Understanding the characteristics of a LED driver

Question

I would like to understand the characteristics of a LED driver I have at home. Here is a photo of the label:

I never studied LED drivers before. This post, this article, and this document helped a little bit.

I understand the following:

• PRI refers to the input, which is AC
• PRI takes 220 to 240 V at 50 to 60 Hz
• PRI consumes a maximum of 0.1A (depending on how much SEC requires)
• SEC refers to the output, which is DC
• SEC is expected to deliver a constant current of 300 mA
• SEC delivers a constant current as soon as the resulting voltage is between 10 and 21 Volts
• Prated 6.3 W is the maximum power SEC can deliver (i.e. 21 V x 300 mA)
• Lambda >=0.5 refers to the power factor of the driver; i.e., PRI may consume up to 6.3 W / 0.5 = 12.6 W.

Is my understanding correct?

I do not understand the following:

• What does the No-Load stand for?
• What is this 32VDC Max?
• What are these figures near the black square symbol of the PRI and LED SEC terminals (left and right sides of the label) ? i.e. 0.75-1.5 and 0.5-1.5

Answer 1

This is a constant-current output off-line switching power supply. In normal operation the output voltage is limited by the LEDs to between 10 and 21VDC. You should not connect it to an LED array with a range of forward voltage that is outside that range, in other words, so a single LED or two in series, in most cases is not acceptable.

If you disconnect the LEDs (a "no-load" condition), a perfect constant-current supply would provide an infinite voltage. Since that would be both impossible and undesirable, this design limits the output voltage to a maximum of 32VDC (in reality you would expect it to be typically less than that, so they can guarantee the 32VDC limit). Perhaps more like 24-30V.

The 0.75-1.5 and 0.5-1.5 are the recommended range of wire cross-sectional areas for the input and output respectively. They sizes are relatively heavy for the required current (minimum about AWG 20 for the 300mA secondary, which is easily capable of carrying several amperes in most conditions) so may have more to do with the terminal connection design or mechanical strength than current-carrying capacity.

Answer 2

The only unanswered question is why 32V.

This design may be using a simple unregulated voltage from which to draw a regulated current.

Thus taking the peak sine rectified voltage with no voltage drop on the diode bridge at 32, the take 0.7 of this for average with barely adequate storage caps to get 22.4V for minimum peak minus 2x0.7V bridge diode drop to get 21Vdc.

The minimum voltage of 10V means the current regulator needs to pass 300mA * 10V = 3W which must be raising some series transistor to the maximum elevated temperature.

Of course this then becomes sensitive to 10% line variations with an unregulated voltage and who knows what if this applies to nominal but ought to be worst case line voltage. Hence 230+/-10Vac is only 4.3% tolerance which excludes many parts of the world. So I would not recommend this cheap design but it may work for you.

Even though the steady input current is small they do not recommend a smaller wire size than the output min size, mainly because of the surge current to charge up the SEC caps which is why the p.f. also so poor at 50%. All the SEC current is rising before the peak voltage then cuts out at peak until the next half cycle so with 10% ripple the peak current is about 10% duty cycle and 10x300mA or a 3A average pulse. Transferred back to PRI, with about a 11:1 ratio means the AC PRI peak current is on average about the same as the CC DC current.