Voltage Regulation Vs Current Regulation.

[Imagineering]

Newbie warning
I've just built a Linear PSU with a rather large transformer (100x100x120mm) with two seperate 20V secondaries. I've run each secondary into its own Bridge Rectifier then paralleled the DC outputs with two 10,000uf 100V Capacitors in parallel for smoothing. It produces 29V DC and gets a 230V 850Watt heating element glowing, so I guess it has Watts to spare.
I've built a Stepper Translator board using 7486 & 7476 IC's which feed TIP122 Power Darlingtons for the Steppers.

My questions are;
As the circuit I'm building has two Stepper Coils energized at a time, do I need to;

1, Calculate the Current Limit Resistor for two coils and place this Resistor in the Common line for each pair, ( 2 per Stepper) ? or

2, Calculate the Current Limit Resistor for one coil and place a Resistor in each Coil Feed, ( 4 per Stepper) ?

3, Do I need to regulate the Output Voltage from my PSU?

Thanks in advance . . .

[ViperTX]

For driving only the steppers you don't need any regulation. What you have will work. You will need to limit the current for each stepper winding (like your 2). As you know the darlingtons use the I(b) to control their conduction. If you look at the current gains of the darlington (hfe...or something like that) you can determine what your I(c) should be based on your I(b). I'm suspecting that you'll use the darlington's in their saturation region.

[gar]

050618-1319 EST USA

Imagineering:

On your question 3. Why would you want to regulate the voltage? Does your input ac line vary from 95 to 135 volts while you are working? Probably not. This is a 42% change. If this change occured, then your drive current to the steppers would change by this percentage and so would motor torque.

You could put a Sola constant voltage transformer at the input and hold this change to about 1%. I do not think you want to do this because it would be more expensive than using chopper drivers.

Further, in our area throughout a day we do not see voltage change more than about 3 volts from about 121 to 124 v. This is totally insignificant. This excludes random times when abnormalities occur.

You certainly do not want to put a linear series pass regulator on your supply output. This would just waste a lot power and create excess heat.

In a standard transformer, rectifier, and capacitor input filter the output voltage will be proportional to ac input line voltage. With a current limiting resistor feeding your motors your motor current will be proportional to ac line voltage. Therefore in the end motor torque is proportional to ac line voltage.

I am not sure what you are saying about motor coils.

Whether you have 4 coils (8 wires), 2 center tapped coils (6 wires), or 2 coils (4 wires) I can not tell from your description.

For one motor that contains two phases and both phases are always on you could use one current limiting resistor in an appropriate circuit. I would not recomend this.

Let us assume you have 2 center tapped coils. Then for each phase use one current limit resistor and connect it to the center tap.

From your description I doubt that you have only one coil per phase.

If you have 4 separate coils, then you can connect them as if they where center tapped coils.

Do you know how to determine the resistance of the current limiting resistors?

.

[Imagineering]

Viper & Gar, Thanks for your comments.
Quote - Why would you want to regulate the voltage? Does your input ac line vary from 95 to 135 volts while you are working? /Quote.

I wasn't sure if I needed to regulate the Voltage which is why I asked before commiting myself to a big BANG. We have a 230V AC supply in New Zealand which is fairly stable, so I can discount major voltage fluctuations.

Quote - Let us assume you have 2 center tapped coils. Then for each phase use one current limit resistor and connect it to the center tap. /Quote.

Yes, this is exactly what I have, which is why I queried the resistor hookup.
Do I really need a resistor on each coil or can I get away with a resistor on the center tap as you suggest?

As an example, I have one Stepper which is rated at; 5.7Volt 1.14 Amp.
By my calculations, using my 29V DC supply, (29V - 5.7V) / 1.14 A = 20.43 Ohms @ 26.5 Watts per coil.
This would mean four 20.43 Ohm resistors per Stepper Motor.

Can I use two 40.86 Ohm 53 Watt Resistors in the Centre Tap leads instead?

Are there any Pros/Cons with either of the above setups?

I can see that One Resistor in each Coil Lead will limit the current in that coil only, I'm fine with that scenario.
But what happens with the 'Resistor in the Common Lead' when only one coil is enegised during the stepping cycle? Does this mean, (because the Resistance is doubled), that when one coil is switched on that the current limiting is incorrect for that coil?

