Hi I am using MT3608 boost converter to boost 18650 3.7v (2600mah) battery voltage to 9v.
how do I know how much max current I can draw after boosting to 9v (battery is rated for 1c discharge) is there any formula to calculate it? I know converter can support max 2A but does that mean I will get 2A at 9V after boost?
here are the details of the converter -
- Input voltage range : 2-24V DC
- Output voltage range : Max 28V DC
- Efficiency : 93%
- Output current : 2A
- The input voltage should not exceed the maximum input voltage.
- The peak current output current is no more than TV university.
- When adjust the blue potentiometer ,please use a multimeter to monitor the voltage
- Rotating Clockwise can rise the voltage while rotating counterclockwise could lower the voltage
- When the module output voltage can not be adjusted ( the output voltage is always equal to the input voltage ) Please adjust the potentiometer counterclockwise 20 laps (20 turns) or more till it works.
bkrenderman:
The peak current output current is no more than TV university
If you can understand that you're doing better than me.
But generally power is fairly constant so to get 9V@2A (18W) out you would need at least 18W in. 18W / 3.7V = approx 5A.
But if your battery can only provide 2.6A (2600mAh @ 1C) then that's only 9.6W (3.7V x 2.6A) in so you'll only get 9.6W / 9V = approx 1A out.
Steve
slipstick:
If you can understand that you're doing better than me.
But generally power is fairly constant so to get 9V@2A (18W) out you would need at least 18W in. 18W / 3.7V = approx 5A.
But if your battery can only provide 2.6A (2600mAh @ 1C) then that's only 9.6W (3.7V x 2.6A) in so you'll only get 9.6W / 9V = approx 1A out.
Steve
Thats great! I now understand how to calculate the output current based on the input voltage.
While that is how to calculate the theoretical current a word of caution.
Those cheap ebay boost / buck converters have a very misleading specification. I have failed to get anymore than 700mA out of one when boosting by times two. Your boost ratio is 2.4 times and I would be very surprised if you can get anything over half an amp.
When it says "Output current : 2A" that is probably the maximum the chip will stand at any boost or buck ratio. With increasing current you need better magnetic saturation characteristics in the inductor and these devices can't cope in my experience.
Grumpy_Mike:
While that is how to calculate the theoretical current a word of caution.
Those cheap ebay boost / buck converters have a very misleading specification. I have failed to get anymore than 700mA out of one when boosting by times two. Your boost ratio is 2.4 times and I would be very surprised if you can get anything over half an amp.
When it says "Output current : 2A" that is probably the maximum the chip will stand at any boost or buck ratio. With increasing current you need better magnetic saturation characteristics in the inductor and these devices can't cope in my experience.
Yes I know they over spec them. I am in need of max only 400ma at 9v so I think I will be good with it. however I will update how much max I can get from it by experimenting with it at 9v.
Normally for boost converters the maximum output current is determined by the switcher chips maximum current rating and the voltage boost needed.
So for 3.7 to 9 v you are boosting by 2.5 times so at a current draw of 400 ma the switcher chip will be switching around 1 amp which should be OK.
Just check that the switcher chip isnt getting too hot as they dont have a heatsink.
You will notice that if you browse through Pololu's excellent selection of boost converters, they take care to specify the maximum input current.
That is a most important limit, imposed by the converter itself, but manufacturers of eBay junk won't tell you that.
