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Offline notgivenTopic starter

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quick driver current question
« on: August 05, 2012, 02:23:52 PM »
Been looking around the forum a lot to find out the answer to my question. Still not done yet actually. But might as well ask this is the mean time:

I have a driver that can only put out 1Amp at 1.5V (called L293D dual H-Bridge). I have a motor that needs 2.8 Amps at 2.1 volts. What happens when I put them together?

Will the driver chip burn out?

How about this combo: driver gives 1 Amp and 4.5V to 36V (called SN754410 quadruple Half H Driver) used with same motor.

I understand that voltage stats are not really important as motors are often run at voltages above their rated voltage. But rather the current is more important. That's what worries me here. I think a motor that takes more current than a driver can give will burn out the driver chip.

BTW thanks for all previous help. I just kinda left my previous thread dead as I found more reading material, and will be back to it eventually to beg more help.
« Last Edit: August 07, 2012, 03:45:51 PM by notgiven »

Offline newInRobotics

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Re: quick driver current question
« Reply #1 on: August 07, 2012, 08:01:47 AM »
Yes, the chip will burn if no protection is used. You can use weaker chip to drive stronger motor, however some current monitoring and adhusting system has to be in place, however that way motor torque/max velocity will suffer. It's best to mach motor and driver in order to get full performance.
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Offline notgivenTopic starter

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Re: quick driver current question
« Reply #2 on: August 07, 2012, 03:53:36 PM »
Thanks for answering.
I don't care about the torque/max velocity.
I don't care about maximizing strength/efficiency. I'm a beginner and I'm almost at the point were I can get a motor to move. I just want that for now, so I can fiddle with it and learn.

Question: so how can I set up a " current monitoring and adhusting system"?
I can use transistors, yes? (I don't know enough about circuits yet. Still reading)

Tell me if the following is true: I just need to increase the current somewhere between the driver and the motor, right?
Then the thing will turn w/o anything burning up? (again, neglect efficiency).

Offline Soeren

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Re: quick driver current question
« Reply #3 on: August 07, 2012, 06:06:42 PM »
Hi,

I don't care about the torque/max velocity.
I don't care about maximizing strength/efficiency. I'm a beginner and I'm almost at the point were I can get a motor to move. I just want that for now, so I can fiddle with it and learn.

Question: so how can I set up a " current monitoring and adhusting system"?
I can use transistors, yes? (I don't know enough about circuits yet. Still reading)

Tell me if the following is true: I just need to increase the current somewhere between the driver and the motor, right?
Then the thing will turn w/o anything burning up? (again, neglect efficiency).

Take a look at this simple controller. If you think you can get this together (MOSFET can be replaced with a power BJT if that's easier for you).
If you think it's within your abilities, I can revise it for your supply, motor etc (just lay down the details). Then you'll have no problems with too little current :)

If you'd rather go with a currnt limiter, you need to detect the motor current and feed it to whatever controller you use to chop the current before it gets critical, but given that you just want to move things however inefficient, another option would be a simple power resistor (calculating value and power rating takes supply voltage/L293D output voltage, motor voltage and whether your L293 is actually the "D"-postfix version
The L293D can only handle 0.6A and both variants drops quite some voltage, so is poor for low supply voltages.


What is your supply voltage? (Battery or mains adapter? Voltage?)

Does your motor come with the specs you mention (2.8 Amps at 2.1 volts), or did you just measure it?

What do you drive the H-bridge with? (Microcontroller, 555 circuit or what?).
Regards,
Søren

A rather fast and fairly heavy robot with quite large wheels needs what? A lot of power?
Please remember...
Engineering is based on numbers - not adjectives

Offline notgivenTopic starter

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Re: quick driver current question
« Reply #4 on: August 07, 2012, 10:17:00 PM »
Thanks for the advise Soeren.

