How to understand DC motor inductor current?

dc prush motor and inductor

Maybe I should simplify my question in a general manner.
First, I know how to calculate the current variation in a stationary inductor under a varying driving voltage source such as PWM controlled voltage source.
Second, I don't know how to calculate the current variation in a mobile inductor which is not only controlled by varying driving voltage, but also affected by surrounding magnetic because of its motion.
I think simply sum up the current caused by these two sources is not correct, but I can't find right solution. So here I am to ask for your help.

inductor current cut off

bittware said:

I agree. However, regardless of what switching style is, I think the phenomenas are the same.

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In my opinion, the switching style will affect the electrical dynamic of the dc motor. For example, during the OFF state of Hard-Switching PWM (S1, S2, S3, S4 are OFF), the Vs term is included in the model equation (Blue line). On the other hand, during the OFF stage of Soft-Switching PWM (S4 is ON, other are OFF), the Vs term is NOT included in the model equation (Green line).

bittware said:

Maybe I should simplify my question in a general manner.
First, I know how to calculate the current variation in a stationary inductor under a varying driving voltage source such as PWM controlled voltage source.
Second, I don't know how to calculate the current variation in a mobile inductor which is not only controlled by varying driving voltage, but also affected by surrounding magnetic because of its motion.
I think simply sum up the current caused by these two sources is not correct, but I can't find right solution. So here I am to ask for your help.

Click to expand...

Let me tries to describe:

Vs = Voltage across power supply terminal
Vm = Voltage across motor winding
If voltage drop across switches is ZERO, then Vm = Vs when switches are ON.

The electrical dynamic of the dc motor is as follows:

Vs = RI + dψ/dt .......... (3)

where ψ is flux-linkage produced by the motor winding, R is winding resistance, and I is the motor current. We can expand (3) to be as follows:

[correction]
Vs = RI + (δψ/δI)(dI/dt) + (δψ/δθ)(dθ/dt) .......... (4)
(Eq.(4) contains partial differential terms)

Vs = RI + L dI/dt + kω .......... (5)

where
L = δψ/δI = constant
back-emf constant, k = δψ/δθ
ω is rotational speed in rad/s
voltage back-emf, Vemf = kω

Frm (5), with little rearrangement, and taking the integration of both sides, we get

I = 1/L ∫ (Vs - Vemf - RI) dt .......... (6)

When switches S1 and S4 are ON, the electrical dynamic model of the dc motor is given by equation (6). When switches S1 and S4 are OFF (S2 and S3 are OFF too, i.e. Hard Switching), the electrical dynamic model of the dc motor is as follows:

I = 1/L ∫ (-Vs - Vemf - RI) dt .......... (7) (Vs becomes negative)

By solving (7), I think you can find the instant when motor current changes direction.

boost converter + regenerative

Hello nicleo,
Great analysis!
Except equation(3) and (4) seem to be not equivalent.
To be honest, I haven't heard of Hard-switching and Soft-switching concept before.
From your description, I conclude Hard-switching means current "fast-decay" and Soft-switching means current "slow-decay". Am I right?
What are the characteristics of these two switching sytles?How to apply them properly?
Does there exist any occasion we can ignore the current contribution taken by BEMF when the switch is off?
P.S.Could you provide some reference which introduce the concept of Hard-switching and Soft-switching?
Thanks in advance.

h bridge chopper

nicleo very nice analysis. bittware I had considered soft switching since it is more generally used (and atleast my power electronics prof. was partial to it ).
Yes you are right about soft and hard switching. Hard switching is generally at a higher frequency than soft switching. This increases the switching losses in hard switching and also causes EMI. However, the operation of the soft-switched converters requires additional active and passive elements. This introduces additional cost and complexity.
So depending on your application specs you will go for either hard or soft switching.

mechanical force caused by currentin inductors

bittware said:

Expect equation(3) and (4) seem to be not equivalent.

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Good catch ! :wink:
Eq. (4) has been corrected. Actually there are 'partial differential terms' in Eq. (4). As I could not find the 'del' (see the circled charactere in the attached figure, is it called 'del'?), I used 'd' for the partial differential terms. In the newly corrected Eq. (4), I used 'δ' for the partial differential terms, as I noted in some mathematical book, 'δ' is used in partial differential equation. My apologize for the 'mistaken' equation.

bittware said:

To be honest, I haven't heard of Hard-switching and Soft-switching concept before.
From your description, I conclude Hard-switching means current "fast-decay" and Soft-switching means current "slow-decay". Am I right?
What are the characteristics of these two switching sytles?How to apply them properly?
P.S. Could you provide some reference which introduce the concept of Hard-switching and Soft-switching?

