Diy brushless esc - Page 2 - RC Groups
Nov 11, 2004, 09:18 PM
Registered User
Here's another circuit layout. Take a look and let me know what you think.

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 Nov 12, 2004, 02:38 AM "Simplify, then add lightness" There are several problems with using mosfets like you show. The first thing is that driving directly from the logic that way, your motor voltage could be no higher than your logic voltage, or the top mosfets would never turn off. Also, mosfets have very high input capacitance, so at a bare minimum, you would need a series resistor to limit the current from the logic to prevent damage to the logic and you would have slow switching times on the mosfets which would cause them to heat up more. Also, you generally would use nmos for the bottom mosfets. The way you have it shown with pmos on the bottom in the common drain configuration, the bottom mosfets would never fully turn on. To turn them on fully you would need to drive the gates below ground.
 Nov 12, 2004, 08:22 PM Registered User To tell the truth, I'm kind of walking in the dark on how you're supposed to wire the fets. Can you help me with this? Any examples would be greatly appreciated.
Nov 14, 2004, 09:40 PM
Registered User
Here's a correct picture of a p-fet. Thanks for pointing out that the fets in my picture were connected wrong.
A few questions:
Does the gate voltage have to be at least as high as the source voltage? For example, if the source voltage is 9v and the gate is 5v, then will the output be ~9v or ~5v?

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 Nov 14, 2004, 10:58 PM "Simplify, then add lightness" For a P-channel mosfet, the source has to be positive and the drain negative. To turn on, the gate has to be negative relative to the source. In the specs for the mosfet, there will be a threshold voltage. When the gate is more than the threshold voltage below the source, the mosfet will turn on, if it is closer to the source than the threshold voltage, the mosfet will turn off. For the low voltage mosfets generally used, the threshold voltage will probably be less than 4 volts. Therefore if the source is at 9 volts, and the gate is at 5 volts, the mosfet will be turned on, and the drain will try to pull up to 9 volts, so with 5 volt logic the high side mosfet will never turn off. With the P-channel on the bottom, the drain will be tied hard to ground and the source will be free to move. This is the common drain configuration which is equivalent to the emitter follower with bipolar transistors. With this configuration, the source will try to stay above the gate by the level of the threshold voltage. If you pull the gate to ground, the source will not go lower than the threshold voltage. Normally you would use an N-channel on the bottom. On the top, you would use either a P-channel, or because N-channels are usually cheaper and better, another N-channel on the top in the common drain configuration with a charge pump driver so the gate voltage would go above the motor voltage. Either way, you need a high voltage driver for the top.
 Nov 15, 2004, 05:24 PM Registered User Thanks for the explaination jeff, it makes more sense now.
Nov 15, 2004, 11:59 PM
Senior Member
Use logic-level MOSFET's and now you are all set to use logic-level control inputs to your N and P channel MOSFET's.

-Matt

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 Nov 16, 2004, 10:09 AM "Simplify, then add lightness" One thing you need to watch for with the circuit Matt posted, is the the maximum Vgs spec for your P-channel mosfets. Vgs is the voltage differential between the gate and the source, and the mosfet will be damaged if you exceed the max Vgs. On many logic level mosfets, max Vgs is only 12 volts, and on some it is as low as 8 volts. Since the npn transistor pulls the gate to ground, the full motor voltage is applied across the gate and source of the P-channel mosfet, so supply voltages higher than max Vgs will damage the P-channel mosfets. It is even worse than that, because the current stored in the windings of the motor will cause a voltage spike when the mosfet is opened as the motor is being commutated. This is recirculated thru the power supply by reverse protection diodes built into the mosfets. That is why you need a very good capacitor with a low ESR(equivalent series resistance) across the power supply right at the mosfets. Section 2.3 of this app note describes why the capacitor is needed, and how to choose it. http://www.st.com/stonline/books/pdf/docs/9214.pdf Jeff
 Nov 17, 2004, 06:17 PM Registered User Zagisrule: thanks for diagram, I appreciate it. Jeff: I didn't know about the voltage spikes, and I'll include capacitors in my next diagram. I should be able to easily get fets that can handle 10 amps, which should be plenty for the small motor on my plane. I'll post a revised diagram sometime in the next day or two. Working on a spanish project for school tonight. Adiós!
Nov 17, 2004, 10:12 PM
Registered User
Well, spanish didn't take as long as I thought it would.
Heres another diagram. If I messed up, let me know. The fets should be wired correctly in this one.

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 Nov 18, 2004, 12:18 AM Senior Member I have no clue as to whether or not this logic-driven circuit would function at all? I am not at all familiar with digital logic devices, my digital experience has been with PIC's and with those I make them do exactly what I want. I guess it would not be hard to try this out, it is quite simple and it would be neat if it worked! -Matt
 Nov 18, 2004, 12:26 AM Senior Member While I am thinking about it... 1. Why bother with switching the rail anyway? Just wire the center tap of the motor to Vcc (the wye connection to the + rail) and switch only the ground to each of the motor leads to accomplish commutation. This means you can get rid of all the P-channels and their inverting transistor drivers. 2. Can you explain the thought behind your digital logic switching? I am thinking that it is not going to work. I have started writing sensorless control code (a failure as of yet...can't get rotor lock) for 3-phase motors and frankly I just do not see the logic solution working unless I am missing something on what exactly it is doing. From the looks of it you are just selectively energizing the coils based on the hall inputs. -Matt
Nov 18, 2004, 05:51 PM
Registered User
This'll be a long post.
To start off with, I left capacitors off in the diagram because my circuit design program is a demo and it doesn't want to put more than 50 components in one picture.

Matt, your question is a good one, I thought of the same thing when I first started this.
Short answer: Your way energizes 1 coil at a time. The standard way energizes 2 at a time, giving greater power.

Long answer: Brushless Motors are wired so that one coil will attract and one will repel when a phase is energized. If you connect the wye connection to ground, then all coils can only be energized in the same direction (+ to -). If you energized 2 coils in the way you specified, then they would be fighting against each other. This makes it so that only one coil can be energized at a time. If the coils are wired in the standard way, then 2 coils will be energized that will help each other; thus making it more powerful. This may not be too clear, so I hope the diagrams help. The first is with the wye connection connected to ground and one phase energized. The second is an example of a standard motor with one phase energized.

As for your second question, that is exactly what this circuit does. Coils are energized based on the position of the rotor (which is detected by the hall sensors). If the controller were to be connected directly to the battery, it would spin the motor as fast as it could. A normal speed controller needs to be connected to this controller's power wires to regulate the voltage. Therefore, the controller will still spin the motor as fast as it can, but this speed is based on the voltage applied to the coils. If the imput voltage is dropped, then the speed is cut also.