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LED driver power supply design need to be done calculations before debugging, measurement when debugging , aging after debugging , the following 5 points for reference
2, the chip heat
This is mainly for the built-in high power modulator driver IC. If the chip consumes a current of 2mA, 300V voltage increases above the chip, the chip's power consumption is 0.6W, of course, the heat will cause the chip. Driver's maximum current from the drive power consumption mos tube, a simple formula is I = cvf (considering the resistance of effective charge, the actual I = 2cvf, where c is the power MOS tube cgs capacitance, v is the power management guide pass when the gate voltage, so in order to reduce the power consumption of the chip, must find ways to reduce the c, v and f. If c, v and f can not be changed, then please think of ways to the power consumption of the chip assigned to off-chip devices, attention Do not introduce extra power consumption. and then simply, is to consider a better cooling it.
3, the power tube heat
On this issue, but also seen some people in the forum sent a paste. Power tube power into two parts, switching losses and conduction losses. Note that most of the occasions in particular electricity-driven application of LED, the switch conduction loss is much larger than damage. Switching losses and power tubes and cgs cgd and chip drive capability and working frequency, so the power to solve the heat pipe can be resolved from the following aspects: A, not one-sided based on the size to select the on-resistance power MOS tube, because The smaller the resistance, cgs and cgd capacitance greater. Such as the 1N60's about cgs is 250pF, 2N60 about the cgs to 350pF, 5N60 about the cgs to 1200pF, the difference is too great, choose the power tube, enough on it. B, the rest is the frequency and chip drive capability, and here only talk about frequency. Frequency and conduction losses are proportional to the power tube heats up, we must first consider the frequency selected is a bit high. To find ways to reduce the frequency of it! Note, however, when the frequency is lowered, in order to get the same load capacity, the peak current is bound to become larger or larger inductance, which may cause the inductor into saturation region. If the inductor saturation current is large enough, consider the CCM (continuous current mode) changed to DCM (discontinuous current mode), thus requiring the addition of a load capacitance.
4, down-conversion on the operating frequency
The users in the debugging process is more common phenomenon, reduced frequency caused mainly by the two aspects. Input voltage and load voltage ratio is small, a large system disturbances. For the former, careful not to load voltage set too high, although the load voltage and high efficiency will be high.For the latter, you can try the following aspects: a, set the minimum current and then a small point; b, wiring a clean point, especially sense the critical path; c, choose a small point of the inductor or inductor selection of closed circuit ; d, plus RC low-pass filtering it, the impact of a little bad, C is not good consistency, bias a little big, but it should be enough for the lighting. In any case the benefits of reduced frequency is not only bad, they must be resolved.
5, the choice of the inductor or transformer
Finally comes to focus, and I have not started, can only point of the nonsense. Response to many users, the same drive circuit, with a production of the inductor is no problem, with the b production of the inductor current becomes small. In such cases, to look at the inductor current waveform. Some engineers did not notice this phenomenon, which directly regulates the frequency of sense resistor or the needs of current work, this may seriously affect the service life of LED. So, in the design of the former, a reasonable calculation is necessary, if the parameters and calculated parameters of poor commissioning a bit far, to consider whether the down-and transformer saturation .Transformer saturation, L become smaller, resulting in transmission delay due to a sharp increase in peak current increment, then the LED will also increase the peak current. In the premise of constant average current, only to see a bad light. |
