Your logic seems sound. Keep in mind, if you measure less than the manufacturer says, he is quoting worst case, and that is what you need to plan for.

But your math is sound.

You laptops will have specifications, those are a safer number to use.

Q2 using a 2N3055 will still burn up (be hot), but it is designed for a heat sink. You need to add one.

 

Ah, finally something I might be able to help with . . .

12.3A each? Are they really that high? Most I've seen are around 5-6A.

In any event, yes, you can use the following:

Vout x Iout / Efficiency = Input Power Required -> Input Power / Input Voltage = Input Current required

So,

19.5VDC x 12.3A = ~240W

I'd guess most switching supplies are about 85% (some higher, some lower), so we'll use that:

240W / 0.85 = ~283W

283W / 220VAC = ~1.3A for each laptop

16 laptops x 1.3A = 20.8A

I'd suggest two 15A circuits (receptacles) each on their own 15A breaker with nothing else on the circuit, e.g., other computers, TVs, lights, etc.

This is a worst case scenario based on the values you gave. 12.3A still sounds awfully high. If the supplies are indeed rated for this, they aren't likely to pull the full current rating 24/7. However, as it is possible all 16 supplies COULD pull the full current rating, better safe than sorry.

You might also look on the back or bottom of the laptops and see if there is a current rating on them as well. That may be lower than the supplies and if so I'd use that number since the laptop is the load.

Where did you measure 350mA, on the DC side (output) or AC side (input) of the power supply? Again, it's unlikely to pull the full amount around the clock, but based on a 12.3A rating, it will pull more than 350mA throughout the day.

If the laptops draw 6A, as an example, then you only need one 10A receptacle for all 16 (though I'd suggest a 15A one).

Hope this helps.

 

Just remember, this is a Australian outlet, 220VAC out, amps unknown (by me).

I didn't mean to ignore you earlier, if I had seen transistors instead of a chip I'm not very familiar with...

 

Just remember, this is a Australian outlet, 220VAC out, amps unknown (by me).

This is why I divided by 220 at the end:

283W / 220VAC = ~1.3A for each laptop

By this, I mean the maximum current that would pass through the outlet being drawn by a single laptop. Sorry for not clarifying.

To my very limited knowledge, countries with 220VAC outlets have 15A outlets? The OP mentioned 10A too, so maybe they have those too? In the U.S., 15A is standard for our 120VAC outlets and we can go up to 20A outlets so long as the breaker at the electrical panel is also 20A.

I didn't mean to ignore you earlier, if I had seen transistors instead of a chip I'm not very familiar with...

<Sniff . . > It's okay, I know you've been busy with the OP on the dimming circuit. Still would love your thoughts on that if time allows later - I figure it is a subject that the OP will come back to and I don't want to interrupt your work on the dimmer.

 

Elec_mech, so good to hear from you again.

Yes, our power supply is actually 240V. Our typical GPO (general purpose outlet) is 10A. Electric stoves, airconditioners, etc use 15A GPOs.

The laptops we ordered were really high spec'd - Dell 6600's. They have really fast processors as well as video cards - hence the extra power made available for them. We wanted them to handle 3D modelling, rendering and animation. AutoCAD, Revit and 3D Max are the major programs we use.

The bottom of the laptop also says 19.5V, 12.3A. We order directly from Dell.

However, I measured the current on the 240V side of the AC adapter whilst idle (i.e. re-charging) and it only went up to 350mA. I'm guessing the battery and AC adapter is large to cater for if we really get the processor and video cards working extremely hard. So during charging, maybe say 500mA x 16 = 8A is all they would draw. Is that correct?

 

Diag 1: Q2 - TIP3055, you need the power package

I've just tried the 1st circuit and all components are keeping cool (including the Q3/Q4 TIP147 Darlington) - so good so far.

Q2 however is getting really warm with only the 7 test LEDs connected. Only when I dim them (by covering the LDR) does the temperature go down.

I know you mention a heat sink, but if it does this with only 7 leds connected, will it be the same (not worse) with 5400 LEDs connected - 10A)? I measured 150mA going through emitter of Q2. This is in line with your note saying it should be 3.5W (i.e. 3.5/24V = 0.145A). Lowest value around 15mA.

I assume when loaded the Darlington will heat up as well??? needing a heat sink.

 

I put the wattages on the schematic, they are worst case numbers. 4W (rounding off) is a heck of a lot of power, and it is the amperage you are pulling that is heating things up. So yes, it will get uber hot. And Q4 is much worse, 10W. It could be used as a minor heater.

The 2nd schematic won't do that. You might blow Q3 if you pull too much power, but it won't get hot. It is a more efficient design.

 

Bill.
I've just made circuit 2.

Works well - everything with the 7 leds at this stage is dead cold (except R1 and R2 - which are just a little warm). I calculate the current through these is 10mA. So this equates to 24x0.01 = 0.24W. Maybe 1/2W resistors would be safer.

By the way, can R1 and R2 be either 1K or even 1.22K (as this is easier to make up out of standard resistors)?

One thing that doesn't work however is that it is in reverse. Covering the LDR makes the LEDs brighter. How come?

