Using a 1K resistor:
V Bat= 7.57 V
V between Dim- and Dim + after the resistor 6.08
and 1.5V between ends of the series resistor
That’s what I wanted to know. So the DIM+ pin is drawing 1.5/1000 = 1.5mA. And that is at an applied 6.08V. It’s probably going into an NPN bipolar, and that would agree with the fact that (according to the ELN spec) you get no output for the first 1.1V input. That being so, the current with 10V applied would be 1.5mA x (10-1.1)/(6.08-1.1) = 2.68mA
That’s too much for a simple integrator, so a buffer will be needed., it would use the 5V PWM signal (to avoid any possible problems of using the 10V PWM) and combine an integrator with a X2 amplifier. The attached circuit should do the trick.
R1 and C1 deliver an analogue voltage of between 0V and 5V - depending on the PW duty cycle - to the non-inverting input of the opamp. The pulse rate of the Typhon is not given, but the 0.5sec time constant given should be OK. The loading on the Typhon is negligible.
R2 and R3 give the amp a gain of X2, so what appears at the opamp output will be an analogue voltage of between 0V and 10V depending on the PW duty cycle.
The maximum output for the CA3140 shown is 10mA so it can cope OK with the current required by the ELN. But in common with most opamps there is an overhead for the output voltage of about 2V. That means the supply voltage must be 12V minimum (you already have a 12V supply feeding the Typhon). In worst cases the overhead may be up to 3V so if you cannot get 10V out of the amp, it is cheaper to get a 15V power supply than to go for a rail-to-rail output opamp at considerably higher cost.
Any equivalent opamp can be used, provided it has a FET input that can accept input levels down to 0V level. JFET or bipolar inputs cannot.
I hope that sorts out your problem.
