1
\$\begingroup\$

I have a question about switching step-down regulators. (As I stated in my previous questions, please consider the fact that that I'm not very expert, so feel free to reply/talk as if I were a student.)

Let's take a practical example of a switching step-down regulator, based on this IC. (I've seen that is largely used and common in various circuits):

We need to feed a device which needs 12V with a power consumption of 200mA. Ok: We'll take a buck converter circuit, and as Vin we'll provide, for example, a voltage of 30V from a batteries pack with a total capacity of 2000mAh, then we will set the Vout of the buck converter to 12V. But If we want to make use of a less number of batteries we can also go with a Vin of 20 or less volt: I have read that for the lm2596 IC, the Vin, should be at least greater of 1,5V than the Vout.

I was thinking: If I reduce 30V (from a batteries pack) to 12V, the difference of 18V could be reason of an higher power consumption from the batteries? Am I right? Eg I know that linear regulators (differently from switching regulators) have a bad efficiency because some of the power will be lost as heat. But what about switching regulators? Some days ago, by a search on Google, I've read of a person which had the needs to get 5V using a Buck converter: someone told him that would be better get the 5V from a Vin of 18V instead of using a Vin of 12V.

So, taking again in consideration my example: when using a switching regulator, is better to start from an higher Vin, for obtaining a same Vout? Why?

I'd also like to see some charts of the switching regulators.

\$\endgroup\$
5
\$\begingroup\$

TI's got a tool, named WEBENCH which can make a lot of charts and calcualtions for you. Here is its output with your parameters in pdf.

Let me highlight the one about the efficiency. The simulations shows that this IC has a better efficiency when Vin is 20V, but this difference is not that much.

enter image description here

It is not just the Vin that matters, if you change the supplied current from 200mA to 3A a different efficiency chart will be shown. In this case the Vin = 30V is the better choice.

enter image description here

Usually, there are similar charts in the datasheets if tools like this are not available.

If you only need 200mA, you should choose converter which is capable of, let's say 300mA maximum current rather than 3A, efficiency is better near the max current. Another converter, which can drive max 300mA, LMR14203's efficiency chart:

enter image description here

It is again the worst at 30V, but it is around 88% while with the LM2596 it is 79% which is a significant difference. On 20V it is above 90% which is pretty good.

\$\endgroup\$
2
\$\begingroup\$

To achieve maximum efficiency, we need to understand where losses can exist, and what measures are available.

I am going to use a more generic circuit as the principles apply everywhere; some circuits offer the freedom to change some parameters to maximise efficiency in a given application and others do not.

To show that, here is a circuit that exposes the power path properly:

Synchronous SMPS Buck

I have highlighted the primary high current paths in red; Q1, Q2/Q3, L1 and D2 and the current sense resistor. Note that the gate drives may have significant current depending on application.

The losses in Q1 are primarily resistive and capacitive, in Q2/Q3 resistive and resistive in the inductor. There is a current sense resistor in this scheme which dissipates some power, obviously.

There are (as always) trade-offs.

For the main switch (Q1), the resistive losses are: \$\frac {Vout} {Vin} (Imax)^2 (1+δ)R_ds(on)\$ where \$\delta\$ is the temperature dependency of \$R_ds(on)\$

The capacitive losses for the main switch is given by: \$k(Vin)^2 (Imax)(Crss)(f)\$

So the resistive losses increase with lower duty cycles which is reasonable as the main switch is on for a longer proportion of the time as Vout and Vin approach each other.

Contrast this to the capacitive term which is directly proportional to frequency. (k is a constant related to the inverse of gate drive current).

There is actually a crossover point; at lower Vin, lower switch resistance is desirable, but at higher input voltages lower total gate charge can be preferable.

I can minimise the inductor size (which minimises windings and therefore DC resistance) by increasing the switching frequency, but this will increase the capacitive losses in Q1.

The losses in Q2 and Q3 are completely due to \$ R_ds(on)\$: specifically

\$ P_(sync) = \frac {V_in - V_out} {V_in} (I_max)^2 (1+\delta)R_ds(on)\$

This shows that at lower duty cycles (higher Vin), the losses increase.

So we like lower duty cycles (higher Vin) for the main switch, but we like lower Vin (lower duty cycle) for the synchronous switch; that said, vast strides have been made in recent years in terms of MOSFET on resistance - see for example, the IRF6718L2 - a very impressive \$1m\Omega\$ at 4.5V \$V_gs\$

Note D1 and D2 - these should be sized for minimum forward voltage at an appropriate current to minimise other losses.

This is an enormous subject (that does not necessarily get enough attention), but with the proper attention, the optimal efficiency for a given application can be achieved.

\$\endgroup\$
0
\$\begingroup\$

@BenceKaulics answer about efficiency is good but doesn't really answer the original question, as I see it.

The question as I read it, is given a buck-converter with an out of 12V @ 200ma consumer by the target device, does the input current to the regulator depend on the input voltage?

A switching regulator roughly maintains power from input to output, less power required by the regulator and other losses, expressed as efficiency.

Your target device is consuming 2.4 watts (12 x 0.2). Therefore, the switching regulator will consume from its power source a bit more than 2.4 watts. If the input is 30 volts, it will consume a bit more than 0.08 amps (2.4 / 30). On the other hand, with 20 volts input, it will consume a bit more than 0.12 amps (2.4 / 20). Both of these figures represent 2.4 watts.

If you look at the curves in the previous posting, you'll see that there is a small change in efficiency (maybe 78% vs. 80%) but this dwarfed by the change in current draw due to changing the voltage but maintaining the power.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.