NiCd charging for dummies?

[uniqorn]

Hello all,

i have some questions about a simple NiCd charging circuit.
adaptor: input 230 V ; 3 W
output 5.5 V ; 225 mA

there are a diode and a resistor in series with the batteries (2 NiCd SC1300).
I'm trying to calculate the energy flow but it doesn't work since there are some things i don't know.

Since there are two batteries the energy in the batteries when fully charged is 3.12 Wh. I read somewhere that there is 30% loss in the batteries while charging (heat dissipation), so the circuit has to deliver 4.46 Wh. My questions...

* what do they mean with the output of the adaptor?? thought there was only a primary and a secondary voltage... what does the 225 mA mean?
* i read somewhere the resistor is to stabilise the current.. how? why?

 

[Klaus]

Hello all,

i have some questions about a simple NiCd charging circuit.
adaptor: input 230 V ; 3 W
output 5.5 V ; 225 mA

there are a diode and a resistor in series with the batteries (2 NiCd SC1300).
I'm trying to calculate the energy flow but it doesn't work since there are some things i don't know.

Since there are two batteries the energy in the batteries when fully charged is 3.12 Wh. I read somewhere that there is 30% loss in the batteries while charging (heat dissipation), so the circuit has to deliver 4.46 Wh. My questions...

* what do they mean with the output of the adaptor?? thought there was only a primary and a secondary voltage... what does the 225 mA mean?
* i read somewhere the resistor is to stabilise the current.. how? why?

Your first post here Uniquorn, welcome.
A few basics about NICAd charging, these require constant current charging as compared to lead acid batteries which require constant voltage charging.
Your charger is a crude limited current charger, the resistor limits the maximum current that can flow into the battery and the diode prevents you from connecting it up backwards. It will work but the current is not regulated (stabilised), just limited. The better chargers have regulation as well.
Your adaptor DOES have a primary and a secondary voltage, it also tells you the maximum current (225mA) you can get from the secondary voltage. At 5.5v this equates to 1.24Watts.

Do read up a bit more on basic electricity to get a better grip on series circuits and their current and voltage relationships. Energy calculations are a bit more tricky and I would leave them until thoroughly familiar with the workings of your battery charger.
have fun
Klaus

 

[uniqorn]

Thank you for the answer, i worked it out a little bit. Actually the charging system is from a dustbuster. So there is also a voltage drop from the contacts between the charger (the adaptor) and the charging circuit. I estimated the voltage drop over this contacts as follows:
i suppose the charging current is 0.1 C so it's 130mA
there's a voltage drop of 2.9 V over the batteries, 0.7 V over the diode and 0.61 V over the resistor. So the contacts have a voltage drop of 1.289 V, it seems a little bit to much, is this possible?
For the energy calculation i made some other assumptions.
First of all, the dustbusters manual says it has to charge for 15 h to have fully loaded batteries (3.12 Wh), for the energy loss in the resistor and other components i divided the 15 hours in two, due to the working of the diode (only the positive side of the sinusoidal voltage). I also read somewhere that the batteries only store 70% of the supplied energy, the rest is lost in heat dissipation.
In this way i have an efficiency of 44% for the charging circuit. Is this a normal value for such a circuit?

The batteries are connected to a PMDC motor (i guess). The manual makes clear that you can handle the dustbuster for 10 minutes. So i have an input power of about 18 W (probably i have to take account for charge/discharge losses also??)
What i actually want to do (or at least show that it's impossible) is replacing the PMDC and the batteries by a constant torque spring motor.
Problem is i can't measure the torque of the motor because it's inside and the armature of the dustbuster influences the torque.
So i thought to handle it like this: since a PMDC has an almost constant rpm with changing torque, i can measure the rpm outside the armature.
If i know the outputpower of the motor (therefore i have to know the efficiency of the motor, something around 50% for such small motors i guess), i can calculate the working point of the dustbuster. Knowing this working point i know the required torque. Since the spring motor has a constant torque i can calculate this motor.... i hope

Maybe there is somebody who can help me with this, maybe telling me in which points i could be wrong?

thanks