This is probably a dumb question, but hey, I'm a newbie. :)
I have a motor controller that requires a logic voltage of no more than
5.5V (but is content with as little as 2.5V). But my motors are happy with 6V, and I was planning to run them off of 4 AA alkaline cells, which total about 6.5V when fresh. Unfortunately this means I can't safely connect the same battery pack directly to the motor controller.
However, I could pretty easily set up a voltage divider (just using a couple of resistors) to cut the voltage to the the controller by, say,
20%. That should put me safely within the range the controller wants. But is it safe, reasonable, and kosher to set things up this way? Or is it going to cause problems, for example, the resistors growing hot or wasting power?
Also, if this is a reasonable thing to do, does it matter what size resistors I use? The division ratio depends only on the ratio of their resistance -- but does their absolute value also matter, or do I just pick whatever's convenient?
I've got some 7085 voltage regulators on order, but since the motor controller accepts such a wide range of input voltage, I'm wondering why I shouldn't just use a couple of resistors instead.
The actual value of the resistor will make a difference. For example is you use a 1K resistor on the high side of the divider you'll be able to deliver max 6ma at 6volts. If you have a 10k you'll only have 0.6ma.
What ever values you use you'll be draining your battery for no reason. Use a voltage regulator or 2 battery packs one of the motor, the other for the controller.
Regards Ian Dobson
Home of the Atmel based UDP mobile web cam
mails scanned with av-filter.pl (F-Prot / perl)
"Joe Strout" schrieb im Newsbeitrag news: email@example.com...
Thanks, that's exactly the sort of insight I was looking for. The educational resources I've found are great at explaining how to make a voltage divider, but utterly failed to point out these realities of actually using them.
that's slightly off. if you have a 1k resistor on the high side it will only deliver 6ma at _0v_, i.e. a dead short. suppose your microcontroller needs between 0ma 50ma at 5v and a 500mv output voltage drop is acceptable under that load. (and that is a pretty huge margin for a power supply) you need a supply with an impedance of 500mv/50ma = 10 ohms. if you have a voltage divider with r1 to 6v and r2 to gnd, you need 1/(1/r1+1/r2) < 10. r1=12, r2=60 gets you that but burns 83 milliamps continuously. and like i said, 50ma is a tiny power supply and 10% is awful. that's why nobody uses voltage dividers for power supplies unless the current in question is really, really tiny.
you want a linear regulator. be careful with 7805s though -- they have pretty awful dropout, on the order of 2 volts or something. you can get LDO regulators with dropout of like 0.5. that is important because your battery voltage isn't much higher than your operating voltage.
using two batteries is sort of silly and i would advise against it.
reason.Thanks, that's exactly the sort of insight I was looking for. The
Thanks, that's a good explanation -- I need to let it sink in again and work through the details, but in broad strokes I think I get it.
True. With the voltage regulator, I was planning on using a separate 9V battery. That's a bit of a nuisance, but not too terrible since I would expect that battery (unlike the motor batts) to last for months even under heavy use.
What does a voltage regulator do when the input voltage drops too low? Does the output voltage simply drop in proportion?
Maybe I could find a 3.3V LDO -- that would give me a fair amount of wiggle room.
Well, I do like the idea of a single set of batteries, mainly because it enables me to have a single on/off switch. But I've already gotten used to having separate batteries on my servo-based robots, where the servos draw so much current (and are so noisy) that they frequently cause the servo controller to reset if driven from shared batts.
I haven't heard of that being a problem with this motor controller, but it might be one of those things I learn about when my robot suddenly freezes. :)
But I have heard that many builders try to keep two separate power busses, a high-current, noisy one for motors (and certain active sensors) and a lower-current, clean one for electronics and control signals. That seems sensible to me... but how would you achieve it with a single set of batteries? Would a voltage regulator on the "clean" line be all that's needed?
A linear regulator is a better choice than a voltage divider.
A LDO regulator is a better choice than a non LDO regulator in this application. (LDO regulators cost more, but not enough to matter.)
There is no free lunch on linear regulators. The key equation is:
Power = Voltage * Current or P = IV
If you have a device that wants to eat 20mA at 3.3V, then the device is chewing up .020A * 3.3V = .066 Watt or 66mWatt. However if the battery is is 9V volts, the power that the voltage regulators is (9V - 3.3V) * .020A = .114W = 114mWatt. Linear regulators are nice and easy, but they can eat up some power.
A switching supply, can be used to drop deliver more battery power to your robot electronics, at additional cost and complexity.
Separate battery packs are reasonable for hobbyist grade robots. A single battery pack can be used with careful engineering, but most of us would rather spend our valuable enginnering time on other aspects of the robots.
