Freewheeling diodes in series - Page 1
Forum section: electronics / beginners

author: fullmoon6661
author_rank: Newbie
timestamp: March 14, 2025, 09:15:36 pm
content: Can I safely use two diodes in series for the flyback path so that my indicator LED could turn on when the motor is freewheeling? The motor draws about 15A when running.EDIT 1:The MOSFET is driven by a simple ON or OFF condition (not PWM or anything like that) by pulling the Gate down like a switch. So the speed is not regulated.The freewheeling LED is just for a fancy visualization to tell the user that there was a flyback current that was successfully dissipated. (There are already also other LED's at the VMOTOR, Gate, and supply power (not visible).)I'm replacing a relay that was originally in the system, which always had its contactors damaged over time, unsure due to the inrush or the flyback arcing.

author: thm_w
author_rank: Super Contributor
timestamp: March 14, 2025, 10:02:48 pm
content: You can try, two diodes probably not enough though only 1.4V. You could use a TVS or more diodes in series.https://www.circuitbread.com/ee-faq/the-forward-voltages-of-different-ledsBut 15A is going to be a beast motor and might draw a lot of current through those diodes, resulting in hard braking not freewheeling.

author: thm_w
author_rank: Super Contributor
timestamp: March 14, 2025, 11:13:45 pm
content: Quote from: fullmoon6661 on March 14, 2025, 10:28:20 pmTurn a motor by hand, with nothing connected to it.Now short the motor out with a wire or diode, turn it again, see it if feels different or not.https://forum.arduino.cc/t/understanding-dc-motor-freewheeling/160654

author: SteveThackery
author_rank: Super Contributor
timestamp: March 15, 2025, 11:49:01 am
content: I might be wrong about this, but.....Delete the two flywheel diodes and change R8 to 1kohm. When the drive transistor is switched off, the motor will freewheel and generate 12V DC with the opposite polarity from the supply, illuminating the freewheel diode.This should then all work, with only the tiniest braking effect from the 10mA or so in the freewheel diode.Won't it?

author: SteveThackery
author_rank: Super Contributor
timestamp: March 16, 2025, 10:08:38 am
content: Quote from: SteveThackery on March 15, 2025, 11:49:01 amOh, wait, I think I am wrong. What I described should work if the load is an inductor. An inductor tries to keep current flowing in it when the drive is removed, and as such develops a reverse polarity voltage across its terminals.But a motor is not an inductor and doesn't behave like one. Unlike an inductor, a motor generates a continuous back EMF when it is running, and this back EMF has the same polarity as the driving voltage. To illustrate more clearly: looking at the OP's original schematic, but substituting a DC motor for L1, when the transistor switches off the motor will freewheel and work like a generator,with the top of the motor at +12V with respect to ground. As the motor slows down, the voltage at the top of the motor reduces, eventually reaching zero when the motor stops.But importantly, the network of two LEDs and two resistorsdoes not see a polarity reversalwhen the motor is freewheeling.The fundamental mistake in the original post is using an inductor to represent the motor.That misled us all, including the OP. The Spice model will produce the wrong results.

author: mtwieg
author_rank: Frequent Contributor
timestamp: March 16, 2025, 02:39:02 pm
content: Quote from: thm_w on March 14, 2025, 11:13:45 pmI would never say that freewheeling diodes create a braking effect (unless an external torque is trying to make the motor spin backwards, that is).When the power MOSFET is switched off, the freewheeling diode will conduct due to the motor inductance. During this time, the motor inductance will apply a voltage to the motor opposing the motor's back emf. This will cause the motor to decelerate, and the motor winding current will drop until it's zero, at which point the diode stops conducting.However this freewheeling period is typically very short, maybe milliseconds at most. Typically its effect on the motor speed is negligible. The motor will still be spinning after the freewheeling is over, and the motor's back emf will appear across the motor's windings, reverse biasing the freewheeling diode.If the motor is being driven with high frequency PWM, the forward voltage of the freewheeling diode will oppose the voltage applied when the MOSFET is on, which will affect the motor's speed (but I would still not refer to this as "braking"). Increasing the forward voltage of the diodes will also cause them to dissipate much more power, of course.Anyways, I'm wondering why the OP wants to detect freewheeling with an LED?You could detect freewheeling by just measuring the voltage across the freewheeling diode with an opamp/comparator. Then there's no need to increase its voltage drop. This is particularly easy if you're using a high side driver like in that schematic.