Click to expand...

Another definition of soft-switching:
Soft switching converters constrain the switching of power devices to time intervals when the voltage across the device, or the current through it is nearly zero.

Some references:
History of Soft Switching
**broken link removed**

History of Soft Switching, Part II
**broken link removed**

Recent Advances In Soft Switching Inverter Technology
http://www.powerdesigners.com/InfoWeb/design_center/articles/SSI/ssi.shtm

Zero-Voltage Switching Quasi Square Wave Converters
**broken link removed**

Switching Characteristics for Power Devices
Some Advantages and Disadvantages of Hard Switching
**broken link removed**

Soft Switching for Power Devices
**broken link removed**

Soft Switching Inverter Technology
**broken link removed**

dc motor pwm current decay

Hello nicleo,
I tried to solve the equation (7) you provided, but I can't not figure out the Vemf=K*w(t) in explicit form because I(t) is no more constant with repect to parameter t.
Could you give me some prompt how to solve such equation?
If I should regard w(t) as constant during PWM off state for simplification?
P.S. Sorry, I can't type out omega, so substitute w for it.

dc motors inductors

bittware said:

.... I can't figure out the Vemf=K*w(t) in explicit form because I(t) is no more constant with repect to parameter t. .... If I should regard w(t) as constant during PWM off state for simplification?

Click to expand...

The magnitude of Vemf does not rely on the armature current. It depends on current in 'field winding'. However, in your case, the field winding has been replaced by permanent magnet already. So, Vemf is just dependent of the motor speed.

Mechanical time constant is usually much longer than the duration of PWM OFF state. So, the motor speed can be considered as constant during the PWM OFF state.

calculate the current in the inductor at t =

nicleo said:

So, Vemf is just dependent of the motor speed.

Click to expand...

Hello nicleo,
Don't you think rotational speed(w) has some relationship with motor rotator winding current(I)? Because motor power is limited, if you wanna get more torque(i.e. more current(I)), you have to sacrifice your speed(w).

24v dc motor field winding resistance

bittware back emf is proportional to motor speed. Beyond that the factors that can effect your back emf depend on the constraints that you impose. For example if you have constant power then as you said speed is inversely proportional to torque or current. However if you have a power that is directly proportional to say torque then motor speed is constant.
Therefore one relationship that always holds is the proportionality of back emf with motor speed. Beyond that back emf will vary with other factors depending on the situation.

all kinds of motors inductores

usernam said:

However if you have a power that is directly proportional to say torque then motor speed is constant.

Click to expand...

Hello usernam,
What if the driving PWM is in off state? How the BEMF varies with current during this period is not simple as it looks like.

negative current in motors

bittware When PWM is in OFF state then the blue line represents current flow.
As I said one relation that always holds is that BEMF is proportional to the motor speed and also armature current is directly proportional to electromagnetic torque. The proportionality constants in both cases are the same.
So first let us apply Kirchoff's laws to the circuit.

L(dI/dt) = -(Vs+Eb) -------- (1)

Now let us look at the mechanical situation of the machine.
Since load torque has been removed
Tem = Bw + J(dw/dt) --------- (2)
Now Tem = KI and Eb=Kw
So substituting for the terms in (2)
KI = B*(Eb/K) +(J/K)*(dEb/dt) ---------- (3)

So given B,J,K,l and Vs we have 2 equations and 2 unknowns which can be solved for. Eb and I will vary such that both equations (1) and (3) are always satisfied.

is a motor an inductor

bittware said:

I am using PWM to control DMOS H-bridge to drive a brush DC motor. When PWM switches off, the current flowing through inductor can't be cut simultaneously. It will find a path through freewheeling diode. But the motor rotor is still spinning, the Back-EMF has a trend to force the current to inverse.
My question is:
Whether the current flowing through motor inductor would reverse or wouldn't reverse at all in a limited switch off time?

Click to expand...

nicleo said:

When switch is OFF, the current flowing through inductor could not reduce to ZERO instantaneously. The direction of the current flow will not reverse.

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bittware said:

Assuming a extreme condition, if a rotor has been accelerated to a relative high speed and the driving power withdraws, the rotor will act as a generator. The kinetic energy will be converted into electric energy, meanwhile the current is reversed. How to explain this case?

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nicleo said:

No. The current will still in the same direction, i.e. not reversed. I assume that in your case, there is no 'external' torque to drive the motor (rotor) after the driving power is removed. The current flows through the motor winding, after the driving power is removed, will not change or reverse direction.

