What happens to the power factor when a lamp is dimmed with a electronic

dimmer?

;q

# Power factor with dimmer

Started by tom, Nov 02 2002 08:41 PM

1 reply to this topic

### #1

Posted 02 November 2002 - 08:41 PM

### #2

Posted 14 November 2002 - 09:51 PM

Hello Tom

What happens to the power factor when a lamp is dimmed with a electronic dimmer?

Power factor is the ratio between KW and KVA or real power and apparent power. A standard incandescent lamp is resistive so when driven by a sinusoidal voltage, it will have a power factor of 1. When you introduce an electronic light dimmer however, you remove part of each half sinewave, distoring the current waveform. This introduces harmonics and drops the power factor. Half power is achieved by allowing current to flow for half or each half sinewave. - full power for half the time. If we calculate the RMS current for this waveform, we find that we have reduced the RMS current to 0.7071 of its full power level.

Now take the RMS voltage input and multiply by the RMS current, and we get 0.7071 of the maximum KVA when we have full power. In other words, the power is half, but the KVA is 70%. To calculate the power factor, 0.5/0.7071 = 0.7071. - The power factor has dropped to 0.7 at half power even though the load is still resistive.

Best regards,

QUOTE

What happens to the power factor when a lamp is dimmed with a electronic dimmer?

Power factor is the ratio between KW and KVA or real power and apparent power. A standard incandescent lamp is resistive so when driven by a sinusoidal voltage, it will have a power factor of 1. When you introduce an electronic light dimmer however, you remove part of each half sinewave, distoring the current waveform. This introduces harmonics and drops the power factor. Half power is achieved by allowing current to flow for half or each half sinewave. - full power for half the time. If we calculate the RMS current for this waveform, we find that we have reduced the RMS current to 0.7071 of its full power level.

Now take the RMS voltage input and multiply by the RMS current, and we get 0.7071 of the maximum KVA when we have full power. In other words, the power is half, but the KVA is 70%. To calculate the power factor, 0.5/0.7071 = 0.7071. - The power factor has dropped to 0.7 at half power even though the load is still resistive.

Best regards,

Mark Empson | administrator

Skype Contact = markempson | phone +64 274 363 067

LMPForum | Power Factor | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | LMP Software | Pressure Transducers | Smart Relay | GSM Control | Mark Empson Website | Soft Starters

Skype Contact = markempson | phone +64 274 363 067

LMPForum | Power Factor | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | LMP Software | Pressure Transducers | Smart Relay | GSM Control | Mark Empson Website | Soft Starters

#### 0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users