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TFT Monitor Input Voltage Tolerance  
User currently offlineManuCH From Switzerland, joined Jun 2005, 3010 posts, RR: 48
Posted (7 years 4 months 3 weeks 21 hours ago) and read 2809 times:
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I'm having a discussion with someone about input voltage tolerance of a TFT monitor. The 15" monitor comes with a 12V DC, 3.5A power supply which is now broken (a condensator leaked the dielectric).

As an IT person who had quite some electronic classes, my opinion is that the power supply needs to be replaced with one having the same voltage, and the same or higher current.

Another person has a spare 16V DC, 3.36A power supply and says it will work with the monitor.

While I can't be sure because I don't know the monitor's internal components, I'm pretty much convinced that 16V will blow up some circuit in the monitor, because some electronic components only have a +/- 10% tolerance (that's 10.8V to 13.2V maximum). But there might not be components with such tight components, and it might as well work.

Unfortunately the monitor belongs to a third person, therefore I can't just go there and blow it up, then say "see? I was right"  Smile ...

Does anyone have any practical experience with the typical tolerance of input voltages on TFT monitors, based on the components typically used in such circuits?

Heck I know, the guy might as well buy a new power supply (or a new monitor), but it's now a matter of principle, as you might have guessed.

Thanks
-Manuel


Never trust a statistic you didn't fake yourself
20 replies: All unread, jump to last
 
User currently offlineQueso From , joined Dec 1969, posts, RR:
Reply 1, posted (7 years 4 months 3 weeks 21 hours ago) and read 2806 times:

I have been doing component-level repairs on monitors and other electronics for about 25 years. I would not try the 16v power supply.

Quoting ManuCH (Thread starter):
I'm pretty much convinced that 16V will blow up some circuit in the monitor, because some electronic components only have a +/- 10% tolerance (that's 10.8V to 13.2V maximum).

I agree with your 10% figure as a working guideline. It has been my experience that it's the practical limit you can work with unless you have specific information directly from the manufacturer that would give you reason to believe the higher voltage is acceptable.


User currently offlineAloges From Germany, joined Jan 2006, 8666 posts, RR: 43
Reply 2, posted (7 years 4 months 3 weeks 20 hours ago) and read 2799 times:

Quoting Queso (Reply 1):
I have been doing component-level repairs on monitors and other electronics for about 25 years.

I haven't, and my opinion is the same:

Quoting Queso (Reply 1):
I would not try the 16v power supply.

That would be a 33% higher voltage than what the monitor is designed to run on - not recommendable!



Walk together, talk together all ye peoples of the earth. Then, and only then, shall ye have peace.
User currently offlineTRVYYZ From Canada, joined Oct 2004, 1369 posts, RR: 10
Reply 3, posted (7 years 4 months 3 weeks 16 hours ago) and read 2786 times:

As per the safety standards for IT UL/EN 60950 the variation for voltage allowed is only +/-10% for dc and -10% +6% for ac voltages normally.

But if you want you can use two resistors with proper rating as a voltage divider and get 12 volts, but thats too much work. I am in this field, so i get crazy ideas to get around things.

[Edited 2007-02-20 04:19:54]

User currently offlineKlaus From Germany, joined Jul 2001, 21382 posts, RR: 54
Reply 4, posted (7 years 4 months 3 weeks 15 hours ago) and read 2778 times:

Quoting TRVYYZ (Reply 3):
But if you want you can use two resistors with proper rating as a voltage divider and get 12 volts

No!!!!

That would be about the second-dumbest thing to do after connecting the overvoltage supply directly!  crazy 

The resulting fluctuations would most probably cause just as much damage.

Get a matching power supply or leave it.


User currently offlineTRVYYZ From Canada, joined Oct 2004, 1369 posts, RR: 10
Reply 5, posted (7 years 4 months 3 weeks 15 hours ago) and read 2775 times:

Quoting Klaus (Reply 4):
That would be about the second-dumbest thing to do after connecting the overvoltage supply directly! crazy

I challenge you, what background do you have? You have a steady supply and if you use a passive network as a potential divider, how on earth can there be a variation?
You are imagining too much. The only problem he has is his second supply has a slightly lower current limit.

Don't think that you know too much. I deal with these things every day. It is only a monitor not a life supporting medical device .Be careful and think twice before calling something dumb. You make yourself one by doing so, but on this forum who knows.

Please check your circuit theory books. These are simple thing that we doing the 1st or 2nd year engineering lab.

