How to Increase DC Current?

[Satoshi Deguchi]

Hello,

I've run into the issue of my batteries not having being able to discharge enough current for my motors to start up (stall current).

Motor​	Voltage​	Stall Current​	Idle Current​
MTB Rhino​	11.1v​	7.8A​	0.3A​

Battery​	Voltage​	Amps / Discharge​
Coolook LFP 14500​	3.7v​	1A​

This battery configuration outputs 11.1v and 2A. It isn't enough current for the motors to draw on startup, I need 16A to do that. There isn't very much space for me to work with either, the battery stock I plan to buy doesn't have enough room for more than this setup with some room to spare.

Using these batteries and this setup, is there any way to increase the current/amperage?

Would it be possible to create a current booster to facilitate this? If so, how can I do this? What parts do I need? Are there any tutorials?

If neither of these are feasible, would it be possible to use a higher voltage battery, reduce the voltage and convert some of the reduced voltage to amperage?

 

[Diver300]

Do you need to start the motor quickly, or against a large amount of friction? If this is for the Nerf Blaster, which I understand uses a flywheel, you could start that more slowly to reduce the power needed. It would also increase the efficiency.

You need a PWM motor speed controller to do that. Although you won't be controlling the maximum speed, a PWM controller can limit the current or limit the acceleration, either of which will reduce the current needed.

 

[Satoshi Deguchi]

Do you need to start the motor quickly, or against a large amount of friction? If this is for the Nerf Blaster, which I understand uses a flywheel, you could start that more slowly to reduce the power needed. It would also increase the efficiency.

You need a PWM motor speed controller to do that. Although you won't be controlling the maximum speed, a PWM controller can limit the current or limit the acceleration, either of which will reduce the current needed.

Yes, this is for a Nerf blaster using flywheels. Two motors, two flywheels.

Preferably I would start the motor quickly. I just need to be able to overcome the stall current when revving the blaster, as the idle current is only 0.6A for both motors.

Wouldn't limiting the current to the motor have negative effects on the motor?

Can you also help me better understand this chart? In what case does the motor reach maximum efficiency or maximum power output?

 

[Papabravo]

There is not much you can do to provide for the stall current on startup except put more batteries in parallel.
There is no practical way to boost the current from a set of batteries without lowering the voltage. The immutable rule of power conversion schemes is that: "The output power will ALWAYS be less than the input power." Sometimes it will be much less.

As for the diagram, it tells you that you can have high speed with low torque or low speed with high torque, but you CANNOT have both high speed AND high torque. The data to the right of the chart tells you what combination of speed and torque produce either maximum efficiency (the magenta peak) or maximum power (the green peak). The blue trace shows possible combinations of speed and torque. No other combinations are allowed or possible.

 

[Diver300]

Preferably I would start the motor quickly. I just need to be able to overcome the stall current when revving the blaster, as the idle current is only 0.6A for both motors.

Wouldn't limiting the current to the motor have negative effects on the motor?

I realise that you want to start the motor faster, but it might be much easier to put up with a slightly longer run-up time than to carry larger batteries.

Limiting the current will not have any negative effect on the motor, as long as it turns and doesn't remain stalled.

 

[Diver300]

Can you also help me better understand this chart? In what case does the motor reach maximum efficiency or maximum power output?

View attachment 142245

The graph shows various characteristics of the motor, plotted against torque, for a fixed input voltage of 12 V.
Black is current, blue is speed, green is power output and purple is efficiency.
At the left, there is no torque, so no power output and 0% efficiency because of zero power. The speed is 36000 rpm, and the current is very small.
On the right, the maximum torque is 21.4 mN-m, and that is stall because there is zero speed and maximum current. With zero speed, there is also no power output and 0% efficiency because of zero power.

Maximum power occours at about half maximum torque and half maximum speed. Maximum efficiency is at about 3.5 mNm and 30000 rpm.

If you reduce the voltage, that graph doesn't apply. The maximum speed and maximum torque will both reduce in approximate ratio with the supply voltage, but the peak efficiency won't change much, but it will happen at a lower speed.

If you reduce the voltage for starting, there will be a lot less torque, so it will be slower. Against that, if you halve the start voltage, that will halve the start torque, but it will reduce the power by about 4 times, so you are getting half the spin-up rate for 1/4 the power.

 

[Satoshi Deguchi]

There is not much you can do to provide for the stall current on startup except put more batteries in parallel.
There is no practical way to boost the current from a set of batteries without lowering the voltage. The immutable rule of power conversion schemes is that: "The output power will ALWAYS be less than the input power." Sometimes it will be much less.

As for the diagram, it tells you that you can have high speed with low torque or low speed with high torque, but you CANNOT have both high speed AND high torque. The data to the right of the chart tells you what combination of speed and torque produce either maximum efficiency (the magenta peak) or maximum power (the green peak). The blue trace shows possible combinations of speed and torque. No other combinations are allowed or possible.