author: thm_w
author_rank: Super Contributor
timestamp: March 18, 2025, 12:52:57 am
content: Quote from: mtwieg on March 16, 2025, 02:39:02 pmGood point, so the LED would not be useful to show anything.How about removing LED_FWL, moving that to the gate to show when the motor is switched on.Then either use the existing LED across the motor to show free-spinning (but it will be on when the motor is powered too), or if more complexity is acceptable, a comparator/XOR gate setup to only trigger the free-spinning LED when motor voltage is seen and motor FET is inactive.

author: SteveThackery
author_rank: Super Contributor
timestamp: March 18, 2025, 10:48:49 am
content: You are right: VMOTOR will be at +12V (approximately) whether the motor is being driven or free wheeling, so you need to introduce another factor. As you say, some kind of logic function involving the gate drive and VMOTOR should do the trick. Alternatively, measure the direction of the current in the motor. It will travel "upwards" when the motor is free wheeling and powering the LED.

author: Siwastaja
author_rank: Super Contributor
timestamp: March 18, 2025, 11:10:44 am
content: Feel free to increase the voltage drop by putting diodes in series, but note this reduces the efficiency of your driver. Assuming CCM, diodes conduct the same motor current during mosfet-OFF time, that flows during the mosfet-ON time. Power loss in diode can be pretty significant already at just Vf=0.7V in low-speed, high torque conditions.This has nothing to do with "braking". As long as the current is positive, motor is generating positive torque. Now with increased Vf, the ramp-down speed of the current is faster (from dI/dt = V/L; L is the same, V is now higher) so torque ripple will increase. If that matters, you can increase f_sw to compensate.And with a diode (not active switch), you are not able to brake. Once current decays to zero, it won't go negative, and no negative torque is ever created.

author: Picuino
author_rank: Super Contributor
timestamp: March 18, 2025, 11:20:01 am
content: With such high currents, you should use an H-bridge or half H-bridge to handle the motor voltages.That complicates the circuit, but you could make the motor short-circuited (freewheeling) or return power to the batteries (dynamic braking).A current sensor wouldn't hurt, to make sure the motor doesn't burn out. From the sensor you can display the signal you want.I know this complicates the circuit a lot, but such a motor deserves to be protected.

author: Picuino
author_rank: Super Contributor
timestamp: March 18, 2025, 11:29:04 am
content: With this low threshold voltage (-2V) and this large resistor (1k ohm) the switch off time will be very high (several microseconds) so the transistor will get very hot if it works in fast PWM.

author: Zero999
author_rank: Super Contributor
timestamp: March 18, 2025, 11:50:35 am
content: We don't know what's driving the MOSFET, because the OP hasn't included the entire schematic.A motor produces a back EMF opposite to the applied voltage. The polarity of the voltage doesn't change, whilst it freewheels, so the reverse connected LED won't illuminate.A brief reverse polarity voltage spike is produced when the motor is disconnected from the power due to the armature's inductance, which is bypassed by the reverse polarity diode and helps keep the motor turning. It's not the same as the back-EMF produced by the mechanical movement of the armature.The results of the simulation cannot be trusted, because a motor cannot be modeled by a single inductor

author: SteveThackery
author_rank: Super Contributor
timestamp: March 18, 2025, 01:30:25 pm
content: Quote from: Siwastaja on March 18, 2025, 11:10:44 amPerhaps I've misunderstood, but this doesn't seem right. The motor is not just an inductor. Those diodes across the motor willnotconduct when the motor is freewheeling.

author: thm_w
author_rank: Super Contributor
timestamp: March 18, 2025, 10:00:13 pm
content: Quote from: Zero999 on March 18, 2025, 11:50:35 amYeah I had to hook up a motor to a scope to correct my mistake above. I would suggest OP does the same, if possible.Its quite interesting, here is a small DC hobby motor running at 10V:Here is the negative inductive spike:OPs large motor will behave a bit differently.