[/quote]
Take a look at this simple controller. If you think you can get this together (MOSFET can be replaced with a power BJT if that's easier for you).
If you think it's within your abilities, I can revise it for your supply, motor etc (just lay down the details). Then you'll have no problems with too little current :)
[/quote]

Actually I don't think I can do that first suggestion. I might not be able to make sense of the schematic. But maybe if I ask a few questions I might be able to, though it might be too much trouble for you to answer all of them:
- What do all the letters stand for? I know PWM and Gnd and I can identify a motor in there, and I guess the R1, R2, etc are resistors, but that's about it. What does UB, BD140, 1N4004, Schottky, etc mean?
-Do you have a more breadboard  friendly representation? Cant solder for this project.
-Would all the grounds end up connecting to the same path to the Gnd on my microcontroller?
- Whats a BJT look like? Would I be using this?: http://www.semiconductor-today.com/news_items/2008/MAY/Transic.jpg

My supply will probably be this: http://www.radioshack.com/product/index.jsp?productId=3932578&gzq=CAT2818121_PartsAndToolsPower

My motor: http://www.automationdirect.com/adc/Shopping/Catalog/Motion_Control/Stepper_Systems/Motors_-z-_Cables/STP-MTR-23055

My drivers:
http://www.hobbyengineering.com/H1384.html
and
http://www.hobbyengineering.com/H1048.html

My microcontroller: Arduino UNO :http://arduino.cc/en/Main/ArduinoBoardUno/
The H bridge must be connected to both the microcontroller and the battery, right?


I think the second proposal would be easiest for me to do (and for you to explain for me). Except, I hope it doesn't involve PWM because I dont know how to apply that.
And when you say "both variants drops quite some voltage, so is poor for low supply voltages", why is it bad to run things at high voltage? Burns things up ?

So yeah, the bottom line is I screwed myself because I didn't have enough understanding before I started ordering stuff in the first place, and now I have to make it work with this stuff or I can't justify ordering more because its not my money.




Offline Soeren

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Re: quick driver current question
« Reply #5 on: August 08, 2012, 03:53:06 PM »
Hi,

Actually I don't think I can do that first suggestion.
[...]
My motor: http://www.automationdirect.com/adc/Shopping/Catalog/Motion_Control/Stepper_Systems/Motors_-z-_Cables/STP-MTR-23055
[...]
So yeah, the bottom line is I screwed myself because I didn't have enough understanding before I started ordering stuff in the first place, and now I have to make it work with this stuff or I can't justify ordering more because its not my money.

You sure did, but hey, call it a learning experience and it ain't all bad ;D

The driver I posted wouldn't work anyway, as you bought yourself a step motor.

I assume you got the 2.1V from the 2.8A and 0.75 Ohm spec(?) but a stepper is sually driven from a higher voltage with a chopper or similar to back off the current when the field winding is near saturation - that's why it hasn't got a voltage spec.

You can still make it move though and with a resistor in each winding (two in total, as it's a bipolar motor) you can keep the 754410 (1A out) alive.

Carbon/zinc cells are the lowest quality of all battery technologies and won't work for long, but aiming for the 6V Lantern battery, a 3.3 Ohm (4W or higher) resistor will keep your driver alive, as the driver itself will "steal" around 2V of the available 6V. The motor will probably not be able to run/drive anything that way, but it will let youexperiment with making it go.
Did you buy a stepper on purpose, or did you mistake it for a regular DC motor?
If so, you might be able to trade it in for a suitable DC motor, but the one you have ain't half bad, provided you build or buy a proper driver for it.

Besides experimenting, do you have a future goal/purpose for this motor?
Regards,
Søren

A rather fast and fairly heavy robot with quite large wheels needs what? A lot of power?
Please remember...
Engineering is based on numbers - not adjectives

Offline notgivenTopic starter

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Re: quick driver current question
« Reply #6 on: August 08, 2012, 05:16:31 PM »
Hi,

You sure did, but hey, call it a learning experience and it ain't all bad ;D

The driver I posted wouldn't work anyway, as you bought yourself a step motor.

I assume you got the 2.1V from the 2.8A and 0.75 Ohm spec(?) but a stepper is sually driven from a higher voltage with a chopper or similar to back off the current when the field winding is near saturation - that's why it hasn't got a voltage spec.