[Edited 2007-02-20 05:03:35]

User currently offlineKlaus From Germany, joined Jul 2001, 21382 posts, RR: 54
Reply 6, posted (7 years 4 months 3 weeks 15 hours ago) and read 2767 times:

Quoting TRVYYZ (Reply 5):
I challenge you, what background do you have?

Design and review of mainly digital but also analog circuits is part of my work.

Quoting TRVYYZ (Reply 5):
You have a steady supply and if you use a passive network as a potential divider, how on earth can there be a variation?

Practically no electronic device is equivalent to a passive resistance. If it was, you wouldn't even need the lower branch of the divider - a simple series resistance would suffice. Unfortunately it doesn't work that way.

The changing load of the TFT display (depending on image changes and backlight setting) would draw a varying current which would lead to a changing voltage in the passive divider.

This would only be unproblematic if the consumer was asymptotically insignificant relative to the resistors in the divider, which unfortunately would be extremely inefficient.

Quoting TRVYYZ (Reply 5):
You are imagining too much. The only problem he has is his second supply has a slightly lower current limit.

No. The Problem would be that a) the passive voltage divider would waste a significant amount of energy and b) any change in the current drawn would lead to fluctuations of the effective input voltage to the TFT which would alternate between under- and overvoltage, most probably leading to malfunctions and even damage of the display.

I don't know what kind of experience with electronics you have, but I'd re-evaluate your theories if I were you.


User currently offlineTRVYYZ From Canada, joined Oct 2004, 1369 posts, RR: 10
Reply 7, posted (7 years 4 months 3 weeks 15 hours ago) and read 2762 times:

Quoting Klaus (Reply 6):

Practically no electronic device is equivalent to a passive resistance.

But the resistive network(to be inserted) is a passive device.

Quoting Klaus (Reply 6):
The Problem would be that a) the passive voltage divider would waste a significant amount of energy a

true not an efficient way to do.

Quoting Klaus (Reply 6):
b) any change in the current drawn would lead to fluctuations of the effective input voltage to the TFT which would alternate between under- and overvoltage, most probably leading to malfunctions and even damage of the display.

This would happen even if you use the rated power supply. If the values are chosen cosidering the loading effect, it won't be much worse than the original supply as the (modification) network is linear.

It was my idea, but not advisable for lay man but you and I know that it will work. Not efficient and not a great solution, agreed. I know that it would be tough to arrive at the resistor values considering the constraints associated with the display.

[Edited 2007-02-20 05:39:42]

User currently offlineDrDeke From United States of America, joined Jun 2005, 830 posts, RR: 0
Reply 8, posted (7 years 4 months 3 weeks 14 hours ago) and read 2741 times:

Quoting TRVYYZ (Reply 7):
Quoting Klaus (Reply 6):
b) any change in the current drawn would lead to fluctuations of the effective input voltage to the TFT which would alternate between under- and overvoltage, most probably leading to malfunctions and even damage of the display.

This would happen even if you use the rated power supply. If the values are chosen cosidering the loading effect, it won't be much worse than the original supply as the (modification) network is linear.

Uhh... Really? I would think that a "12 VDC, 3.5 A" power supply for an LCD monitor would be (actively) regulated to maintain a steady output of 12 VDC at any load between 0 A and 3.5 A, unlike the voltage-splitting resistor scheme.

DrDeke



If you don't want it known, don't say it on a phone.
User currently offlineTRVYYZ From Canada, joined Oct 2004, 1369 posts, RR: 10
Reply 9, posted (7 years 4 months 3 weeks 14 hours ago) and read 2740 times:

Quoting DrDeke (Reply 8):
I would think that a "12 VDC, 3.5 A" power supply for an LCD monitor would be (actively) regulated to maintain a steady output of 12 VDC at any load between 0 A and 3.5 A, unlike the voltage-splitting resistor scheme.

My question is even the 16V supply has the regulation and since the divider is linear won't it be coupled back? I know the resistor values will be critical for matching impedance. I would appreciate if you could do an analysis on that. I know there will be a difference in the current from the rated vaue and how much the variation will be magnified again depends on the resistance. It is not straight forward as the load is dynamic, I am also interested in the current transients caused by the changing picture on the monitor. I know the monitor is high voltage (low current?), but how much variation does it produce on the current drawn from the 12V source?