I've got the PSU, the Logic Circuit Board and the Power Transistor/Heatsink Assembly built, and I'm now ready to hook everything up for the Final Testing, and I need to get the Resistor setup correct as I cannot afford a Meltdown.

Thanks once again & looking forward to your replies.

[gar]

050619-0602 EST USA

Imagineering:

I asked the rhetorical question on voltage regulation to get the many viewers to think about why they should or should not regulate the voltage to their stepping motors. So long as you stay within the motor's temperature limit the only effect is a variation in torque.

But for most integrated circuit logic circuits you must regulate the Vcc and other voltages.

On your current limiting resistors and where to locate. When the resistor is connected to the center tap, then it has a constant dc current because the coil on one side of the center tap or the other is always on. We have to assume that the motor rating of 5.7 V and 1.14 A or 4.65 ohms is for one side of center tap, but check its value. On this assumption, then your calculation for resistance is correct, except that no drop was assumed for your transistor switch. But this does not matter because you will pick a nominal resistance of 20 ohms. Your power calculation is correct and you should use a 50 watt resistor.

It does not matter whether a single resistor is used in the center tap or two resistors are used one in each outer end of the center tapped coil the resistance values and power ratings are the same because you have to design for the steady-state condition of the motor stopped.

Your calculation on the center tapped resistor is wrong on both resistance and power.

Draw the circuit diagram and then look at the current flow for the two different positions of the resistor. The resistor is still in series with the coil whether it is on the outer end of the coil or the center tap end, and in steady-state current only flows thru one side of center tap.

More later.

.

[Imagineering]

Quote - On your current limiting resistors and where to locate. When the resistor is connected to the center tap, then it has a constant dc current because the coil on one side of the center tap *OR* the other is always on.

Your calculation on the center tapped resistor is wrong on both resistance and power. /Quote

Thanks for clearing this one up for me - I made my calculations with the assumption that power was applied to *both* coils of each pair at once.
I can now go & buy some resistors tomorrow & start Blowing things up 8-)

I'll keep you posted . . .

[gar]

050619-0828 EST USA

Imagineering:

Do the following to go to Superior Electric site for step sequence information.

Use Google to search for --- stepping motor theory of operation superior electric ---
.........then pick webmain/DB etc site for motor step sequence.

.

[gar]

050619-0944 EST USA

Imagineering:

Also go to www.silniki.pl/pdf/stepperintro.pdf .

Here a single phase and one coil is energized at one time. Does not change the power requirements on the resistor because you must assume you are in a stopped state with the resistor energized.

.

[Imagineering]

Thanks gar, I've downloaded & printed these and I'm about to set up a cup of coffee & a good read 8-)

[gar]

050620-1937 EST USA

Comments to all:

Imagineering is using the following circuit to make his driver.

http://www.cnczone.com/forums/showth...4&page=5&pp=10

His motor is a five wire unit with both center taps connected together. The desired current limiting resistor is 20 ohms for the parameters he has given. The circuit shown has a back biased diode across each coil. The circuit shows no current limiting resistor.

My suggestion is to use 4 current limiting resistors, one in series with each non-center tapped end. These resistors would be placed between the diode anode and the coil end. This leaves the transistor collector connected to the point between the diode anode and the current limit resistor. By doing this we shorten the decay time constant for the energy stored in the coil when the switch turns off. Also ideally the collector will never go above VS. The time constant of an LR circuit is proportional to L/R. So increasing R reduces the time constant.

This circuit has one coil of each phase always on. The power dissipated in the 20 ohm current limiting resistor is slightly over 25 watts. Therefore I would use a 50 watt resistor. One has to design for a continuous current in each resistor because when the motor is stopped one of the two resistors per phase will have current flowing thru it.

I really do not like the design of the TIP122 base drive circuit. This is very marginal.

Rather I would like to see a pull up resistor that provided 10 ma to the TIP122 base. Then use a 2N7000 to short the pullup to common.

Note the worst case sourcing output of the SN7476 is 0.4 ma @ 2.4 volts.

Extremely marginal to drive the TIP122. Darlingtons are very temperature sensitive by the nature of their design. By using the 2N7000 to short the base to common this sensistivity is reduced and we get high drive to the TIP122 when its base is not tied to common. It is also true if R1 thru R4 are low enough that the 7476 output transistor does a good job of holding down the 7476 base.

The biggest problem with the 7476 is driving the TIP122 solidly into saturation.

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