Temperature stable diode with low Vf. Any suggestions?

[gillestugan]

Im trying to put together some drivers based on the AX2002 chip.
To be able to use different output levels for several drivers Im using a resistor grid. Problem is: I need to use diodes to prevent current taking its own way through the resistor grid, and those I've tried have been awfully unstable in their Vf resulting in uneven output from the drivers.

here are the conditions:
-Low Vf (as it will be battery powered)
-Temperature between -20 and +50 degrees C
-If between 50 and 250 uA

Does anyone know of a suitable diode?

Thanks.

 

[VidPro]

ii use a big fat shotkey diodes, they have .3v drop, and after hearing and seeing all the diodes crack and split over TIME, i always use a big fat one, like 3 amps for <1 amp projects. which is easier with any diode that has minimal voltage drop because there is Less heat to dissapate. so the more efficient it is (in this case) the smaller it can be.

but if voltage fluxuation is the problem, i assume they would have the same issue, because they said here before that the voltage drop is not some specific exact ammount, but varies depending on the power running through it, so being 2x more stable wouldnt help, and that sounds like it might be the problem, the current goes way down, and things flow easier through the diode. could you USE that ? lay out your resistance steps so the diode change was advantageous , instead of fighting? even use different diodes or quantities of diodes in place of resistances?


so that leaves the mosfet rectifyer, which has to be powered, and drawn down with a resister to stop it again. i have pulled it off, but i couldnt tell you how. a mosfet can be turned into a diode with very little ressitance, so very little changes, but i donno if it would work in that situation at all.

 

[Mr Happy]

I am no electronics expert but somehow it seems like you are asking the impossible. Voltage variation with temperature is a fundamental characteristic of normal semiconductor junctions as found in diodes.

Perhaps a more feasible approach would be to change the circuit so you can use a temperature-stable voltage reference and then a voltage divider network of low temperature coefficient resistors? Depending on how stable you want the voltages to be, a 5.6 V Zener might give you a voltage reference with a low enough temperature coefficient, but you might need more than 250 uA through the diode to keep the voltage on spec.

 

[VidPro]

oops nevermind, its one of them teeney chips.

http://www.micro-bridge.com/data/Axelite/AX2002.pdf
did you put all that Crap on them like it shows in the scematic? all them capacitors to stop all the problems that occurs and all.

its JUST like them to tell you this teeney chip magically does everything with a few parts on it.
then you realise you have to build an entire curcuit around it that is twice the components needed to do it from scratch without the chip.
then you spend the next 2 weeks debugging what they dont tell ya.
and THAT is why they want $20 bucks for the working drivers afterwards

how about an audio taper 16ohm wire wound resister instead of a resister array? those are solid , but big. that would be me, get a teeney chip and have parts 10 times the size of it to run it.

well outta my leauge, ohhh Mr AL.

use the force luke

 

[gillestugan]

Thanks for your tips. I will look at the possibility to use diodes only as voltage dividers, but I doubt I will find diodes with suitable voltages. The lamp will be driven from 4 NiMh, so there is much to work with.

Perhaps a more feasible approach would be to change the circuit so you can use a temperature-stable voltage reference and then a voltage divider network of low temperature coefficient resistors?

Im currently using this, but have had to use diodes in the network to prevent current taking the wrong "turns" in the network.

Im aware there are variations in all semiconductors, but I was not prepared for variations in Vf of 100% or more within that range.

did you put all that Crap on them like it shows in the scematic?

Haha, yes, its like the "do not eat" warning signs on the silica gel bags.
The driver in itself is pretty good. I just want to use two of them to drive a MC-E 2p/driver at the same levels while at the same time driving a XR-E at different levels using a rotary switch. Works well if it was not for the unstable diodes.
An im using the KD SKU: S002982 as base, so I don't have to care about all the components. Modding the boards and putting them together in the network.

 

[Mr Happy]

Im currently using this, but have had to use diodes in the network to prevent current taking the wrong "turns" in the network.

That's the bit I don't follow. Maybe you could expand on why the diodes seem to be necessary?

Im aware there are variations in all semiconductors, but I was not prepared for variations in Vf of 100% or more within that range.

The voltage and current relationship in a diode is expressed by the diode law. You can see that the current through the diode for any given voltage is an exponential function of the temperature.

Also see the graph at the foot of the page. Note how it starts out flat with the current close to zero, and then rises steeply with the voltage almost constant. For good results you need to operate the diode in the steep bit (large current). If the current is too small you will be in the flat bit, and then microscopic changes in current will produce enormous changes in voltage.

