Need help with what may seem simple!

Thread Starter

Chris Azer

Joined Sep 23, 2015
8
Hey guys I am stumped with a simple circuit. I am trying to create a small circuit board that brings in a 0 - 10 volt signal
but need to output this same input voltage split to 4 channels with no voltage drop at each channel. The signal out put controls
the dimmer on my LED lights system that needs a 0 - 10 volt to adjust intensity.

Please help I have smoke coming out of my ears.
 

Thread Starter

Chris Azer

Joined Sep 23, 2015
8
it is DC input to DC output but I cannot have any voltage loss between the 4 channels as the will drive each led.
Also it can be digital or analog a can convert each fixture to accept either.
 

AnalogKid

Joined Aug 1, 2013
11,055
Input still is not clear. Is it a signal that can vary anywhere between 0 V and +10 V, or a pulse or trigger signal that has only two values, 0 V and 10 V, a pulse train or PWM, or something else?
Also, assuming that some active circuit will be needed to split/buffer the input to four outputs, what power is available for that circuit?

ak
 

Thread Starter

Chris Azer

Joined Sep 23, 2015
8
Sorry for no clarity guys . Input signal comes from http://www.digitalaquatics.com/downloads/ALC_user_guide_web.pdf
the signal is adjustable between 0-10 volts , It is the actual dimmer control. Each ballast requires the same 0-10 volt
signal to adjust the brightness of the led. What I am trying to do is remove the manual dimmer control on my LED fixture
so my ALC will control the brightness based on timer control. I can control 1 led fixture with this ALC at max 2 but the output voltage drops
and I will only get about 70% of the brightness of 2 led fixture.
 

AnalogKid

Joined Aug 1, 2013
11,055
Source: The ALC has two 10 V outputs. Is splitting each one to two outputs ok, or do you want to split one port to four outputs.

Destination: What about the dimmer? Manufacturer? Model number? Input impedance? I'm surprised that driving two in parallel drops the output voltage so much. That speaks to a relatively low input impedance, near the output impedance of the ALC. Any more information available? Also, how close to 0 V does the lowest output have to be? A simple voltage follower using something like an LM324 quad opamp might work if a) you have a power source for it that is at least 13 V); b) the dimmer input impedance is greater than 1 K (prefer something above 5 K); c) a minimum input of 0.1 V to 0.4 V is ok.

Trying to keep things simple with a single supply voltage rail for the quad buffer, getting the output down to 0 V without needing external buffer circuit added to each opamp means that the opamp must be able to supply all of the output current. Most opamps are not good at swinging close to the power rails, but the 324 is different and probably will work if we can tie down some details.

I've emailed Digital Aquatics for more information about their output.

ak
 
Last edited:

AnalogKid

Joined Aug 1, 2013
11,055
I didn't register just to see the images, but from the text is seems that using an opamp as a buffer/driver is a popular idea. Digital Aquatics tech support is working on my questions.

Here is a pre-pre-preliminary schematic based mostly on wishful thinking.

ak
LED-Splitter-1-c.gif
 

dannyf

Joined Sep 13, 2015
2,197
output this same input voltage split to 4 channels with no voltage drop at each channel.
The simplest would be four wires to four channels. Absolutely no "voltage drop";

The next would be four resistors to four channels, to isolate the channels from each other;

The next would be four followers to four channels, to better isolate the channels from each other;

The next would be some form of isolated amplifiers, to fully isolate the channels from each other;

...

You get the idea.

What you are looking for is called distribution amplifier.
 

Thread Starter

Chris Azer

Joined Sep 23, 2015
8
Yes by definition a distribution amplifier is what I need. But the output device sends a 0 to 10 volt signal to 1 led ballast when I connect 2 ballasts to the same output connection there is a voltage drop. With 2 ballasts connected I only get 6.7 volts at max setting giving me only 70% of the led intensity. The fourth option that you presented is what AnalogKid is trying to help me come up with.

Thanks for chiming in Danny.
 

AnalogKid

Joined Aug 1, 2013
11,055
Actually, I'm hoping for #3. The ballasts/controllers are fully isolated, so a common ground at the DA should not be a problem.

ak
 

dannyf

Joined Sep 13, 2015
2,197
In cases like that, your approach typically starts with

1) nature of the signal: is it providing power or just a control signal? is it digital or analog? ...
2) output impedance vs. input impedance: you need to get a sense how much current drive you have and need?
3) isolation or not.
...

Based on the limited information provided, I would say that something like a emitter follower (a npn + a resistor) per line should work. No reason (has been articulated) to go too fancy.

If the signal is referenced to rail, a pnp + a resistor.
 

AnalogKid

Joined Aug 1, 2013
11,055
I'm still waiting to hear back from the equipment manufacturer (it's been 16 hours; tick tock, people) about the true nature of the signals, impedances, currents, etc. It is a stepped analog voltage. In a private conversation with the TS I mentioned that one emitter follower *might* be able to do everything *if* it is biased upward with a diode to cancel out Vbe and all other signal characteristics play along.

ak
 

Thread Starter

Chris Azer

Joined Sep 23, 2015
8
I have the equipment coming in the mail this week and will test the output to get you guys as much info that I can from the ALC module.
 

AnalogKid

Joined Aug 1, 2013
11,055
The real question is about the input to the light. We know something about the relationship between the ALC output impedance and the light input impedance because of the voltage drop when multiple lights are connected. But that is just a ration between two unknowns. Of the two, the more useful in designing a buffer circuit like this is the actually input impedance of the load. So if you feel like ripping open a brand new light...

ak
 
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