Page top

Building Automation


Industrial Automation


Power Automation & Safety

Bangladesh Distributor

Lead Contents

FAQ02237 of Solid-state Relays FAQ

FAQ No. FAQ02237

Primary Contents

Question

The current is unstable when the power supply is switched using an Solid-state Relay. Why is this?

Answer



Using a smoothing capacitor makes the power supply a capacitive load, but as shown in the following figure, the current output is clearly divided between times when hardly any current flows and times when the capacitor is suddenly charged.

When an attempt is made to turn the Solid-state Relay ON, voltage is applied to the Solid-state Relay, and the Solid-state Relay tries to turn ON, but near point A where there is no current flow the ON status cannot be maintained, so the Solid-state Relay turns OFF and then turns ON when the charge current starts to flow at point B.

Similarly, the Solid-state Relay moves from OFF to ON at point C. If there is a delay in the Solid-state Relay turning ON at point B, the capacitor will no be sufficiently charged, and when the Solid-state Relay turns on at point C it will try to charge the amount insufficient at point B, which will result in a large current flowing. The current is unstable because this process is repeated at points D and E.

As a measure to stabilize the current, connect a bleeder resistor that constantly passes current in parallel with the power supply to keep the Solid-state Relay ON.

100 VAC5 to 10 kΩ3 W
200 VAC5 to 10 kΩ15 W

Recommended Products

MY

New Latching Levers for Circuit Checking Added to Our Best-selling MY General-purpose Relays

LY

Power-switching Compact General-purpose Relays

G3PE (Three-phase)

Compact, Slim-profile SSRs with Heat Sinks. Solid State Contactors for Three-phase Heaters Reduced Installation Work with DIN Track Mounting.

Other Solid-state Relays FAQ