Separate battery packs can introduce their own set of problems such as ground loops. I typically design my electronics to be opto isolated these days so that the "power" battery is completely electronically isolated from the "logic" battery. Spurious resets are a thing of the past for me.
You don't need a voltage divider with two resistors, you only need one resistor. Your motor controller acts as the other resistor and you get the same voltage dividing effect.
The problem with using resistors is that the amount of voltage drop changes based on how much current is being drawn (E=I*R). If your controller has a stable enough current demand it might actually work ok. You would have to determine the maximum and minimum current draw and see if you can find a resistor value (using the simple E=IR formula) that would keep the voltage in the range needed.
If the current dropped to zero at any point however, the voltage drop across the resistor would be zero which means the full 6.5 V would be applied to the microcontroller and could damage it. Dangerous stuff.
This of course is why you would add a second resistor in parallel with the controller - to guarantee a minimal current flow which guarantees enough voltage drop across the first series resistor so the voltage never got over the maximum. Any current flowing through that second resistor, would just be wasted energy draining your battery. But if the current is low enough compared to say what the motors are draining, it might be a reasonable solution.
Now, once you have two loads in your circuit (both the controller and the extra load resistor in parallel with the controller), then the current through the series resistor is the sum of the currents through each load. This means the voltage drop across the series resistor won't change as much the current draw from the controller changes - which means the voltage will be more stable.
So that's two reasons to use the two resistors in a divider circuit as you first thought about. It guarantees the voltage to the controller will always be less than what it is when the controller is not connected (or is drawing 0 ma or current) - and it means the voltage to the controller will change less as the current draw on the controller changes.
You also asked about the resistors heating up.
All resistors heat up when they are working. That's what resistors are - electrical heaters. At normal current levels the heat is so low we don't even notice it. Resistors are rated based on how much heat they can take without damage. It's their wattage rating. As long as the resistor is operated below it's rating, everything is fine. To calculate how much power the resistor is consuming, just multiply the voltage across the resistor, by the current going through it. W=EI. (or W=I^2 * R) If the max current through your resistor is 500 ma and the max voltage drop is 5 volts, then that makes for 5 * .5 or 2.5 watts. A typical 1/4 watt resistor would not be big enough in this case. You would need a special power resistor rated for something greater than 2.5 watts. And it would get warm if it had to deal with that high of a current and voltage. But if the max load was 2 volts, and 50 ma, then the watts would be .1 and a small
1/4 watt (.25 watts) would work just fine).
Another option other than a real regulator is to use a zener diode in place of the second resistor in the voltage divider. This is how you make a cheap regulator. A zener diode conducts when the voltage across it goes over it's voltage rating. So if you use a 5 volt zener diode, the voltage across it will never go over 5 volts (approximately). This means that if the controller is not drawing enough current to make the voltage drop to the needed 5V, the zener diode will draw the current instead. But unlike the second resistor, when the controller was drawing enough current and the voltage dropped below 5V, the zener diode would stop conducting and stop wasting your battery power.
If the currents are low enough, the zener diode solution might be cheaper and easier than a real regulator - but it's only used in low power applications because it does regulate by wasting power.
When you use a real regulator, it works the same way as the series resistor does - it drops the voltage by acting as a resistor. The only difference, is that it automatically changes it's resistance as required to keep the voltage output constant. So, your regulator is going to be producing just as much heat, and wasting just as much battery power, as the series resistor was wasting. So it's no better or worse in that regard. It only saves energy by not needing that second resistor (or zener diode) - that's where those other two simpler systems would waste extra power.
I've not had enough experience with these circuits to know the practical issues with that, but you should be able to solve that with the correct application of capacitors - but maybe the needed capacitors to solve the problem are so large that it's easier to just use separate batteries?
Yeah, it's also nice that the brains of the bot keep functioning even when the power to the servos/motors starts to die because they tend to behave better that way (and sensors and indicator lights can keep working etc).
Yeah. You can just have one buss which is the raw battery buss which is connected to the input of your regulator as well as the servos and/or motors and the output of the regulator is the clean buss for the electronics. You need capacitors on both sides of most regulators (across the power) and you can add additional capacitors at different locations on the "clean" buss to keep it clean (high speed digital electronics can generate a lot of high frequency noise - and the capacitors act to filter that out).
It's typical to use a large electrolytic cap on the input and output of a regulator and to put a small non-electrolytic in parallel on the output. Electrolytic don't act like capacitors at high frequencies so that's why the small cap is also needed. Without the small cap, the output could oscillate at a high frequency.
If you experiment with the voltage divider idea, you should use similar caps as well in the circuit to reduce noise effects.