How to revive dead Nickel-Cadmium batteries

If a Nickel-Cadmium battery is dead (the symptoms are next: battery has zero voltage and it won't take a charge), there's still a chance it can be revived. Of course, there is no guarantee it will work, but at least it's worth to try. Here are several methods to revive a dead Ni-Cad battery:

1. Zapping

This method destroys shorted crystal dendrites. It takes several high current pulses applied to the battery, the strength of them should be at least several Amperes - the current should be 20..40 times more than the battery capacitance. The high current pulses can be achieved using a capacitor (see the Figure 1) which discharges through the battery. It is better to use a capacitor with low equivalent series resistance.

How to revive dead Nickel-Cadmium batteries using high current pulses

Figure 1.

R1 - 1k..10k;
C1 - 200μF x 50 V;
SW1 - Push button;
V1 - 12...50 V

The voltage of the power source V1 may be in range 12...50 Volts. Lower power supply voltage takes a higher value capacitor. The resistor R1 prevents the power supply V1 from overloads, the value of this resistor may vary in a wide range. With parameters of components shown on the Figure 1 the capacitor C1 charges in less than 1 second.

How to use this method. Connect a dead battery (one cell) to the terminals of the device, press the push button SW1. It may take several pushes to fix the battery. If it fails, try to increase the capacitance C1 value and/or power supply voltage, then repeat the procedure.

2. Freezing

If zapping method can't help, try this one - the freezing. Put a dead battery into a refrigerator freezer for one or two hours. Take the battery, knock it several times using a hammer handle. The freeze makes dendrites fragile, so it should destroy them.

3. Charging-Discharging with asymmetrical current

In this method a battery is charging with pulses of a current, and between pulses the battery discharges with a lower current. The charging current is 10% of battery nominal capacity, the discharge current is 10 times lower than the charging current. It takes an AC transformer, a diode and two resistors to create a charging device (see the Figure 2).

Circuit diagram of battery charger that uses asymmetrical current

Figure 2.

D1 - a powerful rectifier diode just like 1N4001, 1N4007;
C - Battery capacity, A*h
Ic - Charging current, A;
Id - Discharging current, A;
Vt - Voltage across the secondary winding of the transformer, AC 50/60 Hz;
V - Nominal battery voltage
Vt = , [V]

V = , [V]

C = , [A*h]

Ic = *C, [A]

Id = *C, [A]

R1 = Ω

R2 = Ω

Let's see how to calculate R1 and R2. As it was said before, the discharging current (Amperes) is

Id = 2*C/100,      (1)

where C is the battery capacity, A*h. Because the ciruit is a half wave rectifier, so the only half of pulses is used, therefore the pulse current should be twice higher to provide the necessary energy, so the current is doubled here.

If a nominal battery voltage is V, by the Ohm's law the value of R2 is

R2 = V / Id,      (2)

Combining formula (1) and (2), we get the value of R2 (Ohms):

R2 = V / (2*C/100) = 50*V / C,      (3)

Next, let's find the value of resistor R1. For the charging current we should also consider the discharging current:

Ic = 2*C/10 + Id = 2*C/10 + 2*C/100 = 22*C/100,      (4)

The pulses of the charging current also should be twice higher in amplitude, so the first term of the formula is doubled.

R1 = (21/2*Vt - V - Vd1)/(22*C/100) = 100*(21/2*Vt - V - Vd1)/(22*C),      (5)

where Vt is the AC voltage across the secondary winding of the transformer. Note, the amplitude is 1.41*Vt;
Vd1 - the forward voltage of the diode D1, it is about 0.8 Volts.

Note also that the half wave rectifier is used in the circuit, so the amplitude of current pulses through the resistor R1 is twice higher than the charging current, because we use only half of a wave, therefore an analog voltmeter will indicate a half of the voltage across the resistor R1. Also the output resistance of the transformer should be considered, but in most cases it is quite low, therefore it can be neglected. A voltage drop across the diode D1 also can be neglected.

Let's consider an example: a Ni-Cad 800 mA cell, V = 1.25V. Using the formula (3),

R2 = 100*1.25/(2*0.8) = 78 Ohms. It is not necessary to use the exact value, a resistor of 80 Ω can be used here.

Now let's find the value of R1. By the formula (5), for the Vt = 10V,
R1 = 100*(1.41*10-1.25-0.8)/(22*0.8) = 69 Ω

This value also may vary in some range (±10%), we can choose R1 = 70 Ω.

Values of R1 and R2 can be calculated by using Online Calculator (see it under the Figure 2).

For the real circuit it would be more convenient to use potentiometers instead of resistors R1 and R2.

How to use this charging device:

  1. Calculate R1 and R2, as it described above;
  2. Connect R2 to the battery to be revived, the current Id through R2 should be monitored, because a dead battery may have zero voltage on it, so replace R2 with a potentiometer;
  3. Turn on power, check out the current Ic through R1, change R1 if it is necessary;
  4. The first half of hour both currents may varies quite a lot, after then it may be better, but anyway the currents should be monitored.
  5. Charge the battery for about 14..16 hours.

By the way, this method can revive not only Nickel-Cadmium, but also Lead-acid batteries.

See also another battery charger-regenerator circuit