You can still make it move though and with a resistor in each winding (two in total, as it's a bipolar motor) you can keep the 754410 (1A out) alive.

Carbon/zinc cells are the lowest quality of all battery technologies and won't work for long, but aiming for the 6V Lantern battery, a 3.3 Ohm (4W or higher) resistor will keep your driver alive, as the driver itself will "steal" around 2V of the available 6V. The motor will probably not be able to run/drive anything that way, but it will let youexperiment with making it go.
Did you buy a stepper on purpose, or did you mistake it for a regular DC motor?
If so, you might be able to trade it in for a suitable DC motor, but the one you have ain't half bad, provided you build or buy a proper driver for it.

So essentially I just need to add two resistors to appropriate places in the circuit to prevent anything from smoking?
Would it be OK for me to draw up a circuit diagram and ask you or someone else here to double-check it for me?

So overall, the idea here is to protect the driver by diverting some of the voltage to resistors?
The driver you posted involved several transistors, right? Is there a way I can use transistors in the circuit to increase the current going to the motors?


Besides experimenting, do you have a future goal/purpose for this motor?

I purposely bought 3 of that stepper model because I originally intended to make robot limbs (arms and legs) and eventually a completely humanoid-shaped bot (though not life-sized). I had to keep scaling back as and putting goals off as I learned more about what skills were needed. Originally though I just bought some stuff to jump in and get my hands dirty, and learn as I go. Things haven't been going as smoothly as I hoped, but I don't care, I sill love doing this kind of stuff.


 
« Last Edit: August 08, 2012, 05:19:00 PM by notgiven »

Offline Soeren

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Re: quick driver current question
« Reply #7 on: August 08, 2012, 06:09:33 PM »
Hi,

So essentially I just need to add two resistors to appropriate places in the circuit to prevent anything from smoking?
Yes. It's gotta be power resistors though.


Would it be OK for me to draw up a circuit diagram and ask you or someone else here to double-check it for me?
Wow, you don't see such politenss all that often these days :)

Sure I'll check it.

It's not that complicated though, If you still have the connector on the motor, one resistor go to either pin 1 or pin2 (doesn't matter which) and the other goes to either pin 3 or pin 4.


So overall, the idea here is to protect the driver by diverting some of the voltage to resistors?
The idea is to limit the current by adding resistance (and that drops some voltage in the resistors, but "diverting" is not a descriptive term here).



The driver you posted involved several transistors, right? Is there a way I can use transistors in the circuit to increase the current going to the motors?
Sure, but then you'd actually build an H-bridge (driven from the H-bridge you have) and that would be a larger circuit than the one I posted (for a DC motor).
I used 3 BjT's and 1 MOSFET (Bipolar Junction Transistor and Metal Oxide Semiconductor Field Effect Transistor respectively) and I'd call that "a few" rather than "several" :)
For a decent bridge driven from one of your chips you'd use 8 MOSFETs and it would be both cheaper and easier to get an appropriate driver chip then.


I purposely bought 3 of that stepper model because I originally intended to make robot limbs (arms and legs) and eventually a completely humanoid-shaped bot (though not life-sized). I had to keep scaling back as and putting goals off as I learned more about what skills were needed. Originally though I just bought some stuff to jump in and get my hands dirty, and learn as I go. Things haven't been going as smoothly as I hoped, but I don't care, I sill love doing this kind of stuff.
OK, have fun with it (it IS), but consider getting those dirty hands on a DC-motor as well, as they are the bread and butter of motion control and is an easy first step to learn.

You can either rip one from a toy car (or similar toy) a broken down VHS recorder, cordless drills/screwdrivers and so on - the size doesn't matter for experimenting :)
Regards,
Søren

A rather fast and fairly heavy robot with quite large wheels needs what? A lot of power?
Please remember...
Engineering is based on numbers - not adjectives

Offline notgivenTopic starter

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Re: quick driver current question
« Reply #8 on: August 10, 2012, 10:01:14 PM »
Here's the schematic.
http://i49.tinypic.com/5b5r1v.jpg

Also, I mistakenly put 3ohm in the picture where it should have been 3.3ohm resistor.

