When the solid state relay is open and there is voltage at the load end, there will be a certain leakage current at the output end, and care should be taken to prevent electric shock when using or designing. When the solid-state relay fails to be replaced, the original model or the product with the same technical parameters should be selected as far as possible to match the original application line and ensure the reliable operation of the system.
superheating
When the SSR is turned on, the component will be subjected to the dissipated power of P = V (tube voltage drop) × I (load), where the V rms and I rms values are the saturation voltage drop and the RMS of the operating current, respectively. The load capacity of the solid-state relay is greatly affected by the ambient temperature and its own temperature rise, according to the actual working environment conditions, strictly refer to the allowable shell temperature rise (75 °C) when the rated working current, reasonably select the size of the radiator or reduce the current use, in the process of installation and use, it should be ensured that it has good heat dissipation conditions, otherwise it will be out of control due to overheating, and even cause product damage.
Generally speaking, below 10A, the instrument base plate with good heat dissipation conditions can be used, the rated working current of more than 10A products should be equipped with radiators, below 30A, natural air cooling, when the continuous load current is greater than 30A, the instrument fan should be used for forced air cooling, and the products above 100A should be equipped with radiators and fans for strong cooling. During installation, attention should be paid to the good contact between the bottom of the relay and the radiator, and an appropriate amount of thermal grease should be considered to achieve the best heat dissipation effect. If the relay works at high temperature for a long time (40°C-80°C), the user can consider derating to ensure normal operation according to the maximum output current and ambient temperature curve data provided by the manufacturer.
Causes of Solid State Relay Heating:
When the solid-state relay is working normally, there is a certain power loss on its internal chip, and this loss power is mainly determined by the product of the output voltage drop of the solid-state relay and the load current, which is consumed in the form of heat. Therefore, the quality of heat dissipation directly affects the reliability of the solid state relay, and the excellent thermal design can avoid failure and damage caused by poor heat dissipation.
Overcurrent and overvoltage
When the relay is used, due to overcurrent and load short circuit, the internal output thyristor of the SSR solid state relay will be permanently damaged, and the fast fuse and air switch can be considered in the control loop for protection (the output protection of the product should be selected for the selection of relay, and the built-in varistor absorption loop and RC buffer can absorb the surge voltage and improve the dv/dt resistance); Fast fuses and air switches are a universal method of overcurrent protection. The fast fuse can be selected at 1.2 times the rated operating current, and the fuse can be selected for small capacities. Pay special attention to short circuits in the load, which are the main cause of damage to SSR products.
For inductive and capacitive loads, in addition to the internal RC circuit protection, it is recommended to use a varistor in parallel at the output as a combination protection. The area size of the metal zinc oxide varistor (MOV) determines the absorbed power, and the thickness determines the protection voltage value. AC 220V SSR, MYH12-430V varistor; MYH12-750V varistor is selected for 380V; For larger capacity motor transformers, MYH20 or MYH2024 varistors with large through-current capacity should be selected. The selection principle is 500V-600V varistor for 220V, and 800V-900V varistor for 380V.
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