# Why does AC input have a parallel resistor and capacitor?

Well, I am trying figure out a situation with images and a circuit diagram.

I am checking all part looks normal, but if I insert this board on a machine the resistor is exploding and one leg of resistor is breaking off.

Actually I don’t understand why a resistor and capacitor was inserted parallel to the input of the AC power supply. Additionally, the resistor is exploding, but the fuse is solid yet.

simulate this circuit – Schematic created using CircuitLab

New schematic as a JPEG image:

• For the circuit, I didn't find a varistor symbol, so I use an adjustable resistance symbol in circuit instead

• Resistance: 470 kilohm

• A 470 ohm resistor connected across 220 V AC will detonate. Are you sure it isn't a 470 kohm resistor because that would be much more likely in that position. Additionally, your photographs appear to show that resistor (R1) as a 470 kohm. – Andy aka Mar 28 '20 at 9:08
• Andy aka is correct, it seems to be a 470kOhm (yellow, purple, yellow, gold). The resistor placed in parallel with the input of an AC Power Supply is usually used to discharge the input capacitance, if you disconnect the power supply from the mains. – vtolentino Mar 28 '20 at 9:13
• One reason for a resistor to be present here would be to ensure the discharge of the X2 capacitor per IEC-950 recommendations: the voltage across the power plug prongs should drop quickly after unplugging the cord from the ac outlet. But a) these resistors are not mandatory for a X2 cap. of 100 nF or less and b) the value is quite high to avoid any useless dissipation, like 1 Megohms for a 1-µF cap. – Verbal Kint Mar 28 '20 at 9:13
• @Andyaka yea you are right it was 470K, I am sorry – mehmet Mar 28 '20 at 9:20
• @mehmet: You have to replace it with a 350V resistor type. Some cheap resistor types are only good for 100V. – Janka Mar 28 '20 at 10:45

The HTCC capacitor is used for filtering and interference suppression of high frequencies as well as the L1/L2 is used for common mode rejection. The resistor is used to discharge a possible load at the HTCC when disconnecting the power supply from main. This HTCC capacitor is also a self-healing type, also typically marked with X2 or Y2 types. They can withstand pulses up to 5 kV. In the schematics the L1 and L2 should be shown as inductively coupled for common mode rejection, they must be on the same magnetic core.

• So Why do you think 470K ohms resistor will burst and one leg break off when 220 AC voltage is applied to the input? – mehmet Mar 29 '20 at 8:46
• @mehmet A 470K resistor across 230V should dissipate about 110mW - this is fairly small and shouldn't pose an issue for a 1/2W resistor. Either your resistor is not 470K or it is not designed to be across 230V. Can you post a picture or a part number? – SomeoneSomewhereSupportsMonica Mar 29 '20 at 9:14
• @mehmet SSSMonica is right, this 470 kΩ resistor dissipates 0.11 W. The damage can be caused due to a mismatch of the resistor or this resistor is outside the specification. Standard types are specified up to 200 V (e.g. docs.rs-online.com/221a/0900766b8072fc9f.pdf). As is well known, the peak voltage at 230 V is 1.414 * 230 V = 325 V. Therefore you often see a series connection of 2 resistors instead of one with a better voltage specification. – Tom Kuschel Mar 29 '20 at 10:16

The 470k resistance is there to discharge any stored charges in the filter capacitors so that the device or disconnected mains plug does not give a surprising shock if touched.

The capacitor directly across L-N is an 'X' capacitor and is used to reduce RF conducted emissions.

Without R1 this capacitor could hold a high voltage so that touching the input terminals may give an electric shock. R1 is chosen to be small enough to discharge the capacitor in reasonable time but large enough to keep power dissipation down to an acceptable level.

This design has a 470k resistor for this a 470R resistor would dissipate excessive power and probably explode.

• "excessive power" = the current would be 0.5 A and the power 100 W (nominally, at room temperature). For 470 kΩ it would be 1000 times lower (0.1 W). – Peter Mortensen Mar 29 '20 at 0:23