Programmable Electronic Load - Analyse the Summing Node Zero Point

Latest measurements, with a 5V DC input:

The DAC outputs 0.980 mV offset. From value 0 to 100, it does not move an inch.

This is measured with the DAC not loaded, the output pin is floating (Behaviour in-circuit is the same).

The DAC is fairly linear from 100 on. Abruptly it forms a sharp knee when the linear ramp rises above the offset (DAC measurements on the right part of the image below).

To confuse you the first point in the graph is the current when the DAC is set to 110, last when set to 100.

For the voltage on the right, it's the other way around. This is the list with DAC set from 100 to 110.

This is to keep you alert.

The load pulls 0.519 mA (you don't see that precision in the graph on the left, but I consulted the data behind it) when the DAC output is 0.980 MV

From then on it rises fairly linear.

When I short the DAC output, the current drops to 0.0016 µA, virtually the same as when I disable the load (0.0010 µA).

It may turn out that the offset of the offset of the DAC is the cause for that low range knee after all....

I've prototyped a "programmable" offset compensation circuit. A differentiator with an OpAmp (I used an UA741 because it's the only DIP I have lying around).

We have 3 spare DACs in our design. I'm using the second DAC to pull the offset of DAC1 to 0 V.

This will also work if DAC2 has a higher offset as DAC 1. You just have to drive DAC1 to an equal or slightly higher value than DAC2's offset and call that 0.

You'll loose that part of DAC1's range.

I have Set DAC1 to 230. That's a point where it is performing good. Then I set DAC2 to 15 to set the output of the OpAmp to 0 V.

Each step of DAC1 from then on nicely increments the OpAmps output.

edit: I could use the free ADC4 to measure and fix that offset automatically, so that it's not a thing that you have to do when building the load?

So I did add capacitors.

The bodge board is attached to the underside of the underside of the driver section.

I repurposed the mounting holes as wire throughputs.

I don't need to cut traces on the original PCB.

I can remove R2 and use the 2 pads as input and output to the little bodge board.

The other wires can be soldered on existing pads.

The compensation circuit is now bodged in.

It's the wires on left and right that disappear through the mounting holes.  The other wires are there to probe the circuit when the LCD is mounted on top.

No success yet.

I am able to drive the output after R5 to 0 by playing with DAC 1 and 2, in esssence taking away the offset.

But any little voltage I go over that voltage fully drives the FET, as if it pushes the integrator off balance.

I 'll have to re-check connections, double check if I used the right resistor value for R5, measure the whole circuit probably.

The fact that I've been working and changing components on the PCB often doesn't help with reliability either...

... I should of course measure before R5 ...

This is very good on why it can make a difference

http://e2e.ti.com/blogs_/archives/b/thesignal/archive/2013/04/23/bypass-capacitors-yes-but-why

Jan Cumps  wrote:

 

No success yet.

I am able to drive the output after R5 to 0 by playing with DAC 1 and 2, in esssence taking away the offset.

But any little voltage I go over that voltage fully drives the FET, as if it pushes the integrator off balance.

I 'll have to re-check connections, double check if I used the right resistor value for R5, measure the whole circuit probably.

The fact that I've been working and changing components on the PCB often doesn't help with reliability either...

I removed the compensation circuit and all works as before again.

On one side that's good news - I don't have to troubleshoot the original PCB.

But something is wrong with my compensation circuit.

R2 gets the original DAC 1 outout,

R1 uses DAC2 output to compensate the offset at 0

R5 is connected to the summing node.

In the original design, the opamp was not there and DAC 1 was connected to the summind point directly via a 100K resistor:

At least that is simple enough circuit to troubleshoot fairly easy ... It worked with a 471 on a breadboard ...

jc2048, do I make a dummy mistake here with the impedances?

When I use the output voltage of a DAC, through a 100K resistor, the eload reacts linear(ish).

When I use that output through a *1 opamp, with a voltage on the - input for zero compenation.

The output of that opamp is a, through a 100 K resistor, to the same point as  in the original design.

As soon as the output of the opamp, measured before the 100K resistor (same point as where I measured the DAC voltage) reaches 0, the load goes to full drive.

I can't see anything obvious.

The output of the op amp is a low impedance (much lower than the 100k). The op amp output open-loop isn't more than 100R at low frequencies and the feedback should bring it much, much lower. If you see a static voltage at the output it shouldn't be any different to what you see with the DAC providing that voltage. Is it static? What does a 'scope show?

I found it: all resistors on my bodge board are 100 Ohm instead of 100K. They overshout the 100K feedback on the summing node.

my envelope witk 100K resistors contained 100 Ohm resistors.

undo

undo