Tuesday, June 29, 2021

Charging capacitors using constant current power supplies

We have all seen defibrillators in movies and on TV. The device is switched on, the paddles are applied to the patient’s chest, the operator shouts “clear” and a jolt of electricity, around (3,000V) is applied to the patient. If a second application is required, a short period of time is needed to recharge the defibrillator. Similarly, professional photographers often use studio strobe lighting, which requires a delay or “recycle time” to recharge after a photograph is taken.

Many pulsed load applications use capacitors to store energy. This enables high levels of current to be available to a load for a very short duration. The capacitor should be situated next to the load to provide a low impedance source. A power supply (or battery for portable equipment) is used to charge the capacitor to a set voltage.  There are two ways of charging a capacitor, using a fixed voltage power supply or using a supply that is capable of providing a constant current.

Lasers are now commonly used in cosmetic surgery equipment, material cutting and additive manufacturing (including 3D printing). Many lasers do not operate in a continuous-wave mode, but are pulsed on and off at extremely high frequencies to control the amount of heat energy they apply to the material.

Using an off-the-shelf constant voltage power supply to charge a capacitor can cause problems. When the power supply is initially connected to the capacitor, it will try to deliver its maximum allowable current and probably go into an overload condition. An uncharged capacitor is effectively a short circuit to a constant voltage power supply and if its protection circuit is the hiccup type, it may remain locked in that state. It is also not wise to repeatedly operate a power supply in an over-current condition for reliability purposes.

One method of avoiding an overload condition is to put a resistor in series with the capacitor to limit the current (Figure 1).


Figure 1: Resistor limiting the capacitor charging current

The main drawbacks to this approach are potentially significant power losses in the resistor and the charging voltage has an exponential response time. This is defined by:

Vcap = Vin x (1-e(-t/RC))

The term RC “time constant” is used. This is the time required to charge the capacitor, through the resistor, from a discharged state to approximately 63.2% of the value of the applied DC voltage.

Figure 2 shows an example of a 10,000µF capacitor (C) charging up to 2,000V via a 100Ω resistor (R).  After 5 time constants the capacitor is approximately 99% charged. In our case the time to charge would be 5RC: 5 x 100 x 0.01 = 5 seconds.


Figure 2: Capacitor charging curve for a 2,000V 10,000µF capacitor via a 100Ω resistor

Another method is to use a constant current power supply. Note, we do not need a series resistor as the power supply will internally limit the amount of current supplied (Figure 3). This current level is usually user adjustable.  Charge efficiency is dramatically improved with no losses in the resistor.


Figure 3: Constant current power supply connection

In this case (Figure 4) we do not have an exponential rise, but a controlled linear increase in voltage until the capacitor is fully charged. The amount of time to charge the capacitor is determined by the power supply. One supply with twice the output current will halve the charging time.

 

 

Figure 4: Charging a capacitor with a constant current power supply

Once the desired capacitor voltage is reached, the power supply will stop delivering current.

For someone who is very familiar with constant voltage power supplies, constant current power supplies are a little like driving on the “wrong” side of the road in another country. TDK-Lambda’s High Voltage Product Line Manager had sent me an application story which took me a couple of reads before I understood the problem!

A customer had a 1,000V rated ALE 802 high voltage supply with a charge rating of 18A. To test it, an 80nF capacitor with a 20MΩ resistor in series was connected across the output and the power supply shutdown. As the power supply can deliver an 18A constant current it would try to generate 18 (A) x 20,000,000 (Ohms) = 360,000,000V (V = I x R). Of course the power supply protected itself and shutdown! Even if the resistor was shorted out, 18A charging up the very small 80nF capacitor would take just 4µs (charge time = C x V/I). That was faster than the power supply’s control circuit.

His words of wisdom to the customer were “First, keep the voltage drop in any series resistance/impedance to around 1% or less of the rated voltage to avoid false overvoltage issues. Second, use a capacitor that charges in 500microseconds or more to give the control electronics sufficient time to respond. There’s always exceptions to the rules, but these numbers are a good starting point.”

TDK-Lambda has a large range of programmable power supplies that can operate in constant current mode. The 4000W TPS4000 3-phase has nominal output voltages of 24V and 48V. At the other end of the spectrum is the ALE series, capable of providing outputs from 0 to 50,000V with power levels up to 1MW. The fully programmable GENESYS+™ series of constant voltage/current power supplies have 0 to 10V up to 0 to 1500V outputs with power levels from 1,000W to many tens of kWs.


Figure 5: TPS4000 and ALE power supplies

There is a great selection of information on high voltage capacitor charging on TDK-Lambda’s website, including a pdf with useful equations.  https://product.tdk.com/info/en/products/power/tec_data/ps_ale.html

Power Guy


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