Okay, I revisited that bypass-capacitor I was planning. I bought a few of these
Cornel-Dubilier 1mF 35V capacitors and originally I planned to go overboard and add up to 6 of these (one per channel).
When one considers that this bypass capacitor (together with the capcitors included inside the power supply) is supposed to smooth the load current almost to a constant DC current one can approximate the current flows over time. FOr this to be true the capacitance needs to be large enough compared to the nominal resistance of the load and the applied frequency:
Code:
capacity >= 10/(resistance_nom * frequency)
In my case:
Code:
resistance_nom ~ 24V / 3.83A ~ 6.266 Ohm
frequency = 1500 Hz
capacity > 10/6.266/1500 ~ 1 mF
When that is fulfilled the calculation becomes a bit easier:
Code:
I_nom = 92W/24V ~ 3.83A # lamp current under full power
duty = 0.0 ... 1.0 # PWM duty cycle
I_average ~ I_nom * duty
I_discharge ~ I_average * (1-duty)
I_charge ~ I_average * duty
Which one can understand principally as well: when the duty cycle is low, the capacitor has a long time charge back up, so the charge current is low. Vice versa, when the duty cycle is large the charge current increases, so I_charge is somewhat proportional to the duty cycle.
Conversely, when the duty cycle is low, the discharge current supplied to the load in the short active time is very large, since the capacitor needs to make up the difference between the average current from the power supply and the much larger nominal current, that the load draws. Vice versa, when the duty cycle increases, the discharge current becomes smaller, since more and more current comes from the power supply until at a duty cycle of 100% the power supply takes all the load and the capacitor is idle.
In this picture the ripple current, described by its RMS (RootMeanSquare) becomes:
Code:
I_RMS^2 = 1/T_period * [ I_discharge^2 * T_discharge + I_charge^2 * T_charge]
I_RMS^2 = 1/T_period * [ I_discharge^2 * duty * T_period + I_charge^2 * (1-duty) * T_period)]
I_RMS^2 = T_period/T_period * [ (I_average*(1-duty))^2 * duty + (I_average*duty)^2 * (1-duty))]
[...]
I_RMS = I_nominal * sqrt(duty - duty^2)
For my case of I_nominal = 3.83A this becomes:
As one expects, the most load on the capacitor happens around a PWM duty cycle of 50%. When one considers the datasheet, the maximum RMS ripple current for my capacitors are 2770mA @ 100kHz and +105°C. I don't know yet how this relates to my conditions of 1.5 kHz at room temperature, but I'm assuming that I'm more or less safe.
To make sure, I took the Dissipation factor of my 35V capacitors, which is tan(delta) ~ 0.12 and calculated the Equivalent Series Resistance (ESR) of the capacitor and from that the electrical power on the capacitor using the maximum ripple current of about 2A:
Code:
ESR ~ tan(delta) / (2*Pi*frequency * capacity)
P_el ~ ESR * I_RMS^2
Which for me equates to about:
Code:
ESR ~ 0.12/(2*3,14159*1500 Hz * 0.001 F) ~ 0.013 Ohm
P_el ~ 0.013 Ohm * (2 A)^2 ~ 50mW
That sounds absolutely manageable.
My much smaller diodes for example had a thermal resistance of about 300 K/W. I'm guessing from
this paper, which gives a thermal conductivity of about 10-20 W/m^2/K for the surface for still air and the dimensions of my capacitor, that I should have something like 50 K/W, which sounds reasonable, since the capacitors are much larger than the tiny diodes. This means my capacitors shoud heat up by about 2,5K (in the worst case maybe something like 10K).