User currently offlineDrDeke From United States of America, joined Jun 2005, 830 posts, RR: 0
Reply 10, posted (7 years 4 months 3 weeks 14 hours ago) and read 2736 times:

Quoting TRVYYZ (Reply 9):
My question is even the 16V supply has the regulation and since the divider is linear won't it be coupled back?

Ahh, I see what you mean. That is a good question. Or a good point, if you meant it that way  Smile. I don't know enough to be able to say whether you are right or wrong, but it certainly seems plausible. It's enough to make me want to do some experimentation once I get my workbench set back up (it's not easy in this cramped apartment).



Quoting TRVYYZ (Reply 9):
I am also interested in the current transients caused by the changing picture on the monitor

Yeah, I don't know what the magnitude of these transients would be. I would think quite small. The current changes I would be worried about would come from the different power states of the monitor: "Off" (which is probably not _really_ off), "power save" mode, and full power.

-DrDeke



If you don't want it known, don't say it on a phone.
User currently offlineTRVYYZ From Canada, joined Oct 2004, 1369 posts, RR: 10
Reply 11, posted (7 years 4 months 3 weeks 14 hours ago) and read 2735 times:

Quoting DrDeke (Reply 10):

I know what you and Klaus said makes perfect sense in the worst case scenario. But as long as it is not so bad my optimism still got a chance.


User currently offlineManuCH From Switzerland, joined Jun 2005, 3010 posts, RR: 48
Reply 12, posted (7 years 4 months 3 weeks 14 hours ago) and read 2732 times:
AIRLINERS.NET CREW
HEAD MODERATOR

Wow, I didn't imagine this could turn into such an interesting discussion. Anyway, in case you're interested, more to the human background of this: the monitor belongs to a kid who doesn't have money to get a new one at the moment. I've found a replacement power supply for him for about $50 at an electronics mail-order place (yes I know, expensive place we live in - but in a store they can't be found under $100). There's a circuit-mount one for $30, but he doesn't have any experience in this field, and I wouldn't want to be responsible for his place burning down in case he messes with it (no, I don't have time to build a box around it).

A 50 year old guy who brags that he's been "studying electricity" (WTF?) for 30 years says the 16V supply will work (probably because you can power a 220V appliance with 224V ).

As I don't want the kid's monitor to break, I'm trying to get more arguments for him to listen to *my* point of view (which I'm glad everyone is sharing here, thanks ).

[Edited 2007-02-20 06:25:01]


Never trust a statistic you didn't fake yourself
User currently offlineDrDeke From United States of America, joined Jun 2005, 830 posts, RR: 0
Reply 13, posted (7 years 4 months 3 weeks 14 hours ago) and read 2730 times:

Quoting TRVYYZ (Reply 11):
But as long as it is not so bad my optimism still got a chance.

Yeah. And now I really am interested in the effects of the regulator in the 16V supply after being passed through a voltage dividing resistor network. There's nobody I can call right now who will just be able to tell me the answer, and I don't have the setup here to experiment with the idea at the moment. I think there are a couple other electrical engineer or technician or "advanced hobbyist" or whatever you want to call them, who hang out in non-av; maybe some of them will chime in.

DrDeke



If you don't want it known, don't say it on a phone.
User currently offlineTRVYYZ From Canada, joined Oct 2004, 1369 posts, RR: 10
Reply 14, posted (7 years 4 months 3 weeks 12 hours ago) and read 2723 times:

ok, i did the analysis the hard way. Assuming different values for resistors(variable load for monitor, 2 values picked) and constant voltage source.

The voltage(through the network) changes with change in (monitor)load current(even if the adapter voltage is regulated). It decreases when current drawn increases and viceversa. But here again the tricky part is the amount of variation can be greatly reduced by choosing a higher value of the second resistor that appears across the load(monitor).I still believe if the percentage variation of current( a rough idea) is known we can come up with values so that the voltage change is with in 10%. However we may need a source with a higher current capacity depending on the R values.

[Edited 2007-02-20 07:55:01]

User currently offlineTRVYYZ From Canada, joined Oct 2004, 1369 posts, RR: 10
Reply 15, posted (7 years 4 months 3 weeks 11 hours ago) and read 2713 times:

Ok, I did a few more calculations based on the rating of the monitor and powersupply and it doesnot seem that a high value of resistor can be used.(for a 16V supply to 12V to get 3.5A).Klaus is right (it was such a simple calculation, lazy me). I might have saved myself it the load was constant (but still I was having a high R value in my mind during the initial post) but very bad arrangement b'cos R is less than the load and it would be taking more power than the load(not sensible at all).
Conclusion: the idea is not good because I cannot use a high value of resistor. might have worked for a supply with much higher voltage (large ratio of the resistors) but not a good way to do.