 

Nope, I calculated current too, each resistor is 1/8W, the total wattage for both is ¼W, right where I wanted it.

The real test is with a full bank of LEDs, at 10A or so.

 

Oh, yes of course. Each resistor is 1/8W - total of 1/4W.

However, the circuit doesn't work. Not sure if I've done something wrong. When I covering the LDR, the LEDs go brighter rather than dimmer as usual. Is something back to front?

 

Try switching the bottom and top resistors (R1 and R3) and moving pin 5 (out) to the top of the LDR. This will invert the signal.

...........................................................Original Schematic

I thought I had looked at that, guess I got it wrong.

 

Bill, I could swap that, however I'm not sure why so far the circuit has been working. With circuit 1 it also worked. It's only circuit 2 that seems to invert it for some reason.

 

It has to do with how the transistors switch.

..................Circuit #1......................................Circuit #2............................................Circuit #3

The analog circuit in Circuit #1 will swing toward ground as the LDR reduces resistance. Its curve is determined by all three resistors, assuming R2 and R3 are one resistor and R4 and R5 are one resistor. Moving the location of out to above the LDR and swapping R2/R3 and R4/R5 will keep the same ratios, and since the circuit is symmetrical, the curve will simply invert, which is what you want.

Circuit #2 when in input to Q1/2 goes high the transistor turns on, which sends a current to Q3/4, turning it on. R2 is meant to turn Q3/4 off when Q1/2 is off.

Circuit #3 uses a P-Ch MOSFET. I was thinking N-Ch MOSFET, which is the source of my original confusion. When the gate of Q3 goes to the same voltage as the source Q3 turns off. This happens when the ouput of the comparator opens, and the inputs to Q1/2 go to 24V. It takes 10V between the gate and source to turn Q3 on, I designed for 12V (which is reduced to 11.6V after Q1/2). The end effect is what turns Circuit #3 on turns Circuit #2 off, and visa versa. In other words it is inverted.

Don't forget, R3 on Circuit #3 should be as physically and electrically close to the gate as possible. This is important.

Since the PWM convertor is symmetrical inverting the signals, both analog and digital, is no big deal.

 

So there is logic then to the 2nd design - not just me making a mistake.

Thanks for the detailed explanation of how it works.
I'll do the inversion as you suggested then.
I am leaning towards this 2nd circuit.
It's really late here now, so I'll test it tomorrow with more current.

By the way, I checked my 24V supply. There is an adjustment (small pot) allowing me to range from around 20V up to 30V. Can I then use this to tweek the voltage up slightly (less than 1V), so that in the end I get 24V at the commons of my digits? This will give me the brightness I was originally trying to design for.

 

Just change the current limiting resistors on the LEDs. I was going to re-recommend circuit 3, just be aware that Q3 itself is static sensitive. If you haven't read my ESD article this is a good time, but don't obsess about it. Just be aware if Q3 dies for no apparent reason this is probably why.

If you want to go over constant current sources after this part is done I think I can give you some ideas to replace a simple resistor. The increase in complexity may not be worth it, but the intensity will be a lot more constant (it may there already though, resistors work).

The reason I like #3 so much more is it is much more efficient. That heat comes from somewhere, and that is the power supply. Not to mention heat is a pain to get rid of, it doesn't go away. I don't know, but I am under the impression OZ is hotter than Texas (maybe not, we ran 30 or so days 115°F (46°C) last year, a record). This year is much better, but we still regularly exceed 100°F (38°C).

For now I would leave the power supply at 24V, maybe 25V to cover wire losses.

I've said this before, but you are coming away with quite a bit of electronics knowledge and experience out of this. I like your stubbornness and stick-to-itiveness, I'm not sure I could do it. This dimmer circuit definitely strikes me as marketable. The rest of the project could be dramatically simplified with PICs or equivalents, but if it works...

 

Just change the current limiting resistors on the LEDs.

I would but unfortunately I've finished building all 15 digits (with 52 resistors each) - 780 resistors to change now. I think I'll go with what I have.

be aware that Q3 itself is static sensitive

I never know what's sensitive and what's not - including how sensative. Intersting, my 40 pin 7107 IC that I'm using for my temperature circuit just suddenly stopped working correctly some time back. I replaced it and the circuit worked again. I'm guessing I ESD was the cause. I'll have to be a bit more careful. PS. Your ESD article was a good read. Thankyou.

If you want to go over constant current sources after this part is done...

Are you steering towards a voltage regulator? I've pre-empted and already bought a few LM317's to enable me to use the one power supply (24V), to feed the other circuits namely:

• 5V for the temperature circuit
• 5V for the PWM circuit
• 9V for the score circuit
• 9V for the count down timer circuit
• 9V for the clock circuit

Maybe all the above can run on 5V? Haven't tested that yet.
I was going to model the voltage regulator on the following circuit:
http://www.circuit-innovations.co.uk/LM317.html

The reason I like #3 so much more is it is much more efficient.

