Voltage drop calculation with a dc motor - Page 1
Forum section: electronics / beginners

author: elefurtronik
author_rank: Regular Contributor
timestamp: January 06, 2023, 03:22:04 pm
content: Hello,Very basic, general question…I drive a dc motor with a n-channel mosfet. I have rated 6v motor which pulls 450mA rated current. I connected 10kohm from Gate to Source. My rds on ia 0.1ohm. Id is 0.5A. I connected the motor do drain. I also have a coil which I drive with another mosfet. How I calculate the voltage drop of motor?

author: wasedadoc
author_rank: Super Contributor
timestamp: January 06, 2023, 03:26:59 pm
content: The voltage drop across the motor is almost constant and is the supply voltage minus the Vds of the FET at the motor current. What will change is the motorcurrent. It depends on how heavily the motor is loaded.

author: elefurtronik
author_rank: Regular Contributor
timestamp: January 06, 2023, 09:42:16 pm
content: Instead of the coil on the 2nd MOSFET when I connect the solenoid valve, it doesn't work. It works minimum with 4.8v. The motor works between 3.5v and 7V. How to make calculations then?

author: wizard69
author_rank: Super Contributor
timestamp: January 06, 2023, 09:50:15 pm
content: This is complex but every motor is a generator. but first:Motor speed is related to applied voltage on a DC machine.Motor current is related to torque demand on a DC machineI tried to find a simplified web page that goes into this but came up with nothing simple. So with a DC motor you set a voltage to get a speed output. The load applied to the motor will determine the motor current. The higher the load the higher the current up until the motor can't handle it. So you don't really get a voltage drop in the normal sense of a resitive circuit, there are some losses but those are tiny.Quote from: elefurtronik on January 06, 2023, 03:22:04 pmI'm not sure what you are trying to do here. I'm not sure why you have a 10K ohm resistor in circuit. Ideally your power supply voltage is rated for the motor. In motor control you generally have to monitor current to protect it. In lock rotor condition you current increases to whatever DC resistance will allow, this is what burns up motors.

author: TimFox
author_rank: Super Contributor
timestamp: January 06, 2023, 10:22:10 pm
content: The total voltage across a DC motor is the sum of two values:(1) Back-EMF, proportional to the rotational speed of the motor.(2) Copper loss, proportional to the current through the motor.In turn, the current through the motor is proportional to the torque, the sum of a roughly-constant term (bearing friction) and the torque applied to the load.For example, if you apply a constant voltage to a motor without shaft load, there is a small current due to overcoming the bearing friction, which makes a small copper loss.The rest of the applied voltage gives you the rpm (rotational speed) of the motor.In general, the back-EMF is also proportional to the magnetic field in the DC motor (which is constant for a permanent magnet motor).When I was working in MRI, a co-worker built a non-ferrous DC motor using the field from the main magnet (far away from the axis) and coils without iron cores.He tested the motor in two systems: one with 0.5 T = 5 kG field, and the other with 1.5 T = 15 kG field.For a given applied voltage, the motor rotated more slowly at 1.5 T, but had more available torque than at 0.5 T.

author: Terry Bites
author_rank: Super Contributor
timestamp: January 07, 2023, 07:21:05 pm
content: This may help.https://omgfreestudy.com/voltage-power-equation-dc-motor/

author: elefurtronik
author_rank: Regular Contributor
timestamp: January 07, 2023, 09:42:35 pm
content: Quote from: wizard69 on January 06, 2023, 09:50:15 pmHello,There is a constant load on the motor. Therefore, I always get a voltage drop. I will also apply an extra load on the dc motor later. However, I need to calculate first with the standard load at the beginning.I pull down the gate with 10k. Should I put a higher-value resistor?Quote from: TimFox on January 06, 2023, 10:22:10 pm1. How can I know what is the Back-EMF value? How to measure it for a dc pump motor?2. How can I know what is the Armature resistance as I can't find it in the motor's datasheet?#############Question again:I assume my two MOSFETs are ON. I am applying 5.5V to those two branches.One branch has a 6V motor which works between 3.5-7V and 0.1ohm rds(on) another branch has Solenoid which works between 4.8-7V and again 0.1ohm rds(on). How to calculate voltage drop on the motor and solenoid?