The graph shows various characteristics of the motor, plotted against torque, for a fixed input voltage of 12 V.
Black is current, blue is speed, green is power output and purple is efficiency.
At the left, there is no torque, so no power output and 0% efficiency because of zero power. The speed is 36000 rpm, and the current is very small.
On the right, the maximum torque is 21.4 mN-m, and that is stall because there is zero speed and maximum current. With zero speed, there is also no power output and 0% efficiency because of zero power.

Maximum power occours at about half maximum torque and half maximum speed. Maximum efficiency is at about 3.5 mNm and 30000 rpm.

If you reduce the voltage, that graph doesn't apply. The maximum speed and maximum torque will both reduce in approximate ratio with the supply voltage, but the peak efficiency won't change much, but it will happen at a lower speed.

If you reduce the voltage for starting, there will be a lot less torque, so it will be slower. Against that, if you halve the start voltage, that will halve the start torque, but it will reduce the power by about 4 times, so you are getting half the spin-up rate for 1/4 the power.

Thank you for the explanations!

Because space is limited and I simply cannot put more of those batteries in parallel, would using a higher voltage battery and converting some of that raw voltage to amperage be viable?

I've heard of, and am researching, the operational amplifier. I feel that may be something I can use. What is your take on this? Can you explain to me what it is, how it works and how I may be able to use this for my build?

 

[Externet]

Try with a healthy automobile battery and come back with observations.

 

[Nigel Goodwin]

Thank you for the explanations!

Because space is limited and I simply cannot put more of those batteries in parallel, would using a higher voltage battery and converting some of that raw voltage to amperage be viable?

A battery only contains so much power, determined mainly by it's size, so as already explained conversion will only reduce the amount of power available.

However, a quick google shows a confusing number of Li-Ion battery packs for nerf guns, what's wrong with those?.

I've heard of, and am researching, the operational amplifier. I feel that may be something I can use. What is your take on this? Can you explain to me what it is, how it works and how I may be able to use this for my build?

An operational amplifier has no bearing on your issue, it could be (and often is) used as part of control systems, such as chargers and battery monitoring.

If you want more power from your batteries, then you need larger more powerful batteries. No doubt you've seen dragsters?, 0-200+mph in a few seconds - why do you think they don't use 50cc moped engines?.

 

[Satoshi Deguchi]

A battery only contains so much power, determined mainly by it's size, so as already explained conversion will only reduce the amount of power available.

However, a quick google shows a confusing number of Li-Ion battery packs for nerf guns, what's wrong with those?.



An operational amplifier has no bearing on your issue, it could be (and often is) used as part of control systems, such as chargers and battery monitoring.

If you want more power from your batteries, then you need larger more powerful batteries. No doubt you've seen dragsters?, 0-200+mph in a few seconds - why do you think they don't use 50cc moped engines?.

I'm firmly against using Lio-Ion / Li-Po batteries in Nerf because majority of them are unprotected cell and require a very high level of maintence to prevent them from having a catastrophic failure (explosion, fire, etc). I don't want to burn my house down or kill myself using them.

If I can find a Lithium battery that is protected cell and can be used in Nerf, then I will use them. But that is a challenge, I haven't been able to find any.

 

[audioguru]

On your other thread I showed that the Coolook batteries are low voltage and low current LiFeP04 type and are not a Name Brand. They are low power batteries.

My electric Radio Controlled model airplanes are Name Brand and are purchased at my local hobby store, not on-line junk. The small but high power Li-PO batteries are high quality without a protection circuit. The motor controllers in the model airplanes never overload the battery and disconnect the motor when the battery voltage becomes low. The chargers charge safely and properly.

 

[Satoshi Deguchi]

On your other thread I showed that the Coolook batteries are low voltage and low current LiFeP04 type and are not a Name Brand. They are low power batteries.

My electric Radio Controlled model airplanes are Name Brand and are purchased at my local hobby store, not on-line junk. The small but high power Li-PO batteries are high quality without a protection circuit. The motor controllers in the model airplanes never overload the battery and disconnect the motor when the battery voltage becomes low. The chargers charge safely and properly.

I have made the decision to not use Li-Po because the batteries are unprotected cell and can be quite volatile. The maintenance for those batteries is quite high as well, which is another reason I will not be using them.

In regards to Coolook LPFs not having enough power, I'm considered using protected cell 18650s with a corresponding smart charger. There are some that the ability to discharge the current needed and with them being protected cell it gives me piece of mind.

If I use 18650s, as a precaution, I will store them in a fireproof container. Do you know of any that come from a reputable source and are not bags coated in fire retardant?

What is your take on using 18650s for this project? With them being higher power it seems like a much better idea then using LFPs in parallel, I wouldn't need to wire as many.

 

[Papabravo]

I guess you are well and truly hosed!