Yes. It's gotta be power resistors though.


Like this ?: http://www.radioshack.com/product/index.jsp?productId=12554561&filterName=Type&filterValue=5-watt+resistors

Why the hell are these things so hard to find? Radioshack sucks. I'll probably have to order those two little resistors.
Can I instead put 3 10ohm resistors in parallel to get 3.3 ohms? They are more likely to have  10 ohm resistors in store.


If you still have the connector on the motor, one resistor go to either pin 1 or pin2 (doesn't matter which) and the other goes to either pin 3 or pin 4.

What is the "connector" you refer to? Is it one arm of the resistor?

OK, have fun with it (it IS), but consider getting those dirty hands on a DC-motor as well, as they are the bread and butter of motion control and is an easy first step to learn.
You can either rip one from a toy car (or similar toy) a broken down VHS recorder, cordless drills/screwdrivers and so on - the size doesn't matter for experimenting :)


Could run it with the other driver I have on hand (L193D)? Would experienced people normally use a driver like that for running a DC motor? Or a simpler driver instead?
« Last Edit: August 10, 2012, 10:02:48 PM by notgiven »

Offline Soeren

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Re: quick driver current question
« Reply #9 on: August 10, 2012, 10:40:41 PM »
Hi,

Here's the schematic.

I have to stall the check till I get some sleep (it's 6:40 a.m. here and I need a little sleep).


Also, I mistakenly put 3ohm in the picture where it should have been 3.3ohm resistor.

Repenting sinners are always forgiven :P ;)


Like this ?: http://www.radioshack.com/product/index.jsp?productId=12554561&filterName=Type&filterValue=5-watt+resistors

Exactly!


Why the hell are these things so hard to find? Radioshack sucks. I'll probably have to order those two little resistors.

It's a test... to expel the unworthy :P
(Did I watch too many medieval flicks this evening/night? ;))


Can I instead put 3 10ohm resistors in parallel to get 3.3 ohms? They are more likely to have  10 ohm resistors in store.

Sure and then each would only have to be 1.5W (or more) then.


What is the "connector" you refer to? Is it one arm of the resistor?

Oh, I found a datasheet on the motor and the wires from it was terminated in a connector. If the motor you have, hasn't got a connector, you have to find the two coils with an ohm-meter or a continuity checker (or just an LED a 1k resistor and a 9V battery in series.


Could run it with the other driver I have on hand (L193D)? Would experienced people normally use a driver like that for running a DC motor? Or a simpler driver instead?

Yes, as long as it doesn't take more current than the driver can supply.

The L293D consists of 4 "half-bridges" (push-pull drivers) and two of them can be coupled as an H-bridge, so you can control two DC-motors bidirectionally with one L293D, or you can pair them up in parallel and get (almost) twice the current capacity (i.e. up to around 1.2A) driving a single DC-motor.

Each push-pull driver can either pull its output to ground (0V) or to V+, depending on its input. It can also be disabled (=high impedance) by the "ENABLE" pins, meaning it isn't connected to either, but "floating", which is important if you want a motor to coast (slow down over time, like if you just removed the battery). If both terminals of a DC-motor is pulled to either ground or V+, the motor will brake fairly hard.


Some datasheets on the L293D show how to couple them for DC-motors , both for uni- and bidirectional use.
Regards,
Søren

A rather fast and fairly heavy robot with quite large wheels needs what? A lot of power?
Please remember...
Engineering is based on numbers - not adjectives

Offline notgivenTopic starter

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Re: quick driver current question
« Reply #10 on: August 15, 2012, 08:21:58 AM »
Could anyone tell me if the schematic posted above would work?
Also, if I used a higher voltage battery, would I need higher ohm/ watt reistors? Or would the answer be no because the situation is that the motor is pulling current through the circuit as opposed to the battery pushing current through the circuit?

