Digital potentiometer over heating

Other Hardware Displays
1

Hi,
I am trying to use a digital potentiometer with my graphic LCD, I wired my AD5220 digital potentiometer as follows:
Vdd --> 5v power
B --> Gnd
CS --> Gnd
W --> Wiper pin 3 on graphic KS0108 LCD
GND --> Gnd
A --> contrast out pin 18 on graphic KS0108 LCD
U/D --> pin 1 set as output low
CLK --> pin 0 switches high and low to toggle wiper steps

With this wiring setup the digital potentiometer heated up quickly to the point I fear it might not work anymore. The data sheet for this potentiometer does not say to use resistors anywhere that I can see.
My question is how do I wire this potentiometer so that it does not heat up?

2

A --> 5v

3

Are you sure the digital pot still works OK? Pin 18 on the display is the negative voltage output, according to the datasheet linked from the adafruit page.

4

OldSteve:
Are you sure the digital pot still works OK? Pin 18 on the display is the negative voltage output, according to the datasheet linked from the adafruit page.

Hi oldsteve! I haven't used that lcd..... but assuming that contrast adjustment is similar in style to 16x2 lcds.... the usual style is an internal contrast pin voltage governed by an internal dc voltage source .... with a pullup resistor. The contrast pin hangs off the end of the pullup resistor..... therefore defaulting to a dc source voltage on that pin..... even without anything connected.

In order to change contrast... an external resistor to ground is required only. Adding the external resistor puts the contrast pin at the middle of the voltage divider created by the internal pullup resistor and the manually added external resistor.

Just took a look at the KS0108 datasheet a moment ago. But the datasheet I'm looking at is for the KS0108 chip itself. As for the whole LCD module (LCD plus chip), would be nice to find a reliable site that describes each pin (on the LCD module) - eg. input or output pin, etc.

For a {KS0108 chip + LCD} module, Vee appears to be a pin that will have a default voltage on it, set by (and INTERNALLY generated) the factory (or manufacturer). And that default voltage (which could be measured with a multimeter) will be NEGATIVE, with respect to 0V (GND).

If the LCD contrast pin is Vo, (assuming pin #3 of LCD), then connecting an external variable resistance across this pin (#3) and the stupidly-named Vss pin (which oddly happens to be the GND pin) allows the Vo pin voltage to be adjusted.

5

Southpark:
Hi oldsteve! I haven't used that lcd..... but assuming that contrast adjustment is similar in style to 16x2 lcds.... the usual style is an internal contrast pin voltage governed by an internal dc voltage source .... with a pullup resistor. The contrast pin hangs off the end of the pullup resistor..... therefore defaulting to a dc source voltage on that pin..... even without anything connected.

I'm fully aware of how a standard LCD and it's contrast adjustment works. You don't need to give me a lesson on it.

I was referring to the fact that this particular display has a negative output on pin 18, which should not be applied to the digital pot and so might have damaged it.

6

ardfar:
Hi,
I am trying to use a digital potentiometer with my graphic LCD, I wired my AD5220 digital potentiometer as follows:
Vdd --> 5v power
B --> Gnd
CS --> Gnd
W --> Wiper pin 3 on graphic KS0108 LCD
GND --> Gnd
A --> contrast out pin 18 on graphic KS0108 LCD
U/D --> pin 1 set as output low
CLK --> pin 0 switches high and low to toggle wiper steps

My question is how do I wire this potentiometer so that it does not heat up?

Just seeing if I can picture what's going on.

Wiper pin is connected to pin #3 LCD (aka contrast pin). That's fine.
One end of the pot (ie. A) is connected to pin 18 (aka Vee pin of LCD) ....and that pin might have a default negative voltage on it...such as -8 Volt. That's fine too.

Now.... the other side of your pot (B) is connected to GND. You connected it.

I think what I really want.... is to just have A connected to pin 18, while 'B' should be connected to nothing. ie....don't connect B' to GND.

Otherwise, if you keep 'B' connected to GND, then there's going to be some sort of 3-way circuit going on.

diagram.jpg
diagram.jpg1600×868 140 KB

Assuming.... just assuming that Vee has some internal default value of -8V. And assuming...just assuming that the internal ladder resistors are 2.2KOhm.

Then applying Kirchoff's Current Law to find Vo (contrast pin voltage) ...would be like...

(Vo - 5V)/(11.0 KOhm) + (Vo - 0)/(R-R1) + (Vo + 8V)/R1 = 0

Can solve for Vo, and then figure out how much power is being dissipated in each resistive portion of the potentiometer.

I don't know what your total potentiometer resistance (R) is. If R = 10000 Ohm, and if one side of the resistor is set to R1 = 2 KOhm..... then the other side of the potentiometer would just be 8 KOhm.

This circuit is just what I 'think' it might look like. Maybe you can try disconnecting side 'B' from ground....so keep 'B' totally disconnected.