I agree completely, hence the reason I'm pushing to go with this option. In summer our high temperatures tend to range up to 35°C'sh. Our highest ever too was 46°C in 2009 - but rare in Melbourne. I'm envisioning this scoreboard enclosure to get fairly hot, so I'll be building in ventilation at the base, top and maybe a fan as well.

For now I would leave the power supply at 24V, maybe 25V to cover wire losses

25V to account for wire losses - sounds good. Done.

I think the dimmer circuit is brilliant in its simplicity with such common components - hopefully it proves robust. As for the other circuits - yea, a PIC could have replaced stacks of ICs, but this way my circuits can be built by anyone (without programming knowledge or equipment).

I'm getting closer every day. Thanks tonnes for your continual help (and mech_elec as well)

 

Side trip (not about the score board).

There are two fundamental kinds of regulators, a voltage regulator and a current regulator. I think you are familiar with a voltage regulator. The link you pointed to looks good.

A current regulator tries to keep the current in the circuit a constant value, perfect for LEDs. Lets say you have a 20ma regulator. If you short the output you will get 20ma out. If you put a 27Ω on the output the voltage will adjust to whatever it needs to be to provide 20ma (in this case 0.54V). With 1KΩ you would get 20V out.

Of course, where the regulator breaks down is when it can not go high enough voltage to regulate the current.

With LEDs this is a perfect environment, it eliminates all variation in components, Vf simply does not matter any more.

One of the simplest current regulators is a chip you already have, the LM317.

If you go to Chapter 2 of my LEDs article it explain how to use them, along with other designs.

I have started an album just on LED drivers.

LED Drivers

BJT are fundamentally current regulators to begin with, you can use this to your advantage. One of my favorite designs for driving a large bank of LEDs looks something like this...



What ever current you set through Q1 is duplicated in the rest of the transistors. This design is analog, and analog generates heat (something you have a handle on).

There are digital constant current sources. A digital regulator of either variety is a SMPS (Switching Mode Power Supply). The are a bit more complex, but the advantage is they use all the power, they don't throw it away as heat, except for accidental losses that always come with things man made. So if you are feeding 12W to a bank of LEDs it may only use 13W. Their complexity denies them common use however, but I have seen them used in flashlights (where battery life matters).

BTW, my personal cost for ¼W resistors is 2¢ each. If you ever need any just ask.

 

Yes, our power supply is actually 240V. Our typical GPO (general purpose outlet) is 10A. Electric stoves, airconditioners, etc use 15A GPOs.

Okay, that is helpful.

The laptops we ordered were really high spec'd - Dell 6600's. They have really fast processors as well as video cards - hence the extra power made available for them. We wanted them to handle 3D modelling, rendering and animation. AutoCAD, Revit and 3D Max are the major programs we use.

The bottom of the laptop also says 19.5V, 12.3A. We order directly from Dell.

However, I measured the current on the 240V side of the AC adapter whilst idle (i.e. re-charging) and it only went up to 350mA. I'm guessing the battery and AC adapter is large to cater for if we really get the processor and video cards working extremely hard. So during charging, maybe say 500mA x 16 = 8A is all they would draw. Is that correct?

Are you asking if the max draw will be limited to 8A and thus if you can get away with a single 10A circuit for all 16 laptops? If so, then no.

You said these were purchased with the intent to run power-hungry programs and include high-end processors and video cards. These are going to consume some serious power. I would not assume the battery charging current to be the largest power draw, otherwise Dell would've used 2A chargers and laptops which would cost them a lot less than 12A ones.

In this case, I'd refer you back to my previous calculations and use two dedicated 15A receptacles, each with 15A breakers and not shared with anything else. Eight laptops per receptacle. This is your safest bet.

 

Sorry, I think I wasn't clear regarding the charging.

At the end of the day, these laptops will be placed in a cabinet and left to re-charge overnight. This will be when they are off. I measured one drawing around 350mA when off and re-charging. The laptops will never be in operation when in the cabinet. Hence I'm assuming 350 x 16 = around 6A.???

When they're being used however, then they will draw alot more. The GPOs they'll be run off will be a completely different set and size appropriately.

 

If you go to Chapter 2 of my LEDs article it explain how to use them, along with other designs

I did take a look at this before buying the LM317's (just by chance). Your article is what made be decide to go that way to regulate the voltage.

My 24V supply is a "switching" power supply. Is that classed an SMPS?

Your cost for 1/4W are really cheap. I did however buy 1000 units for my digits for only $10 from rockby electronics. Reason I think is because I bought a full box.

Back on the PWM for a minute if I can. I wired up circuit #2 with a few adjustments to POTs and got the intensity changes I was after. No heat anywhere so far. Excellent. I did however pop an electrolytic experimenting a bit - boy what a bang!

I've only hooked up 2 digits so far (so max of 1.4A) - via my temperature sensor circuit. The problem I'm seeing is that when the LEDs are "dim", then the ones that are supposed to be fully off, are "just glowing". Just as a reminder, the digits are common anode. The PWM feeds the common. The segments are fed directly via the ULN2003a's. It "feels" as if I need to use a pull up or pull down resistor somewhere. Any ideas?