Offline Soeren

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Re: quick driver current question
« Reply #11 on: August 17, 2012, 08:26:17 AM »
Hi,

Could anyone tell me if the schematic posted above would work?
It wouldn't, you have mixed up some pins.
I have modified and attached a page from the datasheet to show where the resistors should go.


Also, if I used a higher voltage battery, would I need higher ohm/ watt reistors? Or would the answer be no because the situation is that the motor is pulling current through the circuit as opposed to the battery pushing current through the circuit?
'You would do yourself a huge favor by learning Ohms Law and Watts Law. You don't need to keep it all in the head, as long as you know the connection between V, A, Ohm and W - you can find a circle "diagram" that combines all the relevant (easy) formulas.

You need to limit the current in the stepper's winding to 1A.
The stepper winding is 0.75 Ohm.
The chip can drive the outputs to 1.4V from the positive voltage and 1.2V from ground (typical values at 25°C). This means that the voltage for your calculation is: VCC2-2.6V
To find the total resistance needed for limiting it to 1A (what the driver can handle), you use Ohms Law, stating that resistance is equal to voltage divided with current.
So:  (VCC2-2.6V)/1A

For a "12V" lead-acid battery (which is around 12.6V average, it would be:
(12.6-2.6) / 1 = 10 Ohm

To find the value of the limiting resistor, you just subtract the value of the motor winding: 10-0.75 Ohm = 9.25 Ohm

Since the aim is to protect the driver, you can not choose a smaller resistor, but a 10 Ohm resistor is only 8.1% higher than needed and would be fine.

You can (should) always do the reverse math, to find the current with the standard value resistor selected and by rearranging Ohms Law to current equals voltage divided by resistance, you can find the actual current:
(12.6-2.6) / 10.75 = 0.93A which would suffice for your experiments.

Then you need to find out how much power the resistor need to handle and for that, you need Watts Law (P = U*I), where P, U and I are the symbolic notation for watt, volt and ampere respectively). The resistor will have to drop 9.3V, as the motor winding, regardless of the battery voltage, is limited to: 0.93A*0.75 Ohm = 0.7V
With Watts Law, the power dissipated in the resistor is found to be: 9.3V*0.93A=10W

And the point to note is, that the more you raise the voltage, the more you increase the power in the resistor, while the motor power will stay the same (which is the inevitable result of limiting the current).

The solution to more power in the motor remains to be getting a driver that can handle the 2.8A (with a current limiter or better to avoid melt down if a winding is continually powered for several seconds).
Regards,
Søren

A rather fast and fairly heavy robot with quite large wheels needs what? A lot of power?
Please remember...
Engineering is based on numbers - not adjectives

Offline notgivenTopic starter

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Re: quick driver current question
« Reply #12 on: August 17, 2012, 11:24:05 PM »
Thanks again for helping me out Soeren. I have a few more questions about reading the schematic. A lot of the things I'm going to ask are for double-checking/clarification purposes as I've never done something like this before.

-The schematic specifies 24V supply at the top. Does that mean I MUST use 24V battery for this project? Or "12V" battery? Or am I still using 6V like I was originally going to?

-Why are we talking about 10 ohm resistors now when we use to be talking about 3.3 ohm resistors?  Is it because we went from a 6V battery to a 12V battery? Or was that post just a very thorough example for a different voltage from what I'll be using?0 (if so, thank you so much again for going through the basics for me like that. You can probably guess how enlightening it is for me)

-Where do the 4 wires labeled "phase" lead to?

-When you said before that "If you still have the connector on the motor, one resistor go to either pin 1 or pin2", the pins you refer to are the 4 motor connector pins right?

-Are the black triangles on the right side of the picture diodes? Shouldn't they be white triangles, or am I mistaken?

-So besides the obvious (microcontroller, driver, battery, motor, breadboard and hookup wires), according to the schematic I also need 8 diodes, a 10 kilo-ohm resistor, and 2 3.3 ohm resistors? Am I missing anything?