author: SteveThackery
author_rank: Super Contributor
timestamp: March 19, 2025, 10:19:34 am
content: Quote from: fullmoon6661 on March 19, 2025, 01:13:40 amNo!! You said the freewheeling LED was to show when the motor is freewheeling, NOT to indicate "flyback current that was successfully dissipated". They are separate and different things and you really need to be clear about this.The inductive pulse that happens when you disconnect the drive to the motor is short (milliseconds) and of the opposite polarity to the drive voltage. Youmightdetect a brief flash from the LED. The voltage produced when the motor is freewheeling is the SAME polarity as the drive voltage and will not illuminate the "freewheel" LED with the circuit as shown.

author: mtwieg
author_rank: Frequent Contributor
timestamp: March 19, 2025, 06:29:43 pm
content: Quote from: SteveThackery on March 19, 2025, 10:19:34 amHmm I think you're defining freewheeling as the motor spinning when the driver is disabled, while I was defining freewheeling as when current conducts through the diode due to the motor inductance. To be honest I'm not sure which definition is really correct (if either). From a mechanical standpoint, I think your definition makes more sense (the wheel spins freely while disconnected from the drive). But the "freewheeling diode" as it's often called has nothing to do with that sort of operation, it's only relevant to the motor inductance. I guess we should refer to it as a "catch diode" or "flyback diode" instead to avoid confusion.

author: Zero999
author_rank: Super Contributor
timestamp: March 19, 2025, 09:51:49 pm
content: It's a bit tricky sensing the current through the flyback diode, so just design a circuit which turns on an LED, when the voltage across the motor drops below 0V.V_MOTOR is the voltage across the motor and flyback diode.

author: SteveThackery
author_rank: Super Contributor
timestamp: March 20, 2025, 12:41:47 am
content: Quote from: mtwieg on March 19, 2025, 06:29:43 pmActually I was using the terminology used by the OP, when he clearly used the term "freewheeling" to mean the motor freewheeling between drive pulses.His mistake was that he thought the motor behaved like an inductor (as per his original schematic) and as such would produce an opposite polarity EMF while freewheeling, thus illuminating the diode labelled 'LED_FWHL'. In reality, as we apparently all agree, only a brief spike of negative polarity will appear, giving a brief flash of the LED. The remainder of the time, while the motor continues to freewheel, the voltage generated will be the same polarity as the supply. Thus LED_FWHL will not light, and LED_MOTOR_ON will illuminate, just as it will when the FET is switched on.By the way, I totally agree with you about the confusing nature of the terminology. I actually prefer to use the term "freewheel" for diodes connected across an inductive load, but your "catch" term makes sense, too. Personally I don't understand "flyback" for diodes in that role. I thought flyback was a thing with CRT TVs.

author: SteveThackery
author_rank: Super Contributor
timestamp: March 20, 2025, 12:53:15 am
content: Quote from: Zero999 on March 19, 2025, 09:51:49 pmAren't we just going round the same loop here? The voltage across the motor only drops below 0V very briefly when it's supply is cut off, due to the inductance of the motor. Once that has been dissipated the motor generates apositivevoltage (ieabove0V) while it freewheels. So you can't use the voltage across the motor to detect whether it is freewheeling or driven.When the OP was talking about the motor freewheeling he meant it continuing to rotate after the drive has been removed. And he wanted to illuminate an LED to signal that.If I have misunderstood your point,@Zero999, I apologise.

author: Zero999
author_rank: Super Contributor
timestamp: March 20, 2025, 08:08:47 am
content: Quote from: SteveThackery on March 20, 2025, 12:53:15 amI agree. It's of no practical use. Heck, you might not even see the LED blink. My point was merely that it can be done and was purely for educational purposes.

author: mtwieg
author_rank: Frequent Contributor
timestamp: March 20, 2025, 12:34:41 pm
content: Quote from: SteveThackery on March 20, 2025, 12:41:47 amI see what you mean. I assumed they wanted the LED to turn on when the diode is conducting, since that's how their schematic actually behaves. Some clarification would help.QuoteYeah engineering is full of these semantic traps. Like "saturation", "bootstrapping", etc.