I'm designing a power supply for a small robot. I will use a 6V NiMH battery pack and I want to deliver 9V at 2A to drive the motors. I will use a step-up regulator to reach 9V but I don't know how to deliver 2A. Can anyone give me some hints about what should I study to project this circuit?

  • \$\begingroup\$ Hey, I am confused about the use of a transformer/step-up regulator here. When we step up either the voltage or the current in a circuit, the other decreases (Power P=VI), so if your initial circuit does not support 2A current at 6V, then I don't think that using a boost convertor would help. If it is not the case, then I see no reason why it cannot be done. \$\endgroup\$ – Anshul Nov 19 '12 at 17:31

What you would be looking for is a boost regulator, e.g. TI's LMR61428 or SemTech's SC4501.

An easy way to identify potential parts that meet your requirement is to search for "boost regulator" on a site like DigiKey, then apply filters in the parametric search for the characteristics you require.

Also useful is some exploration of online tools for power management design, such as TI's Simple Switcher tools. Each manufacturer who offers such a tool, would propose parts and designs built around their respective product ranges, so it would behoove you to use several such tools to understand the types and prices of parts involved, before making a decision.

This exploratory exercise will also help gain some understanding of the simplest ways to implement a boost regulator for your project, if needed. Many of the free design tools go all the way through to providing reference schematics based on your inputs, plus costing for the bill-of-materials involved, and PCB area budget.

  • \$\begingroup\$ Thank you for the help. I found a CI that can deliver 2A. This tools will really help me to find the CIs that I need. \$\endgroup\$ – Eduardo Nov 19 '12 at 22:55

This isn't a direct answer to your question, but I think addresses the actual problem.

What you are doing sounds backwards. The motors are presumably by far the highest power devices in the robot. It therefore makes sense to tailor the battery back to them and convert the voltage as needed for the other purposes. In this case, arrange the battery pack to put out 9V or more. The step up regulator you propose will lose some power and will be fairly big. All this can be avoided by running the motors off the battery pack directly, then using much smaller buck regulators to power the lower voltage electronics.

Most likely you already have a motor controller in there that can effectively throttle back the motors as needed to attain the required position or speed. If that is the case, then the battery pack can be safely made to be a bit higher than the maximum effective voltage you are allowed to deliver to the motors. The controller will automatically lower the PWM duty cycle to compensate. For example, you might use a nominal 12V battery pack so that you still have the minimum 9V at the end of battery life. The buck regulators making 5V or 3.3V for the electronics will do that easily over a range of 9-12V.

  • \$\begingroup\$ I understand that I will lose power raising the voltage, but I think I will save space, since I would have to add more battery cells and I would have to do a circuit to connect the motors to the battery anyway, right? I found a CI that can deliver 2A. Thank you for the attention and I learned a lot with your observations. \$\endgroup\$ – Eduardo Nov 19 '12 at 23:01
  • \$\begingroup\$ The point of higher voltage battery pack is that don't need a converter circuit between the battery and the motors. The motor controller would be powered directly from the battery. I don't know what a "CI" is, but keep in mind that if you want 2A at 9V and you are starting from 6V, it would draw 3A from the batteries without accounting for conversion losses. With losses it will be closer to 4A. \$\endgroup\$ – Olin Lathrop Nov 20 '12 at 13:34
  • \$\begingroup\$ Doesn't I lose speed control connecting motor drivers directly to the battery due to "fluctuation" (I don't know if this is the correct the term, I mean: "not exactly constant")? Sorry, "CI" is integrated circuit (IC) in portuguese. I made a mess with the acronym. \$\endgroup\$ – Eduardo Nov 20 '12 at 16:15
  • \$\begingroup\$ @Eduardo: Presumably the motor is inside a control loop that is trying to regulate its position, speed, torque, or whatever. The controller will automatically draw more current at lower input voltages to compensate. \$\endgroup\$ – Olin Lathrop Nov 20 '12 at 18:07

NiMH comes in 1.2V cells, so if you must use 9V motors, then you ought to choose 8 cell batteries and factor aging effects on capacity when considering size.

Here is one source of 9.6V battery packs.

Words of advice.

  • Note many are not recommended for motors.
  • This has to do with the battery's ESR and current drain specs.
  • Get the best you can afford for low ESR or max current rating.
  • i.e. if your load is 2A then battery ought to be spec'd at 10A or more.
  • Caution: Batteries cells must be extremely well matched to be packed in arrays. (<1%)
  • 1
    \$\begingroup\$ A common recommendation I've seen for small robot powerpacks is to cannibalize cordless powertools of the appropriate size because even after throwing away the chassis and tool head the manufacturer of a home depot drill's volume discounts mean it's still cheaper than buying separate batteries, chargers, and motors from a hobbyist supplier. \$\endgroup\$ – Dan Neely Nov 19 '12 at 18:09

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