Controlling DC Motor using DC-to-DC Converters

[beyondlight]

Homework Statement

We have a small car that is powered by a battery that feeds a DC-motor. The system consists of two DC-to-DC converters that controls the rotor and stator current. The DC-motor i separately magnetized.

The purpoce of the DC-to-DC converters is to modulate the time the transistors are on and off so that the voltage over the motor can be regulated.

When transistors T1 and T4 are on T2 and T3 are off.

http://oi46.tinypic.com/10xhkpd.jpg

Homework Equations

$$U_1=\frac{t_{p}}{T}U_2$$

The Attempt at a Solution

I can't decide between two alternatives.

The first is that the current through the DC motor has the same direction (rotor-curent I_a) Which means that during the time t-->T the current I_a must go in the negative direction, over the capacitor instead of the battery.

In this case if it is right, how does the voltage vary over the motor when transistors T2 and T3 are on?

http://oi48.tinypic.com/rib3hi.jpg

The second is that the current I_a has different directions according tho which transistors are open. If the second alternative is correct, does it mean that the motor works in both directions and thus the wheels can't roll or does the commutator make the current go in the right direction so that the weels can spinn in the right direction?

http://oi50.tinypic.com/2eyl8gn.jpgWhen i know how the currents and voltages actually do vary in the circuit, then i need to calculate how the average value of

u_x1 and u_x2

vary depending on the pulseratio t/T. And how U_a (DC-voltage) vary as a function of the pulseratio.

 

[NascentOxygen]

Is this a toy car?

If the stator field remains fixed, then reversing the direction of the current through the rotor reverses the direction of rotation of the rotor. If the rotor reverses, then so do the wheels.

The capacitor and battery together serve as the power supply. The capacitor just gives it a better performance by lowering the impedance and smoothing out spikes. You can ignore it when explaining how the circuit works. It functions a bit like a parallel battery, but charged by the real battery. It doesn't do anything special like you are imagining it does, sorry.

 

[beyondlight]

Is this a toy car?

If the stator field remains fixed, then reversing the direction of the current through the rotor reverses the direction of rotation of the rotor. If the rotor reverses, then so do the wheels.

The capacitor and battery together serve as the power supply. The capacitor just gives it a better performance by lowering the impedance and smoothing out spikes. You can ignore it when explaining how the circuit works. It functions a bit like a parallel battery, but charged by the real battery. It doesn't do anything special like you are imagining it does, sorry.

Its a small car with a 180 V battery with energy 4kWh.

Now i think i know how it works, by changing the time t you can make the average voltage over the DC-motor either positive or negative. This determines the direction the current goes. And it will always go in one direction depending on the voltage over the DC-motor.

Since the current can't change the direction instantaneously because the motor is inductive, the current will have to flow trough the capacitor when the current of the battery is going in the opposite direction. (when the transistors swich)

I guess that the capacitor makes sure that we get the desired average current?

You get it?

 

[NascentOxygen]

So you can fit 2 people into it? What is its cruising speed? What distance can you travel on a full charge?

You seem to be referring to a diagram that you have not included. But in general, if you vary the duty cycle of switched DC, you allow more or less of it through to the output, meaning you vary the average voltage making it somewhere between 0 V and 180 V. Polarity reversal of the voltage to the motor is not involved until you want to throw the car into reverse.

The capacitor voltage can't be different* from the battery voltage. The diodes allow a path through the battery/capacitor to take care of the reverse voltages due to inductive kicks when the transistors switch off. For example, when T1-T4 are conducting and then switch off, the rotor current continues to flow for a brief time via D3, the capacitor and D2.

Yes, I think I get it now.

 

[beyondlight]

So you can fit 2 people into it? What is its cruising speed? What distance can you travel on a full charge?

You seem to be referring to a diagram that you have not included. But in general, if you vary the duty cycle of switched DC, you allow more or less of it through to the output, meaning you vary the average voltage making it somewhere between 0 V and 180 V. Polarity reversal of the voltage to the motor is not involved until you want to throw the car into reverse.

The capacitor voltage can't be different* from the battery voltage. The diodes allow a path through the battery/capacitor to take care of the reverse voltages due to inductive kicks when the transistors switch off. For example, when T1-T4 are conducting and then switch off, the rotor current continues to flow for a brief time via D3, the capacitor and D2.

Yes, I think I get it now.

But when T1 and T4 are switched of, what is the voltage over the motor? Isnt ux2 +180 in that case?
The current over the capacitor must go in both directions during the modulation time because the average voltage must be the same, right?

So when T1 and T4 are on the current is flowing upward trough the capacitor and then trough T1 and T4.

And when T1 and T4 are off the current must go trough diodes D3 the capacitor and D2. So this time the current over the capacitor is the opposite.

This is a home assignment task. I am actually not making the car.

 

[beyondlight]

$$u_{x1}=\frac{t_1*180 + (T-t_1)*0}{T}=\frac{180*t_1}{T}$$

$$u_{x2}=\frac{0*t_1 + (T-t_1)180}{T}= \frac{180*(T-t_1)}{T}$$

$$U_a=u_{x1}-u_{x2}=\frac{180*t_1}{T}- \frac{180*(T-t_1)}{T}=180*\frac{2*t_1 - T}{T}$$

Seems correct to me.

Put in tp=T and we get U=180V

tp=T/2 and we get U=0 V

am i right?

 

[NascentOxygen]

But when T1 and T4 are switched of, what is the voltage over the motor? Isnt ux2 +180 in that case?

Sorry, that's a convention I guess I've forgotten.

The current over the capacitor must go in both directions during the modulation time because the average voltage must be the same, right?

I don't know what you mean by modulation time. The capacitor current due to inductive kicks is always going to be "downwards", the diodes ensure this.

So when T1 and T4 are on the current is flowing upward trough the capacitor and then trough T1 and T4.

Well, there would only be current through the capacitor if it were charging or discharging. Because the battery is holding it constant, I wouldn't say there is capacitor current; the current to run the motor is coming from the battery.

 

[beyondlight]

Sorry, that's a convention I guess I've forgotten.

I don't know what you mean by modulation time. The capacitor current due to inductive kicks is always going to be "downwards", the diodes ensure this.

Well, there would only be current through the capacitor if it were charging or discharging. Because the battery is holding it constant, I wouldn't say there is capacitor current; the current to run the motor is coming from the battery.

By modulation time i mean switch-time. The total period T.

Of course there is no capacitor current in that sense that it is running the motor. But the current trough it must be both upward and downward i guess, because you can't charge it all the time. It has to be discharged? But over one period T, the average voltage over the capacitor should be 180 V?

 

[NascentOxygen]

Of course there is no capacitor current in that sense that it is running the motor. But the current trough it must be both upward and downward i guess, because you can't charge it all the time. It has to be discharged? But over one period T, the average voltage over the capacitor should be 180 V?

Fair enough, no nett capacitor charge over period T.

Earlier I wrote:

Polarity reversal of the voltage to the motor is not involved until you want to throw the car into reverse.

I could also have written: Reversal of current in the motor is not involved until you want to throw the car into reverse. I prefer this wording because voltage tends to do what it wants, and the inductive kick will reverse the voltage polarity in an attempt to maintain a constant direction of current. The diodes constrain the magnitude of this reverse voltage swing to protect the transistors.

You do appreciate that for steady forward driving, only two transistors will be operating, the other two will be off, fully off, and staying off. If you didn't ever need to reverse* the car, you could omit one of the transistor pairs during construction.

* I think they may come into play when braking, but that's a separate issue