7

Southpark:
One end of the pot (ie. B) is connected to pin 18 (aka Vee pin of LCD) ....and that pin might have a default negative voltage on it...such as -8 Volt. That's fine too.

He actually has 'A' connected to pin 18 on the display, not 'B'. And see below regarding the red bit.

Now.... the other side of your pot (A) is connected to GND. You connected it.

No, see above.

I think what I really want.... is to just have B connected to pin 18

Does he? The display outputs a negative voltage on pin 18. You cannot connect a negative voltage to the ends, (or wiper), of the digital pot. The voltage must be between Vdd and 0V.

From the digital pot datasheet:-

ABSOLUTE MAXIMUM RATINGS*
(TA = +25°C, unless otherwise noted)
VDD to GND . . . . . . . . . . . . . . . . . . . . . . . . . . . . –0.3 V, +7 V
VA, VB, VW to GND . . . . . . . . . . . . . . . . . . . . . . . . . . 0 V, VDD

Southpark:
Assuming.... just assuming that Vee has some internal default value of -8V.

Not just assuming. From the KS0108 LCD datasheet linked from the adafruit page:-

18 VEE Negative voltage output

8

My diagram was correct. And I merely forgot to re-edit my post after drawing the diagram. It doesn't matter which side of the pot you connect to pin 18.

Just keep the 'other' side disconnected.

9

Southpark:
My diagram was correct. And I merely forgot to re-edit my post after drawing the diagram. It doesn't matter which side of the pot you connect to pin 18.

Just keep the 'other' side disconnected.

You can't connect any part of the digital pot to a negative voltage.

10

To Whom It May Concern,
The cocept of the forum is that people who reply to an OP's post have some actual EXPERIENCE and actually KNOW what they are talking about, and not just posting because they have nothing else to do. Posting with advice when you have neither read a datasheet nor know how to read a datasheet (so far you have never quoted a datsheet) is a waste of everyone's time. So far you have yet provide any useful input on anything in any of your many posts, often stating the obvious and never showing any math or linking any reference documents . Often (as in this case), your "advice" is bad. If you have no experience then maybe you should find another hobby and let the "real" experts advise the newbies. The fact that you CAN post does not mean that you should. You have yet to prove you are an "expert" at anything. You may fool a newbie or two but you are not fooling the rest of us. Please try to support your statements with math or documentation or links in the future instead of just babbllng away with unsubstantiated opinion. Thank you for your understanding and cooperstion in this matter.
@OP,
TURN OFF THE POWER AND DISCONNECT THE DIGITAL POT FROM THE REST OF THE CIRCUIT IMMEDIATELY AND DO NOT RECONNECT IT UNTIL YOU HAVE IDENTIFIED THE CAUSE OF THE OVERHEATING.
I need some time to look into this. In the meantime test the pot to see it it is still hot after disconnecting it from all other circuitry. It it is still hot it is damaged and will need to be replaced.
I'll get back to you after looking into this.
You didn't tell us what value the pot is. What is the resistance value of the pot ?
What arduino are you using ?
Why are you using pin-0 ? Don't you have any other digital outputs available ?
D0 is the serial Rx pin

11

OldSteve:
I was referring to the fact that this particular display has a negative output on pin 18, which should not be applied to the digital pot and so might have damaged it.

Possibly. Can only see what happens later.

If those pot terminals can't handle negative voltages, then the OP could always try this: leave 'A' disconnected, and connect 'B' to Vss (0V). Wiper remains connected to pin #3 LCD (contrast pin Vo)

12

If those pot terminals can't handle negative voltages,

Really ? What do you mean by "IF" ?
Don't you KNOW ?
Did you read the datasheet ?
Have you ever used a digital pot before ?
What does it say on page 3 of the datasheet under Absolute Maximum Ratings ?

@OP,
Your mistake was connecting pot pin "A" to the LCD negative output voltage.
Pot pin-A should have been connected to +5V.
The pot is probably damaged.

NOTE: In your OP , you DID NOT explain your OBJECTIVE (PURPOSE) for using the pot. You simply stated you were using it, but not WHY. Obviously, if your objective was to control the contrast, you could have easily done that using a PWM pin output with a 4.7k ohm/4.7 uF RC Low Pass Filter .
If you insist on using a digital pot , the pot terminals MUST BE as follows:

Dimonira nailed it when he said:

A --> 5v

A- +5v
W- LCD contrast pin-3
B- GND.

The only advantage of using this digital pot to control the LCD contrast is that the pot has 12-bit resolution whereas AnalogWrite is only 8-bit resolution but the human eye is not capable of distinguishing the difference in contrast between 8-bits and 12-bits because 6-bits would probably be sufficient for a contrast sdjustment. Since you already have the chip , it it isn't already damaged, you can use , wired as shown above. Please disregard anything else said about this topic previously.
Make sure the LCD and pot share a common ground. Do not use the original digital pot chip until you have verified is not damaged.

