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Topic: Digital potentiometer with op amp (Read 2132 times) previous topic - next topic

borillion



Hi I'm to take a 5V 500mA power supply and pass it  through a digital potentiometer AD5235 and then step it up through an op-amp with a gain 2 to get up to 10V output and put that into voltage based meanwell dimmer controller.

I'm a little confused on how to handle grounding with this. They are circled blue. Can I "ground" the -5V from the pot, and the -10V from the dimmer and the opamp together without damaging the pot by connecting them on the power supply -5V terminal of the wall wart l'll be using?

Also I have a couple questions about the pin meanings:
WP - Optional Write Protect. When active low, WP prevents any changes to the present contents
PR - Optional Hardware Override Preset. Refreshes the scratchpad register with current contents of the EEMEM register.  PR is activated at the logic high transition.
RDY - Ready. Active high open-drain output

what is active low vs just low, why the verb active?
what do Logic high transitions and active high open-drain output mean exactly?  :smiley-red:

Thanks you for your help!

dc42

#1
Jan 17, 2013, 01:43 pm Last Edit: Jan 17, 2013, 01:45 pm by dc42 Reason: 1
If you are driving that from an Arduino, then there is a much simpler solution. Take a PWM output pin, pass the output through an R-C filter, then use exactly the same op amp configuration to double the output from 0->5V to 0->10V.

To answer some of your questions: "active high" means the signal does what it is supposed to so when it is HIGH (i.e. voltage neat +5V or whatever Vcc is), and doesn't when it it LOW (near 0V). Active low is the reverse. So "RDY - Ready. Active high open-drain output" means that the output is HIGH when the device is ready.

"Open drain" means that the output pulls the output low, but not high. So you need a pullup resistor to Vcc to actually get a high on the pin.

"Refreshes the scratchpad register with current contents of the EEMEM register.  PR is activated at the logic high transition." means that when this input changes from LOW to HIGH, it refreshes the scratchpad register.
Formal verification of safety-critical software, software development, and electronic design and prototyping. See http://www.eschertech.com. Please do not ask for unpaid help via PM, use the forum.

SirNickity

Quote
WP - Optional Write Protect. When active low, WP prevents any changes to the present contents


That is simply bad wording in my opinion.  As already stated, the phrase "active low" means that the pin does whatever (i.e., it's active) when the input is low.  So there's no state called "active low."  There are only high and low states, and the pin's function is either activated by being high or low.  To put it one more way:  It can't be active low, it can only be activated when low.

borillion

That's what I figured it activates a function being high or low but wasnt sure.

As for the rc filter would it look something like this? Course I would be replacing Vs here with the pwm signal source and connect the end grounds to the -5V line on the wall wart Ill be using for power? Can I leave out the resistor and cap I don't understand why would need a filter?

Sorry if I'm being stupid but this is the first time I've dealt with pwm and not put it directly to and led or servo.


majenko

Just a note on your terminology...


A 5V wall wart doesn't have a -5V line.

You have a +(whatever)V connection, and a GND connection.  The same goes for the Arduino, the op-amp power supply, etc.  You have + and GND, not + and -.

For a 5V supply you have +5V and GND.  For a 10V supply, it is +10V and GND.

If you had +5V and -5V, then the difference between them would be 10V!  (ouch).

Also, I assume you have a 10V power supply to run the op-amp from?  You do realise that you can't power the op-amp from 5V and expect a gain of 2 to give you 10V out of it...?

dc42


As for the rc filter would it look something like this? Course I would be replacing Vs here with the pwm signal source and connect the end grounds to the -5V line on the wall wart Ill be using for power? Can I leave out the resistor and cap I don't understand why would need a filter?


Yes, it would look like that. You need an op amp whose input common mode voltage goes down to ground, and whose output goes close enough to ground as well. Best is a rail-to-rail op amp, but 1/2 a LM358 or 1/4 a LM324 will do if you use a 12v supply. Making Ra and Rb quite low (say 1K each) will help ensure that the output goess close enough to zero.

The RC network is needed because the dimmer expects a voltage between 0V and 10V, not a 10V PWM digital signal. Values of e.g. 10K and 22uF should be OK using the Arduino default PWM frequency.
Formal verification of safety-critical software, software development, and electronic design and prototyping. See http://www.eschertech.com. Please do not ask for unpaid help via PM, use the forum.

retrolefty

#6
Jan 18, 2013, 01:48 am Last Edit: Jan 18, 2013, 01:54 am by retrolefty Reason: 1

Just a note on your terminology...


A 5V wall wart doesn't have a -5V line.

You have a +(whatever)V connection, and a GND connection.  The same goes for the Arduino, the op-amp power supply, etc.  You have + and GND, not + and -.

For a 5V supply you have +5V and GND.  For a 10V supply, it is +10V and GND.

If you had +5V and -5V, then the difference between them would be 10V!  (ouch).

Also, I assume you have a 10V power supply to run the op-amp from?  You do realise that you can't power the op-amp from 5V and expect a gain of 2 to give you 10V out of it...?


I don't know if there is any really accurate terminology one can use with DC wall warts that doesn't either make assumptions or needs context to be understood correctly.

 Certainly a 'wall wart' type DC voltage source has both a positive and negative terminal (just like a battery) usually terminated into a coaxial DC connector, but no standard as to if center pin is positive or outer shell is positive, they come in both flavours. And one is free to connect either the negative terminal to one's circuit common and that results is a positive 5vdc available to power the circuitry, but of course there is nothing wrong with connecting the positive terminal to one's circuit common when they do require a -5vdc voltage source as for perhaps a op-amp's negative rail. But it's certainly true that a 5 volt wall wart doesn't have both a +5vdc terminal and a -5vdc terminal, at least not at the same time.  ;)

And the term 'ground' is such an overloaded term that can have several meanings and is a major pain in the butt to determine what one really means without the specific context being explained. I prefer the term circuit common or just common, over the term ground. And calling it GND is no help as that is almost impossible to pronounce. But then again I'm a crusty old hardware type that has his idiosyncrasies.  ;)

Lefty

SirNickity

Sure a wall-wart can have both positive and negative pins!  XD  I have one originally terminated on mini-DIN with +12, -12, +5, and gro--- uh, common.  ;-)

Granted, that's not what one typically imagines when thinking "wall wart", although I think discarding preconceptions when dealing with 'warts is probably not a bad thing.  After all, is it center-positive?  Center-negative?  Regulated?  50% voltage rise under no load?  AC??

Plugging in some random black box and expecting 5.0vDC on the center pin of a barrel connector is a good way to produce magic smoke.

borillion

Thanks Guys, I got this down and understand it now!

Ill use the circuit here


retrolefty


Thanks Guys, I got this down and understand it now!

Ill use the circuit here




Nice drawing. You missing a series resistor wired between the PWM input and the cap and + op-amp input to form a proper low pass filter. 10K ohms should work.

Lefty

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