author: TimFox
author_rank: Super Contributor
timestamp: January 07, 2023, 10:25:11 pm
content: Back-EMF: the proportionality constant between EMF and rpm can be estimated from the no-load rpm at rated voltage, since the copper loss from only the bearing friction should be small.Armature resistance: either measure it (with the shaft clamped) or find a motor with a decent spec sheet. You could probably estimate it from an algebraic calculation starting with the loaded rpm at rated voltage with specified current.Whenever I used DC motors, I always started from the resistance and the two proportionality constants: torque vs. current and EMF vs. voltage.Simple data sheets may give only the rated voltage, rpm, torque, and current for the maximum mechanical-power output operating condition. No-load rpm (at rated voltage) and stall torque (at specified current) may be specified, but the stall torque condition may damage the motor if operated with a clamped shaft for too long, due to overheating the armature.

author: Terry Bites
author_rank: Super Contributor
timestamp: January 08, 2023, 05:45:17 pm
content: I recall that in an ideal machine, back emf is prortional to torque, no?

author: wasedadoc
author_rank: Super Contributor
timestamp: January 08, 2023, 05:53:00 pm
content: @elefurtronikEither you did not read my first reply at all or you did not read it carefully enough.

author: TimFox
author_rank: Super Contributor
timestamp: January 08, 2023, 06:05:13 pm
content: Quote from: Terry Bites on January 08, 2023, 05:45:17 pmIn an ideal machine, thecoil currentis proportional to torque, and the back-EMF is proportional torpms.

author: elefurtronik
author_rank: Regular Contributor
timestamp: January 08, 2023, 09:51:49 pm
content: Quote from: wasedadoc on January 08, 2023, 05:53:00 pmI am a little bit confused. Because in parallel to the motor driving MOSFET, I have the second MOSFET which drives the solenoid valve. When I connect two loads in parallel, normally two branches must have the same supply voltage. However, when the motor runs my solenoid doesn't work. But when I connect one cable of the solenoid to the first MOSFET, both work. I really didn't understand what is happening.I have 6v motor but I don't know the voltage drop here. Because when there is an inrush current, I have a big voltage drop in the beginning.Then here, I understand when the motor loaded more, VD also increases.I think I am having here the difficulty with voltage drop on the battery in the beginning due to the motor. How can I solve this? Is there any recommendation?

author: wasedadoc
author_rank: Super Contributor
timestamp: January 08, 2023, 10:08:43 pm
content: If you have one MOSFET for the motor and one MOSFET for the solenoid then the motor and solenoid are not in parallel.If you want to have both the motor and solenoid taking current at the same time then the supply has to be capable of supplying the total current. It is usual that when a motor starts it takes appreciably more current than after it has got up to speed.Provided you are applying adequate turn on voltage to the gates of the MOSFETs they act as almost perfect switches. They drop very little voltage with the result that almost all of the battery voltage is given to the motor and the solenoid. Your repeated asking how to calculate the motor voltage drop indicates that you have not grasped that. Same for the solenoid voltage. Its steady state current will be almost the "on load" battery voltage (see next paragraph) divided by the coil resistance.Batteries are not perfect voltage sources. Their voltage will drop as more current is taken from them. How much the voltage drops depends greatly on the battery and the current. Also long thin wires will drop voltage. We know no details of what you are actually doing there.

author: elefurtronik
author_rank: Regular Contributor
timestamp: January 08, 2023, 10:47:28 pm
content: Quote from: wasedadoc on January 08, 2023, 10:08:43 pmI didn't understand the first sentence. When the MOSFETs are on, the drain is directly connected to the ground. For two MOSFETs, if the motor and SV are connected to the Vdd. Then the loads are parallelly connected to eachother, aren't they? How? I am controlling two MOSFETs from different digital pin of arduino.

author: WattsThat
author_rank: Frequent Contributor
timestamp: January 08, 2023, 11:22:00 pm
content: Your problem is likely a construction error and text will not work.Post a photograph of the mosfets with all wiring shown to the motor and solenoid.

author: RJSV
author_rank: Super Contributor
timestamp: January 08, 2023, 11:27:02 pm
content: You need to revert to OHMs LAW,...the first point when analyzing. Plus have to realize, at first glance you have two separate circuits, (driven by Arduino). That is because, first thing, those two output driver circuits only have the battery in common; that does NOT mean parallel, as that's only the plus or supply.Your ground or minus sides are distinct, or should be, as each component is separately brought to the minus, via the MOSFET, on each....separate driver circuit.As far as some interaction there, that's just a bit of sag, as you've loaded the battery down.Both devices, in the two little separate circuits, will see that battery voltage sag, like 9V will go down a bit, to 8.2 V with a bigger load, and smallish battery.Sounds like a schematic, or more verbage needed, as you tend to add more relevant info, over the posts.For measuring DC pm motor current, a 0.5 (half) ohm resistor, inserted into the (+) side, of either component will allow current reading directly. Like other posts, you should do the motor shaft clamp very very briefly, as that is a stall, could get smokey.