 

[gillestugan]

Thanks, that explains the huge variations in Vf. Problem is I need huge variations in current to be able to regulate between 30 and 1500mA by voltage division at the feedback pin. (0-0,25V)

Here is a portion of what Im working on. (only with 2 drivers, so more resistors and diodes are to come)

The resistor in the top (R5) is about 12k, the lower ones (R3,R4) in series are about 1k. The higher output with higher resistance.

Only the first driver should be used in I place the switch in level 2. If I don't use the diode, driver 2 (not in picture) will get the resistance from R4 and R10, and give a higher output than driver 1 instead of being off.
The same for level 5 were only driver 2 should be active.
At level 3-4 both drivers are at use at the same time, but at different levels. This is because I have one wide reflector and one medium, where the medium is weaker to create a softer hotspot.

An alternative way is to put all the resistors in parallell, but then I will need one resistor/level/driver as well as potentiometers for finetuning as the resistors doesn't come in that many values. Too many compontens...

 

[VidPro]

ok i know i still dont belong in this thread, but i am still thinking.

the web says silicoln carbide shotkeys are far less resistant to termperature issues. see i was right mabey. they also are for high voltage apps?

also if you use mondo parts, the variations in temperature dont effect things as much, think about the diff between the tiny glass diode , with thin contact areas, and a big fat diode with metal and big potting around it, the bigest variation in temps would be via the current.

of course now your going to say that hand in hand the current and temps would add to eachother in making things even worse in the big insulated package.

and on lots of drivers you can choke off the driver itself, at the input, instead of after the driver has done the current limiting , that helped me with some stuff. depending on the driver, it can make it more efficient, or less.

the solution to the above puzzel is to use 2 DPDT switches with center off, center off double pull switches have 3 positions , and properly placed they can be smacked with a glove, and are easier to operate than some other switches. and they come in tiny little packages . that could give you 3 levels for the secondary light, and 2 levels and off for the primary, and no interconnection of feeds crossing between the 2 chips. but most arent to water resistant. so i spray silicoln in and around them, and they lasted 5 years Outside , without boots on them, i also use 2x dpdt on my headlight.

 

[Mr Happy]

I think there may be a different way to organize the circuit, but I have not time to really think about it right now.

 

[gillestugan]

I think there may be a different way to organize the circuit, but I have not time to really think about it right now.

Thanks for the thought.
There are different ways. most I have come to think of includes a 2 pole switch where one of the poles is used for selecting main power to the various drivers.
Im looking for a solution where I use the Enable pin to switch the drivers on and off. That would be very neat as my switches only can handle 200mA and the MC-E takes 2000+.
I have a version that uses a 2 pole switch, but it has more resistors and trimpots than I have room for in my lamp.

 

[gillestugan]

the solution to the above puzzel is to use 2 DPDT switches with center off, center off double pull switches have 3 positions ,

Thanks, it really feels like a puzzle, I think I have found a solution only to find a problem somewhere else.
2 switches would make it easy, a 4 pole rotary switch as well, but unfortunately they would need a larger hosing if I would be able to fit them in the housing together with the optics. (about 6X4X3cm).

 

[VidPro]

its just that isolating with diodes as you have there, would really strain my brain, you have to determine the FLOW from the one chip to the other chip, to STOP that flow, and as your resistance changes, doesnt the "polarity" of the diodes position VRSES the polarity of the other chips location change too?

ummm picture time
if your voltage dividing with resistance (which i can only assume that is how the chip understands your changes in resistance), then the voltage is different at each resistance locations.

Chip A 1V + ---{~}-|--.7v-----{~}----.5v--{~}--.2v----{~}-- -

Chip B 1V + ---{~}---.7v-----{~}--|--.5v---{~}-.2v----{~} -

if your diode goes between the 2 red lines there, dont you have to determine what the ACTUAL polarity difference (direction) is at those locations , vrses assuming that one side is positive and one negative, when the FLOW your creating, or stopping, is a voltage difference between the 2 chips (the 2 red lines) , NOT thier original polarity.

so if there was confusion as to why the diodes are not isolating the flow, that would confuse me enough

 

[gillestugan]

You are right. The voltage is higher on the Chip B side. This is not so hard to calculate as the voltage is divided on the resistors in series.
The diodes isolates the two chips from each other, but the variance in Vf of the diodes made the output levels far from calculated and also changed a lot at different levels.

I have been looking at lots of data sheets now, think Ill try these diodes.
As I can calculate current through each diode, I should be able to calculate Vf (approximately) and compensate voltage drop with the resistance values.
Going to sleep now. Time is 01:30am.
Thanks for your help

 

[VidPro]

good to know that somebody knows what they are doing, because it isnt me