Click to expand...

bittware said:

Althogh after driving power is removed no external torque drives the motor, the residual kinetic energy instored in rotor itself will also be converted into electric energy. Right? So I believe a part of kinetic energy will cause revesed current eventually. What I am concerning is when the reversed current appears after the PWM switches off.

Click to expand...

nicleo said:

When Stage 2 (i.e. swtiches (S1, S2, S3, and S4) are OFF, just after motoring current decays to ZERO, and rotor still rotates), and if the stator has permanent magnet or the residual magnetic field on stator is strong enough:
Current path: -PS -> D4 -> motor winding -> D1 -> +PS
In this case, both the current at PS terminals and in motor winding are reversed (based on Fleming's rules) because the rotor is still in 'clockwise' direction.

Click to expand...

username said:

bittware are you refering to some sort of regeneration. If you are then yes eventually the current in the inductor will reverse. But not instantaneously. It will first decay to zero and if you are still in regenerative mode the current will go negative. Ofcourse you have to provide a path for the reverse current so you must be using a Hbridge or a Class C chopper.

Click to expand...

username said:

Now suppose inductor current has finally come to zero while motor is still rotating in the same direction. We turn S2 on and S1 and S4 off. Voltage across the inductor becomes negative (using previous refernce directions) so current through the inductor falls i.e it becomes negative. Current flow is D4 Dc motor and then S2 then back to D4. Now regenerative process has begun. Now we are using the Hbridge in boost converter mode. Now turn S2 off and D1 begins conducting. Voltage across the inductor is positive so the inductor current rises i.e becomes less negative. Current flow is D4 then DC motor then D1 then source and back to D4. So now the average inductor current is negative. Average power flow is from motor (now a generator) to the source.

Click to expand...

bittware said:

What I care about is the current variation in inductor when PWM swithes off. Based on common sense, the motor resistor will consume up the energy stored in indutor during recirculation phase. And as for PWM off time is enough long, the current will alternate its direction finally. Bigger the motor resister is, quicker the current changes its flowing direction.

Click to expand...

It looks to me that, at least "bittware" and "nicleo", agree that the current in motor winding, if the rotor is still running during PMW OFF state, will reverse direction eventually if the OFF state is long enough. In my opinion, this comment is correct only at certain operating points, or proper control strategy is employed.

Case 1
Power Converter Topology: H-bridge (pls refer to nicleo's diagram).
Control Strategy: PWM, Hard-Switching (i.e. ON-State -> S1 & S4 are ON; OFF-state -> ALL switches are OFF)
Motor Operating Point: Below No-Load speed
Description:
During OFF state (before current decays to zero), the current path is -PS -> D2 -> Motor -> D3 -> +PS.
After the current decays to ZERO, D2 and D3 will be OFF. The voltage appears across the winding will be the Vemf (as motor is still rotating).
As the Vemf is LOWER than Vs (power supply), the current in motor winding will stay at ZERO and will NOT reverse.

Case 2
Power Converter Topology: xxx
Control Strategy: xxx
Motor Operating Point: Above No-Load speed
Description:
Say the motor is driven above No-Load speed (by external force, for example, an Electric Vehicle is travelling downhill).
The generated Vemf is HIGHER than Vs, the current in motor winding will reverse and flow back to power supply through the following path: -PS -> D4 -> Motor -> D1 -> +PS

Case 3
Power Converter Topology: H-bridge.
Control Strategy: PWM + usernam's regenerative switching sequence
Motor Operating Point: Below No-Load speed
Description:
Say it's in PWM OFF state (all switches are OFF) and the motor current has finally comes to zero while motor is still rotating in the same direction. Then S2 is switched on. Voltage across the motor armature winding reverses (because Ri = 0, Ldi/dt = 0, and Vemf does not change because motor still rotates in the same direction) so current through the armature winding falls i.e it becomes (more) negative. Current flow is D4 -> Motor -> S2 -> D4. As the result, energy is built up again in armature winding. Then, S2 is turned OFF, and the inductance of the armature winding forces D1 to conduct. The current flow is -P2 -> D4 -> motor -> D1 -> +PS, and the power flow is from motor (now a generator) to the source. In this example, the H-bridge, at latter stage, works as boost converter to enable the regenerative process.

Conclusion
If the Vemf is higher than Vs after current decays to zero during PWM OFF state, the current in motor winding will reverse. Higher Vemf than Vs can be achieved:
1) If motor is driven (by external mechanical force) above No-Load speed.
2) If the H-bridge is controlled as 'boost converter' as described by 'usernam'.
3) If variable Vs is used so the Vs can be reduced to be less than Vemf.