Anyway a nice exercise to actually do some drill.

[Edited 2007-02-20 08:51:39]

User currently offlineKlaus From Germany, joined Jul 2001, 21382 posts, RR: 54
Reply 16, posted (7 years 4 months 3 weeks 7 hours ago) and read 2707 times:

Quoting TRVYYZ (Reply 14):
But here again the tricky part is the amount of variation can be greatly reduced by choosing a higher value of the second resistor that appears across the load(monitor).I still believe if the percentage variation of current( a rough idea) is known we can come up with values so that the voltage change is with in 10%. However we may need a source with a higher current capacity depending on the R values.

Actually, by increasing the "lower" resistor in the divider, you're approaching the case of a series resistor to the consumer which would have to be dimensioned so at the current drawn by the display it would just see the excessive 4V at its leads.

In effect, the series resistor would have to have 1/3 of the apparent resistance of the consumer so it would see 12V and the resistor 4V.

To achieve that with a dynamic load, you need an active regulation which basically varies the series resistor (in the form of a transistor) or with a 4V zener diode instead of the resistor. In all those cases, however, the resistor, transistor or zener diode would turn 1/3 of the TFT's power consumption into heat, so good cooling would be required to prevent a fire hazard.

This is why power supplies usually resort to more efficient regulation methods which only waste very little energy with (ideally) non-resistive "energy pump" switching methods.

Which brings us back to the practical solution for the problem:

A reduction from 16V to 12V would be possible with an intermediate active regulator (electronics stores may have such regulator modules available).

But you would usually have to do some installation work to put it into a case and fit the proper connectors.

If the display should use its own regulator internally to derive its actual working voltages from the 12V supply, it could possibly cope with 16V as well (if the internal regulator didn't run into its thermal limits by doing so); But I suspect that at least the backlight supply will probably work directly with the 12V input, so an overvoltage will most probably cause damage to the circuit and possibly the CCFL backlight itself, if not to the digital circuits as well.

So if the display says 12V+-5%, this limitation must not be exceeded.

The by far simplest solution would be just getting a fitting 12V power supply. They should be in abundance in every electronics store, even at the power rating required. Be sure to take the original power supply with you and ask an hopefully competent sales person if they have matching models.


User currently offlineWSOY From , joined Dec 1969, posts, RR:
Reply 17, posted (7 years 4 months 3 weeks 6 hours ago) and read 2700 times:

Quoting Klaus (Reply 6):
No. The Problem would be that a) the passive voltage divider would waste a significant amount of energy and b) any change in the current drawn would lead to fluctuations of the effective input voltage to the TFT which would alternate between under- and overvoltage, most probably leading to malfunctions and even damage of the display.

The easiest way to dissipate the volts is to use a number of silicon diodes in series with the load. Each diode will drop 0,7V, irrespective of currrent. At over 3 Amps, you also have quite a bit of thermal energy to lose so you'd need to build a heatsink. Another possibility is to use a single 3,9V Zener diode (25W type, not available in real life). I used to use zener diodes a lot to drop off a few wolts and clamp down on the noise of case fans etc. Simple and clear-cut!

[Edited 2007-02-20 13:35:00]

User currently offlineKlaus From Germany, joined Jul 2001, 21382 posts, RR: 54
Reply 18, posted (7 years 4 months 3 weeks 6 hours ago) and read 2693 times:

By the way, it's quite conceivable that the original power supply could be repaired by replacing the capacitor. Of course it may have killed other components as well on its way out, but then again, it might not have...

User currently offlineTRVYYZ From Canada, joined Oct 2004, 1369 posts, RR: 10
Reply 19, posted (7 years 4 months 3 weeks 6 hours ago) and read 2691 times:

Quoting Klaus (Reply 16):

well explained

Quoting WSOY (Reply 17):

Are there practical values of IC regulators for this?


User currently offlineWSOY From , joined Dec 1969, posts, RR:
Reply 20, posted (7 years 4 months 3 weeks 5 hours ago) and read 2684 times:

Quoting TRVYYZ (Reply 19):
Are there practical values of IC regulators for this?

Having leaking electrolytics is a very common fault. You can try replacing the cap, but it needs to be a switching power supply low-inductance type.
For more info try www.badcaps.net .

For a linear mode IC see: http://www.national.com/pf/LM/LM1084.html


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