author: BillyO
author_rank: Super Contributor
timestamp: January 09, 2023, 01:38:25 am
content: What are you using as a power supply?Why do you have resistors on the gate?Can you please post a schematic?

author: elefurtronik
author_rank: Regular Contributor
timestamp: January 09, 2023, 10:08:36 am
content: Quote from: WattsThat on January 08, 2023, 11:22:00 pmI attach the circuit schenatic that I created.Quote from: RJHayward on January 08, 2023, 11:27:02 pmMy ground is not distinct. I connected arduino, battery and mosfet grounds together.I measured motor current. In the beginning for 45ms, it pulls 450mA and after that in steady-state it pull 250mA.Quote from: BillyO on January 09, 2023, 01:38:25 amI am using LiMnO2 6V battery.I have gate source resistor to pull down the gate.The schematic is also attached.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: January 09, 2023, 10:42:25 am
content: Having a pull down resistor on the gate in that way is a good thing. It makes sure the fet does not open accidentally during reset of the microcontroller, when all I/O pins are usually floating. The value of this resistor is not critical. Anything between a few kOhm and 1MegOhm will work.The motor voltage is simple.The Fet works as a switch and is either on or off. When the fet is off, there is no voltage over the motor, when the fet is on, the full power supply voltage is over the motor. When you PWM the fet then the (average) motor voltage follows the PWM duty cycle. That means that if you have 6V power supply and a duty cycle of 25%, then the average motor voltage will be 6*25/100 = 1.25V.When you are modelling a DC motor there are a few main things to recognize. Voltage controls RPM, and current controls torque. The voltage over a motor consists of two main parts (and a few smaller corrections) The biggest is the back EMF. (ElectroMotoric Force) When a brushed DC motor rotates, it acts as a dynamo (with output voltage proportional to RPM). If you apply an external voltage to the motor, it will go spinning at (nearly) the same RPM as the RPM needed to generate the motor voltage when it's used as a dynamo.Second consideration are losses in the motor, and usually modeled as a single resistor. (In practice it's mostly a combination of the resistance of the copper wire and the brushes). A resistor has a voltage over it that is proportional to the current. In the model this resistor is in series with the back EMF of the motor.When there is a load on the motor, it has to deliver torque, and to deliver torque there has to be a current though the motor (Direct relation between torque and current). This current causes a voltage drop over the motor resistor, and thus there is less voltage left over for the back EMF and the motor will run at a lower RPM.Those are the main factors of a brushed DC motor and it's all you need to know to get started. Later on you can refine the model if needed with more things that cause losses.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: January 09, 2023, 10:51:29 am
content: Quote from: elefurtronik on January 08, 2023, 09:51:49 pmYes, that seems likely.In the beginning, the motor is not running yet, so there is no back EMF, and all the motor voltage will be over the resistor. There will be a very high current through the resistor, and it is likely this pulls down the battery voltage.The simplest solution is to slowly increase the duty cycle of the PWM signal to ramp up the motor speed.

author: elefurtronik
author_rank: Regular Contributor
timestamp: January 09, 2023, 09:04:36 pm
content: Quote from: Doctorandus_P on January 09, 2023, 10:51:29 amEven though I increase the motor with PWM at the middle, I am having that voltage drop which I see if I don't slowly increase the PWM.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: January 13, 2023, 02:30:09 pm
content: What sort of battery are you using, and how much does the battery voltage drop?You can also do a test with a 10 Ohm resistor, and post the results. When you do a measurement with a resistor we have a common reference to compare with, as motor current varies wildly.

author: Picuino
author_rank: Super Contributor
timestamp: January 13, 2023, 07:34:53 pm
content: Equivalent circuit of a motor in steady state:https://www.eeeguide.com/types-of-dc-motor/* In a small motor the flux is produced by magnets and is constant.Φ = constant* You can measure the Ra resistance by stopping the motor shaft and measuring the motor resistance with a small current applied.Ra = V / I* The Ke constant can be calculated if you can measure the rotational speed, the applied voltage and the current flowing through the motor. Thanks to the formula:Ke = (V - Ra · I) / ωm