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In one of my circuit, the incoming variable voltage(8.5-24V DC) from the source will be first stepped up to 17V using a boost converter. The same output shall then be stepped down to different voltages for driving PMDC Motors:

  1. Motor: 15.5V, current consumption 1.25A

  2. Motor: 11V, current consumption 800ma

    Following options are available:

(1). Single boost supply followed by two buck regulators.

(2). Two boost supplies followed by their respective buck regulators.

(3). Single boost supply followed by single buck regulator that steps down to 15.5V, and then followed by the other linear regulator that steps down to 11V for driving second motor.

The voltage drop from 17V to 15.5V is unavoidable due to diode and other losses. Considering the cost and durability, which option is the best?

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  • \$\begingroup\$ Option 3 wastes a lot of power making heat in the linear reg. Try to only use linears for signal circuits; they're quite inefficient for all but the most minor currents (and voltage drops), and their precision/stability is quite unnecessary for driving a motor. \$\endgroup\$
    – Robherc KV5ROB
    Commented Mar 27, 2016 at 13:54
  • \$\begingroup\$ What about the other options? Which one to choose? \$\endgroup\$
    – Akash
    Commented Mar 27, 2016 at 14:02
  • \$\begingroup\$ Use LTM8055.... Fully integrated buck boost 8.5 Ampere rated presentable output voltage. Not cheap but excellent device in terms of efficiency and reliability. \$\endgroup\$
    – soosai steven
    Commented Mar 27, 2016 at 14:37
  • \$\begingroup\$ Generally #1 will cost less, but either 1 or 2 are completely valid options \$\endgroup\$
    – Robherc KV5ROB
    Commented Mar 27, 2016 at 14:44
  • 2
    \$\begingroup\$ Have you considered using a flyback or forward converter topology? That way you could step the input voltage up or down in a single stage and produce both supply voltages at the same time using separate secondaries. Could you just run the 11 V motor on 15.5 V? If you limit the driver PWM duty cycle to max 71% the effective motor voltage will not exceed 11 V. \$\endgroup\$
    – jms
    Commented Mar 27, 2016 at 16:28

1 Answer 1

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Using converters in "cascade" is a waste of power. A simple solution is to use a single separate converter for each motor. In this way you resolve the problem noted by @AdamHaun in the comments. As your energy supply range can be lower or greater than the desired output voltages, you need a converter with step up and step down capabilities.

Some non isolated topologies can achieve this: "Non Inverting Buck-Boost", "SEPIC" or "Inverse of SEPIC". If you are ok with negative voltages you can use also "Buck-Boost" or "Ćuk".

Another advantage of using separate converters is that you can just design one single converter (and build it twice), since the specifications are not very different between the two in your application, this saves you money and time. Even another advantage is that if one converter fails the other may operate ok.

Isolated converters like the ones suggested by @jms are desirable if you need electrical isolation. But if you don't need it, it's better to avoid those converters as their design is more complicated. Also there are power losses in the transformer.

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    \$\begingroup\$ The fact that transformer-based converters can be isolated doesn't force you to do so. Without isolation, they are neither more difficult to design nor more complex. I'd say that e.g. a SEPIC is considerably more difficult to implement than a non-isolated flyback converter and that most boost converter ICs will drive a flyback converter just fine. The only real disadvantage is the need to choose or custom design a transformer. Regarding losses, the same power losses (magnetization, winding resistance, eddy currents) are found in inductors as well. \$\endgroup\$
    – jms
    Commented Mar 27, 2016 at 17:55
  • \$\begingroup\$ Yes, transformers are subject to the same physical phenomena as inductors. In fact, transformers can be seen as inductors that share magnetic flux, so they have the same losses as inductors plus the magnetic coupling losses. So they have a tendency to more losses compared to just inductors. \$\endgroup\$
    – berto
    Commented Mar 27, 2016 at 18:31
  • \$\begingroup\$ The only effect that affects a transformer but not an inductor is leakage inductance. Leakage inductance doesn't cause a "magnetic coupling loss" per se since any energy stored in the leakage inductance isn't inherently lost and can be recovered. A flyback tends to be less efficient than a SEPIC at low conversion ratios for a whole bunch of reasons, but "magnetic coupling losses" are not the deciding factor, with switching losses (in the transistor and diode) being vastly more important. \$\endgroup\$
    – jms
    Commented Mar 27, 2016 at 20:12
  • \$\begingroup\$ Your link does not work without subscription. Is it the same as this one? "Component count is similar for the two designs, but the flyback has the disadvantage of requiring snubbers." I believe @Akash has got all the help he needed in the first place. \$\endgroup\$
    – berto
    Commented Mar 27, 2016 at 22:26
  • \$\begingroup\$ It's the same article. You forgot that you would need two SEPICs in this application VS a single flyback converter, it was the entire point of my recommendation. Engineering is about tradeoffs, and efficency is a consideration, not the consideration. How much weight 3% reduced efficiency and slightly worse regulation carries compared to having twice the board area and a higher component cost is up to OP to decide, not you. \$\endgroup\$
    – jms
    Commented Mar 28, 2016 at 5:20

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