-Do I really need those diodes? Would it be a big mistake to leave them out considering that the motor is not running full capacity/strength? I ask because dealing with the diodes is an element of uncertainty for me and I would rather avoid them this time if I could.

-If I do need the diodes, do I connect them in series? Is that how the schematic is read?

-What kind of diode should I get? Will any do? How about this one :http://m.radioshack.com/radioshack/product/detail.do?itemId=13015070&categoryId=&path=
« Last Edit: August 18, 2012, 11:13:54 PM by notgiven »

Offline waltr

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Re: quick driver current question
« Reply #13 on: August 18, 2012, 12:44:10 PM »
Quote
-Where do the 4 wires labeled "phase" lead to?
They come from the controller. See the chips data sheet for a description of all of its pins.

Offline notgivenTopic starter

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Re: quick driver current question
« Reply #14 on: August 18, 2012, 04:33:21 PM »
They come from the controller.

The microcontroller right? if so, then yeah, I see one of the mistakes I made in my first drawing up there. It was a late night. Thank you.

Also, When it says "Phase 1" and then "/Phase 1" below it, does that mean the microcontroller should be connected to one driver pin OR the other? For a total of 4 outgoing wires from the microcontroller to the breadboard? Or should it be 6 total such that "Phase 1" and "/Phase 1" BOTH get a microcontroller connections?
« Last Edit: August 18, 2012, 11:06:24 PM by notgiven »

Offline Soeren

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Re: quick driver current question
« Reply #15 on: August 19, 2012, 08:28:34 AM »
Hi,

-The schematic specifies 24V supply at the top. Does that mean I MUST use 24V battery for this project? Or "12V" battery? Or am I still using 6V like I was originally going to?

It's just whatever motor voltage you use - 6V is fine.


-Why are we talking about 10 ohm resistors now when we use to be talking about 3.3 ohm resistors?  Is it because we went from a 6V battery to a 12V battery? Or was that post just a very thorough example for a different voltage from what I'll be using?0 (if so, thank you so much again for going through the basics for me like that. You can probably guess how enlightening it is for me)

You asked " if I used a higher voltage battery, would I need higher ohm/ watt reistors?", so I made you an example with a 12V lead-acid battery, to show you how you find the result for any given voltage.

I would recommend staying with 6V though, as I tried to convey in the last two paragraphs


-Where do the 4 wires labeled "phase" lead to?

To output pins on your microconroller.
Phases A and /A controls one winding of your motor and B and /B controls the other.
To make it move in pseudo code:
Begin:
Set PhaseA high and /PhaseA low
Short pause
Set PhaseB high and /PhaseB low
Short pause
Set PhaseA low and /PhaseA high
Short pause
Set PhaseB low and /PhaseB high
Short pause
Goto Begin

The speed of rotation will depend of the pauses between each step.
To reverse the direction, play them in reverse order.
This is just the basic full step operation, but learn this before even looking into half- or microstepping.
Diodes are important, as reversing the current through a coil gives you kick back voltage spikes in large amounts - the diodes keep your driver alive by shunting these spikes to safe levels.


-When you said before that "If you still have the connector on the motor, one resistor go to either pin 1 or pin2", the pins you refer to are the 4 motor connector pins right?

Yes. It's unimportant which you see as A and B though.


-Are the black triangles on the right side of the picture diodes? Shouldn't they be white triangles, or am I mistaken?

They could be rainbow colored and still represent the same thing, but luckily it was not a hippy making the diagram ;)
Yes, they are the diodes that keeps your driver alive.


-So besides the obvious (microcontroller, driver, battery, motor, breadboard and hookup wires), according to the schematic I also need 8 diodes, a 10 kilo-ohm resistor, and 2 3.3 ohm resistors? Am I missing anything?

The 10k resister is not needed in your case and a direct connection will actually be better, as it will tie harder to the positive rail, keeping noise solidly away.

You need the diodes and they need to be fast acting, a regular diode is way too slow to catch the spikes.
So called "(ultra)fast recovery" diodes, some schottky diodes and similar would work.


