DC to DC converter question

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jchittle

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I am using a DC to DC converter, model number cc3-0505sf-e from tdk lambda. On the primary I am supplying 9V from a lithium ion battery which is supposed to have 1200mah of life. The primary is also drawing 90mah of current. I thought the battery would last about 13.3 hours(1200/90). However, when the circuit is on the voltage contiunously drops about 0.01V/s. The battery only lasts about 10 minutes. Does anyone know what the problem is? The current being drawn is continous 90mA.
 

mdshunk

Senior Member
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Right here.
Sounds like your battery vendor lied about the battery spec. Got anything like a calibrated load that you can put on it to see for sure?
 

dereckbc

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jchittle said:
I am using a DC to DC converter, model number cc3-0505sf-e from tdk lambda. On the primary I am supplying 9V from a lithium ion battery which is supposed to have 1200mah of life. The primary is also drawing 90mah of current. I thought the battery would last about 13.3 hours(1200/90). However, when the circuit is on the voltage contiunously drops about 0.01V/s. The battery only lasts about 10 minutes. Does anyone know what the problem is? The current being drawn is continous 90mA.
Converting 9 V to what? What you have is a linear regulator which can be as much as 50% ineffecient. So what voltage is th eload side at what current?.

As noted you cannot just divide the AH rating by the load current and get the number of hours the battery should last. When a battery is rated at say 2000 mAH it is specified at a discharge rate like 20 hours (.1 amps). If you discharge at a faster rate you have to apply a correction factor which will always be less than 1. If discharged at a slower rate then the correction factor is greater than 1. So when you see a AH rating you also have to know at what discharge rate it is specified at.

Also Lithiun Ion batteries degrade from 10 to 20 % per year from the moment they come off the assemble line. So if you buy one and it has set on the self for 2 years, you have already lost 20 to 40% capacity.
 
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brian john

Senior Member
Location
Leesburg, VA
Isn't there some issues with deep cycle discharging where the low voltage shut off should be at a higher voltage as the discharge time is extended?
 

ELA

Senior Member
Occupation
Electrical Test Engineer
Looks like a switching regulator to me.
Measure the input current with a current probe and you may see current spikes higher than 90 ma.
 

dereckbc

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brian john said:
Isn't there some issues with deep cycle discharging where the low voltage shut off should be at a higher voltage as the discharge time is extended?
Only if a LVD is installed
 

tallgirl

Senior Member
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Controls Systems firmware engineer
90ma on a 1200mah battery isn't so much of a load that it would be dragging the battery voltage down too far due to excessive load, so I think that rules that out 1200 / 90 = 13 and I don't think C/13 is too high, especial for Li-ion which is widely used in high-output applications.

You're going to see a decline in voltage over time -- that's the discharge curve of a battery, output voltage reflects battery state of charge. With Li-ion batteries there is a relatively fast initial drop, then the voltage stays somewhat flat, then there is a rapid drop. I forget the range of voltages, but it's about a 30% total drop in output voltage from "full" to "empty".

My recollection of Li-ion chemistry is that they are a 3.65v per cell technology. So, a 9v nominal battery would be three cells for a total of 10.95v at full charge. If you're charging to less than that, you have less capacity, and if you've got an older battery, or one exposed to high heat, you again have less capacity. And finally, they age faster that other technologies.

What's the application?
 

ELA

Senior Member
Occupation
Electrical Test Engineer
Max input for that converter is 9 V so if your battery volts is high that might also be an issue.
 

jchittle

Member
Thanks for the replies everyone. I had a general idea that 1200/90 would not give me an exact value, but I thought it might give something close. I was unaware that there was a correction factor. I will contact the battery manufacturer tomorrow and see about this correction factor.

The converter is a 9V to 5V dc to dc converter. The output is isolated. The output is drawing 40mA continuous. Very inefficient. I have ordered another chip which is more efficient for the current draw I need but it is on backorder.
The problem with the chip I have now is it draws a no load continuous current of 90mA. Even when I get the new chip, it will draw 36mA no load continuous current on the primary. Then the battery won't last much longer it seems.


The function of the circuit is to display the value of force on 2 small 7 segment LED's. The load cell which measures the force requires a constant 5V across it in order for it to operate properly.
 

tallgirl

Senior Member
Location
Great White North
Occupation
Controls Systems firmware engineer
jchittle said:
The problem with the chip I have now is it draws a no load continuous current of 90mA. Even when I get the new chip, it will draw 36mA no load continuous current on the primary. Then the battery won't last much longer it seems.


The function of the circuit is to display the value of force on 2 small 7 segment LED's. The load cell which measures the force requires a constant 5V across it in order for it to operate properly.

What's the primary power source? How does that Li-ion battery get recharged? And what is the open circuit voltage on the battery at the end of the charging cycle? The battery should output between 10.8 and 10.95vdc when fully charged.

Also, since you mentioned this is a force measuring circuit, is there a way to turn the circuit off when there is no force?
 
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