Bob I'm a novice who who's profession does not normally involve electronics. Due to inexperience I assumed I could place 3 in series and one in parallel to the series. I see now that the result would differ from what I expected. Now you know.
This puts me in a bad place as far as power management as you pointed out. Being that this application would normally be idle and operate for short amounts of time, perhaps it could be designed as to perform a certain amount of work when idle and less at run time.
You know those wildfires in California? They aren't wild.
Just kidding, fortunately/unfortunately I've only been testing with a simulator so I never actually tried it.
Is it out of the question to use a standard 9V? That would take maybe twice the space of 3 CR2032s.
BYW, I tried to design a boost converter. With the 40 Ω internal resistance and could no achieve your voltage requirements without using a large inductor (1mH or so) which would be larger than your battery.
Bob
BYW, I tried to design a boost converter. With the 40 Ω internal resistance and could no achieve your voltage requirements without using a large inductor (1mH or so) which would be larger than your battery.
Bob
Yes it would be physically out of the question to use a 9V, but I just tested the output of two 2032s in a series on a 10 ohm 5 watt resistor(not in simulator) and the internal resistance I'm getting is around 10-13Ω, and the current is several hundred ma. I don't know how I get these results being that the data sheet specifies 10-40Ω each, and I assumed that putting them in a series would increase this.
If these results are in fact accurate, then my desired results would become achievable from a power perspective. The question would be the best method.
This 1:50 transformer has a 300V isolation voltage, so could I use +6 and -6 sources to achieve aprox. +300 -300?
That would be good news
Yes I realize that, for this application internal resistance rising over time is irrelevant. It will only be used for a couple of seconds at a time and batteries will be replaced when it stops functioning. I was surprised to see that two new batteries together were only 10Ω.
TI has an excellent tool (that they inherited from National Semiconductor when they bought them out), called Wbench Designer. It is a free web based app that will design a boost converter (among other things) based on your input of requirements and their parts.
I used it to design a boost converter that could produce 200V output, 2mA current from an input of 5 to 6V. The lower value is because, with your high internal resistance, the voltage is going to drop. It is a guess, and I think an optimistic one.
They are able to do it, with 47% efficiency. But, they call for a custom transformer with the following parameters:
443uH inductance
31:1 ratio
0.9Ω internal resistance (I presume this is the primary)
I expect that that transformer would be a lot bigger than you can tolerate.
They also estimate the area required for a PCB at 790mm^2, a little over a square inch. But I doubt that includes the transformer since they would have not info on a custom transformer.
Just letting you know what you are up against.
Given the effiiciency at less than 50%, let's go through the power calculations again.
Your output is 200V at 2mA which is 0.4W.
With 47% efficiency, that means 0.86W from the battery, or an average current of 142mA. Using your 10Ω for internal resistance, that would be a voltage drop of 1.42V, which brings your terminal voltage down to 4.6V, so my 5V assumption does look optimistic.
Executive summary: The best TI can do with its line of boost converters is borderline to achieve the parameters you are trying for.
Bob
I used it to design a boost converter that could produce 200V output, 2mA current from an input of 5 to 6V. The lower value is because, with your high internal resistance, the voltage is going to drop. It is a guess, and I think an optimistic one.
They are able to do it, with 47% efficiency. But, they call for a custom transformer with the following parameters:
443uH inductance
31:1 ratio
0.9Ω internal resistance (I presume this is the primary)
I expect that that transformer would be a lot bigger than you can tolerate.
They also estimate the area required for a PCB at 790mm^2, a little over a square inch. But I doubt that includes the transformer since they would have not info on a custom transformer.
Just letting you know what you are up against.
Given the effiiciency at less than 50%, let's go through the power calculations again.
Your output is 200V at 2mA which is 0.4W.
With 47% efficiency, that means 0.86W from the battery, or an average current of 142mA. Using your 10Ω for internal resistance, that would be a voltage drop of 1.42V, which brings your terminal voltage down to 4.6V, so my 5V assumption does look optimistic.
Executive summary: The best TI can do with its line of boost converters is borderline to achieve the parameters you are trying for.
Bob
Well I think I just got it to achieve the output I desired in the OP.
Using this transformer (1:50) in a flyback converter:
Aprox 4ma on the load. When I replace the -6V with ground I obtain around 2.7ma.
This is not using any ideal components, and when I replace the wave generator with a 555 output is 3ma.
I guess now my question would be if I can improve the output even more. I thought placing the -6 instead of a ground would provide a superior improvement...
The encasing in which this circuit would go is aprox 6mm. at its largest, I can fit around 4 coin cells. Also, I can probably fit 1-4mH inductors if I need to, as I have found several small ones.
Red is the secondary output between transformer leads, blue is primary. Doesn't look that funny to me, but then again what do I know...
Don't know why the primary is shorter above the zero than it is below...
In regards to your second question unfortunately I can't say. Its running on default multisim settings.
I should have been more clear. The image in post #30 has the voltage across the 100K load resistor. Or is it the voltage from the top of the 100K resistor to the primary side system ground? Either way, what I'm asking for is a scope shot of the voltage across R2 only. I want to see the ripple.
Thanks.
ak
Thanks.
ak
This is across the load. There is little to no ripple, it can hardly be seen in the scope.
When I put a ground symbol on the transformer lead or connect it to primary ground it makes little to no difference.
I don't really understand why these results are surprising now that we lowered the internal resistance.
It's not really that efficient comparing whats going in to whats going out. I would think there is a way to get better results.
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