-Do I really need those diodes? Would it be a big mistake to leave them out considering that the motor is not running full capacity/strength? I ask because dealing with the diodes is an element of uncertainty for me and I would rather avoid them this time if I could.

-If I do need the diodes, do I connect them in series? Is that how the schematic is read?

-What kind of diode should I get? Will any do? How about this one :http://m.radioshack.com/radioshack/product/detail.do?itemId=13015070&categoryId=&path=

Yes, you absolutely need them.
You connect them as shown - each side of the coils need one diode to ground and one diode to Vcc2 (your motor voltage positive side).
Your link doesn't work (always check links in the preview), but the RatShack might not be the best place to shop anyway - they don't carry neither the BYV27(/100) which I prefer, nor the 1N5819 which would be another candidate.

Edit: Just realized that I used the terms PhaseA and -B here, just read them as Phase1 and -2 respectively (they're just names anyway).
« Last Edit: August 19, 2012, 08:59:15 AM by Soeren »
Regards,
Søren

A rather fast and fairly heavy robot with quite large wheels needs what? A lot of power?
Please remember...
Engineering is based on numbers - not adjectives

Offline notgivenTopic starter

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Re: quick driver current question
« Reply #16 on: August 19, 2012, 04:40:08 PM »
In place of the 3.3 ohm 4 watt resistor, if I connect 3 10 ohm 1 watt resisitors, will they be completely useless for protecting the driver? Or only midly useless? I know you said The 10 ohm ones ought to be 1.5watt or higher.

Also I couldn't find the diodes you specifically recommended (you're right, ratshck doesn't have them at all) but I did find this 1N914 silicon "switching diode" (http://www.oup.com/us/pdf/microcircuits/students/diode/1N914B_philip.pdf). What do you think of it?
It does not say ultra-fast recovery, but I think it should be OK since the datasheet says max switching speed = 4ns whereas the data on 1N5819 you mentioned was listed (http://www.chiplook.com/stock/1N5819.html, in Parameters table) as having Speed : Fast Recovery =< 500ns, > 200mA (Io).

Are the two datasheets talking about the same thing? (being fast-acting / fast recovery)
« Last Edit: August 19, 2012, 04:50:17 PM by notgiven »

Offline Soeren

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Re: quick driver current question
« Reply #17 on: August 19, 2012, 07:07:36 PM »
Mouser has got both BYV27 and 1N5819 (and the needed resistors) - 1N914 is good for only 100mA.

If you were to bake a cake, would you substitute the ingredients and expect the same result?
Regards,
Søren

A rather fast and fairly heavy robot with quite large wheels needs what? A lot of power?
Please remember...
Engineering is based on numbers - not adjectives

Offline notgivenTopic starter

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Re: quick driver current question
« Reply #18 on: August 19, 2012, 07:21:03 PM »
Nope, but I'm asking if you think it would be edible ;D
Yeah well I guess all I can do is wait till the order comes...
Look man, thanks so much for all your help, especially for helping me smooth over my screw-ups, and going through basic stuff with me. This is my first project and I hope you'll check out my working humanoid robot (much) later down the road.

Offline notgivenTopic starter

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Re: quick driver current question
« Reply #19 on: August 20, 2012, 07:22:59 PM »
I know its troublesome but could anyone take a look at this breadboard picture and tell me if I made the correct connections that the schematic called for?

I made the breadboard on Fritzing and tried using the schematic version of it to check agaist the correct schematic, and I think the connections are indeed correct, but they're too convoluted for me to be sure.

My breadboard is here:


Compare with:

« Last Edit: August 20, 2012, 08:25:51 PM by notgiven »

Offline notgivenTopic starter

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Re: quick driver current question
« Reply #20 on: August 22, 2012, 08:59:30 PM »
I'm updating my breadboard because someone pointed out that the way I had it, half the motor was being powered by logic +5V and the other half by motor supply.

I just moved one red wire on the right:
http://i50.tinypic.com/5xlnxz.jpg


Anything else wrong with my attempt to put the schematic on bread?

 


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