DC supply. Because there is only ever very small voltages across the switch and diode power loss is kept to a minimum.

A Boost converter is used when we require more output voltage than input and its basic circuit is as follows.

schematic

simulate this circuit

The Boost converter also uses an inductor to store energy and a capacitor to smooth the current. When the switch is on current builds up in the inductor and when the switch is opened this current has to go into the output capacitor charging the output above the input voltage. This can also be used in CCM or DCM depending on whether the current is allowed to decay to zero or not.

The Buck-boost is capable of operating in Buck mode or Boost mode and its basic circuit is.

schematic

simulate this circuit

In practice the switches for any of these converters are implemented using transistors and there are lots of control ICs available to help you control the exact switching times.

These are not the only types and there are many others. Which is best for you? If you ever need more output voltage than input a switcher is the only way to go. If your output is always less than your input a Linear regulator may be OK if your power requirements are small enough but at higher powers you would want a switcher.

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I think you will find it difficult (if not impossible) to find a single IC type solution for this task.

I would probably use a switchmode tracking pre-regulator followed by a linear output stage; such an approach may be found in this journal (scroll down to page 12).

Although that particular setup does not meet your specific requirements, they could be met by using this technique, but that requires a significant effort in the design, not least in choosing appropriate parts.

I do not know your skill level, so it is impossible for me to comment on how hard that would be for you (if you chose to go that way).

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Linear regulator will dissipate a lot more when the input voltage and the output voltage has significant difference or if your load is high.

For example if your input voltage is 9V and your desired output is 5V with an 1A load, then a linear regulator will dissipate:

$$ P_{d} = (V_{in}-V_{out}) \times I_{load} $$ $$ P_{d} = (9V-5V) \times 1A = 4W $$

This means heat and usually a general linear regulator needs additional heat sink to deal with it.

The second problem is efficiency, which in this case would be:

$$ \eta = \frac{P_{out}}{P_{in}} $$ $$ \eta = \frac{5V \times 1A}{9V \times 1A} = 55.56 \% $$

This isn't very good, it almost wastes half of the energy. But in certain circumstances linear regulators are much preferable.

A linear regulator is usually used when the \$ I_{load} \$ is low or if \$ V_{in} - V_{out} \$ is low. They usually require only an input and an output filter capacitor so the BOM is quite small.

On the other hand, DC-DC switching regulators have around 80% efficiency on average. There are models near 94%. Most models can handle great input output differences without heat sink but switching regulators require more external components like inductors, filter capacitors and diodes.

Now, given your 50V input voltage with a desired 1-50V output voltage with loads up to 10A, both \$ V_{in} - V_{out} \$ and \$ I_{load} \$ are high so you would definitely need large heat sink, and you would waste a lot of power. In this case a switching regulator is recommended.

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You will need at least 500W of input capacity to be able to get 10A at 50V out.

Working at 10A to 20A is begging for a switching regulator in the modern era. There may be some noise, cost or complexity issues that would make it less beneficial but the heat sinks on any linear design that must dump 49V at 10A should cost more than any reasonable 20A 1V to 50V DC/DC switching regulator.

Unless you need multiples you will save time, money and frustration if you locate an off the shelf part.

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