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AlanM

Interested in Apex/RK controlled d120's?

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Sorry, had to head home. Just catching up. Where does pin 2 of the header go?

 

(Sent from my phone)

 

Here.  I tried to annotate my sketch a bit better.  I drew in the header and labelled the pins.  Pins 1-2 go to the two sides of the switch built into the pot and don't connect to the dimming circuit at all. 

 

The driver supplies 10V on pins 3-4 and I can dim the lights by putting a 20k pot across pins 3-4 to drop that 10V.  Max current across there is around 2.3mA at low resistance.  I posted the curve of what that looks like up higher in the thread.

 

post-2633353-0-77728000-1372818113_thumb.jpg

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Alan, that diagram helped. The markings on the transistor helped, too. The HF marking on an SOT-23 package corresponds to a 2SC3124 silicon NPN transistor. Typical DC current gain is 100 for this device.  Looking closer at your original drawing, I failed to notice the very, very small current in the "max bright" mode. The transistor is apparently in cutoff mode in this scenario. It is active in the "max dim" drawing. It appears that the circuit is acting as a buffered variable resistor swinging between nearly 493.5 kOhm and 3.65 ohms on the low end.

 

I'm not sure if the transistor circuit is really that important to the response to tell the truth. With a manually-adjusted circuit like this, linearity isn't all that important as we will adjust brightness as we see fit, not by the knob position, but by what we see.

 

Looking back into the VDIM+ and VDIM-, the circuit sees a load of about 4.3 to 5.0 kOhm.  You can safely drive this with an inexpensive op amp with good rail-to-rail performance. Given that the APEX variable outputs routinely drive dimmer circuitry, I don't think it's necessary to use an isolation amp here but can probably tie the circuit grounds together or through a very low resistance if concerned about common mode problems. I am guessing that VDIM- can be used as the circuit ground and VDIM+ would be the output of the unity gain op amp. You need access to a 10V supply from the main board. I'm guessing, but suspect that if you were to measure the voltage between pin  2 and pin 4 on the header, you would get a consistent 10 volts given the way the board is marked. Can you verify this with a measurement? Calling this V24, can you measure it with the switch open and with it closed? If V2 is a good 10 volt source, then it would be used as the positive rail voltage for the amp. Add a decoupling cap across the op amp power supply and maybe a clamping diode between the input and the positive supply (if one's not already built into the amplifier), and it should be able to adequately buffer the Apex from the LED driver.

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wow, you gentlemen know quite a bit about this stuff, eh? thank you so much for your time involved in this project - a lot of folks will benefit!

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Yes, I get like 20 microAmps at full on and 2.33 milliAmps at fully dim. 

 

OK.  So between pin 2 and 4 on the header I get 9.1V when the switch is off.  It's nice and stable. 

 

At minimum brightness with the switch on I get 11.83V across 2 and 4 and 0.652V across 3 and 4.

 

At maximum brightness with the switch on I get 11.98V across 2 and 4 and 9.85V across 3 and 4.

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wow, you gentlemen know quite a bit about this stuff, eh? thank you so much for your time involved in this project - a lot of folks will benefit!

 

Tom knows the things.  I'm just doing trial and error here and trying not to fry my VDM, heh.

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Try pin 1 to 4 also.

 

When you say "switch off" you mean that the switch is open, right? Or is it closed?

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Switch off means the lights are not on, meaning the switch is open.

 

Pin 1 to 4 is a bit weirder.  When you turn off the switch on the pot (open it), it flips from 11.83V which corresponds to min brightness to around -6V and bounces around there a bit between -6.2 down to -6.8.  Also, when checking it with the voltmeter the fan starts moving slowly.

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OK, so pin 2 sounds like it is the supply side and pin 1 is the load side. The supply side has some source resistance and that's why it's voltage drops a little at max dimming (when it's supplying current to the auxiliary board). So, all seems adequate enough to use.

 

Get a small op amp - a 741 from Radio Shack costs about $1.50. You'll run it with a single supply. Normally it runs cleaner, with fewer common mode issues using a dual supply, but let's ignore that for now. Connect the V+ (pin 7) of the 741 to header pin 2. Connect the V- (pin 4) of the 741 to header pin 4. Connect the output of the 741 (pin 6) to header pin 3.

 

Now, we want to connect the amplifier up as a unity gain amplifier. The simplest way to do this is to tie the output of the 741 (pin 6) back to the inverting input of the 741 (pin 2).  Apply the variable voltage control signal (10 volt range) from the Apex to the non-inverting input of the 741 (pin 3). Tie the Apex ground reference to header pin 4 (which we'll assume is the ground reference).

 

This would just be a test circuit to check the dimming when under control of the Apex. I would also consider adding a high-valued pull-up resistor (e.g. 100K) from the source (header pin 2) to the 741 non-inverting input (pin 3) and another (e.g. 20K) from the non-inverting input to header pin 4, so that when the Apex was not connected, the lights are on full-bright. This potentially sources a tiny bit of current (about a 0.1 mA) back to the Apex, but there is no reason to think that it can't deal with that since it routinely drives dimming inputs on other driver circuits which source a small amount of current.