13

My opinion is..... if that digital potentiometer has switches that switch ladder resistors in and out....and the two ends are merely 'two ends', then I don't see any problem with connecting any end of the digital potentiometer to any voltage (regardless of whether it is a positive one or a negative one). The main thing would simply be to not have the device over-heat and become destroyed.

14

Southpark:
My opinion is..... if that digital potentiometer has switches that switch ladder resistors in and out....and the two ends are merely 'two ends', then I don't see any problem with connecting any end of the digital potentiometer to any voltage (regardless of whether it is a positive one or a negative one). The main thing would simply be to not have the device over-heat and become destroyed.

You might see no problem, but the digital pot datasheet does. As I posted earlier:-

ABSOLUTE MAXIMUM RATINGS*
(TA = +25°C, unless otherwise noted)
VDD to GND . . . . . . . . . . . . . . . . . . . . . . . . . . . . -0.3 V, +7 V
VA, VB, VW to GND . . . . . . . . . . . . . . . . . . . . . . . . . . 0 V, VDD

Note that it says absolute maximum ratings. The low voltage for A, B and the wiper is 0V, not a negative voltage.

Anyway, here's a link to the adafruit page for the KS0108 LCD:- Graphic KS0108 LCD

And this is the equivalent datasheet that they link to:- http://www.vishay.com/docs/37329/37329.pdf

15

I don't see any problem with connecting any end of the digital potentiometer to any voltage (regardless of whether it is a positive one or a negative one). The main thing would simply be to not have the device over-heat and become destroyed.

My opinion is....

That's the problem. It is an uninformed opinion, as already pointed out by OldSteve.

Please do not post on this topic anymore. You obviously have no digital pot experience and it is questionable if you have any digital experience. The pot terminals cannot be connected to any negative voltages and for that matter, pin-18 is NOT the contrast adjustment so there is no reason to connect to it. If you do not know what you are talking about (from experience), please do not say anything.

3 V0 Contrast adjustment

18 VEE Negative voltage output

(see Reply#13)

16

I don't know anything about digital potentiometers but even with the self appointed forum police present I am going to offer some insight that may be of interest.

The typical way to connect a regular potentiometer to deal with the contrast on most character mode LCDs is to connect one end to the positive supply, typically +5 v, and the other end to GND. The wiper goes to pin 3.

Some extended temperature range character mode LCDs require a negative voltage (with respect to GND) at pin 3. This requires a second supply that has it's positive side connected to GND. The 'lower' end of the potentiometer is connected to the negative side of this second supply instead of to GND.

Since so many (all ??) graphical LCDs require a negative voltage at pin 3 the manufacturers of such displays frequently provide this second supply on the pc board with it's positive end already connected to GND and it's negative end at pin 18. Thus you connect your potentiometer between +5v and pin 18 and connect the wiper to pin 3.

If your digital potentiometer cannot deal with negative voltages then you can't use it in this situation.

One more point. Note that the datasheet specifications for the voltage at pin 3 are generally given with respect to Vcc, not with respect to GND. If you are using a 5v supply and this specification is more than 5v then you need a negative voltage at pin 3 when measured with respect to GND.

Don

17

That explains the need for the negative voltage and confirms that a 5V digital pot isn't going to work gor this. Now the question is how to get the negative voltage to the cotrast pin using a transistor or mosfet with a voltage divider. I don't have any experience doing this particular circuit but this post might help:
Using arduino to generate negative voltage

This post seems to be relavant

Here's another circuit

18

Why does anyone want to use a digital pot with a T6963C or even a KS0108 ?

The chip generates its own negative voltage. You adjust the contrast with a manual pot as a one-off operation.

Modern controllers might have voltages controlled by software registers. You still need a human to select the best settings for a particular display. These settings will be written when you initialise the display. And will be the same for a whole batch of displays.

I suppose that you could generate your own negative voltage for a KS0108 with the Arduino PWM instead of using its own negative generator. And adjust it in software.

This approach could also be used with a 16x2 running at 3.3V. Note that you need external diodes, capacitors, ... for the "voltage pump". Surely it is easier to just buy a suitable modern display in the first place.

David.

19

The chip generates its own negative voltage. You adjust the contrast with a manual pot as a one-off operation.

While it's true I did not specifically ask, I am hoping that the OP was aware that the contrast was meant to be controlled with a manual pot connected between GND and pin-18 (negative voltage output with the wiper connected to pin-3 (contrast adjust) but to be honest, I still haven't received an answer from the OP about what value digital pot he has or why he wanted to use a digital pot so at this point there are still some unanswered questions.

20

Although, level shifting can be done with the proper
divider, zener diodes can also be used, with less
loss of range.
One does wonder why he needs the digital contrast
control?
Dwight