 

A simpler way is to just connect the Apex 10V variable output to pin 3 of the header and the complementary side to Pin 4, set the light to max bright, and drive the sense circuit directly. In this case, 2.5 mA would flow back into the Apex control circuit. I have a hard time wrapping my head around something in the Apex driver circuit that couldn't put up with 2.5 mA so I think that this would still be safe. But, there is a (slight) risk in not knowing....

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I now get why it is called a unity gain amplifier. It isnt giving more current. It is just isolating the Apex side from the driver sense circuit side. I will risk the VDM by just hooking it up across the snipped wires which used to go to pins 3 and 4 to see what happens since I really did only see a couple milliamp through there. In that case I dont have to set it to max bright at all because the dimmer board is not even in the picture.

 

I dont think pin 1 and 2 are supplying current to the auxiliary board at all, by the way. When I look at the traces those two pins go right in and out. Unless it goes back in to the driver and out to the dimmer board.

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I think that it's connected to a supply on the driver side, not on the auxiliary side. The point was to get a supply voltage to drive the amplifier.

 

Personally, I think a direct drive is the easiest and probably still safe. But that's just a guess.

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I think that it's connected to a supply on the driver side, not on the auxiliary side. The point was to get a supply voltage to drive the amplifier.Personally, I think a direct drive is the easiest and probably still safe. But that's just a guess.

Will try it this evening.

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Well, we had a drawer of 741 chips, and lots of resistors, so I will try that first and then do apex directly.

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It works!!!!1!!1!!  The easiest way possible, like Tom thought it would.

 

Video below.

 

 

I just decided to risk sacrificing my VDM in the name of scientific progress.  I did what Tom suggested to just try hooking the darn thing up to the VarSpd ports on the Apex and drive them.  It works just fine.  In the video you can see my little multimeter measuring current on the LED string.  I flipped the outlet that activated a ramp profile to go to 0 to 100 in 1 minute.  Max current is 580mA (I hooked it up backwards in the video), which is right.  The min current is 170mA, so it doesn't dim as low as it does with the pot, probably because the Apex, has a little bit of internal resistance even when you're not supplying a voltage on the VarSpd ports. 

 

I have the d120 plugged into outlet 1 of my EB8 and called that one d120s and turned it off to turn it off all the way.

 

So it seems all you need to do to control it is to get a little 4 pin header, unplug the pot board, short out pins 1-2 to power it on, then hook up pins 3 and 4 to the + and - wires coming out of the Apex VarSpd port.  Or just snip off the connector and solder the wires together.  Ooh, or cut a little square in the metal panel on the top and put a keystone RJ45 jack, then you could just use a Cat5 patch cable to go between the Apex and the d120.

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Well done.

 

I like the Ethernet connector idea. Matches the aquabus using USB connectors with 12V to destroy standard USB devices.

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Alan, if you want to play around, try the op amp and see if you can drive the light down to dark.

 

(Sent from my phone)

 

 

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Alan, if you want to play around, try the op amp and see if you can drive the light down to dark.

 

(Sent from my phone)

 

will do. Good idea.

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If you mean how much current is on the dimmer sense line for the light, it's 2.5 mA or less. In typical notation, -2.5 mA to 0 mA, meaning the sense line sources a small amount of current to determine how much resistance is present on the line.

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The Photon/IT20** lights use the same drivers, so in effect this should possibly work for them too.

 

What's the safest way to short the 2 wires? Can we get a pic of the 4 pin connector from the driver with labels?

 

If it is that easy, I may possibly be able to splice into, or separate the DIM+ and DIM- wire while this is plugged into the IT2080 controller, and power it with the Reef Angel 0-10v dimming expansion.

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I will post pictures when I get back from my in laws house tomorrow. But some of the other pics on the thread here may help. There are pics of the dimmer boards at least.

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Tom,

 

With the op-amp hooked up as you describe above the minimum current is 0.25A on the LED.  It does ramp up to 0.58A similar to the direct hookup.  With it hooked up directly to the Apex the minimum is 0.17A.  I tried it with the 100k and 20k resistors and without them, and there was no difference either way. 

 

Here is a pic attached of how I hooked it up.  The red wire you see coming from pin 7 on the 741 goes to pin 2 on the d120 header (which is shorted to pin 1 to get the lights to turn on).  The red wire from pin 6 of the 741 goes to pin 3 on the d120 header and the black wire from pin 4 goes to pin 4 on the d120 header. The blue/white is 10V from apex, the green is ground from apex.

 

Do you see something wrong?

 

post-2633353-0-90048700-1373301530_thumb.jpg

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0.25 Amps? Or 2.5 mA? A quarter amp is unexpectedly high.

 

It could be offset current. The 741 is typically a split-supply device and may not work as nicely on a single-supply.

 

I see several wires coming in from different directions. Some old telephone cord... a black wire and a couple of red ones. What are they?

 

You could try pulling those two resistors and seeing if that changes anything off hand.

 

The output voltage on pin 6 should follow the input voltage on pin 3 of the IC. Do you see that behavior?

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