Capacitor sizing for IR2104 (half-bridge driver)

Question

I am planning to use the IR2104 half-bridge driver in a project to control the intensity of a fairly powerful LED array using the PWM output from a Raspberry Pico.

The main question I have is how to size the capacitors that are present in the IR2104's typical connection schema.

I have made a diagram of my full circuit as it is planned for now, taking into account the comments I have received so far. My computations for the capacity C_B of the B capacitor are detailed at the end of this post.

The LED panel is composed of 4 LEDs in parallel, with a power rating of 50 W at 12 V each, but this is largely overestimated as I measured the 4 LEDs to draw only 3 A all together in parallel at 12 V constant voltage. I am planning to overdrive these LEDs, but with a very finely tunable PWM (hence the two potentiometers). I might also add a thermal probe later if this works. This is just an educational project for me.

I would like to use a fairly high frequency for the PWM (the Pico has PWM between 7 Hz and 125 MHz), I am hoping I can reach 100 kHz.

I guess that the capacity of the capacitor that is connected between V_CC and COM (on the left on the manufacturer's diagram, D on my diagram) is not very important as it is just a decoupling capacitor. Am I correct?

I am more unsure about the capacitor between V_B and V_S, that I denoted B on my schema. Does its capacity have an impact on the maximum achievable switching rate?

I am not totally sure I understand the purpose of the capacitor that I denoted C either, nor which capacity it should have.

I am planning to use an IRF3205 N-channel MOSFETs.

Main edit of this post, computation of C_B

Based on the comments I received and using the formula detailed in the an answer that was suggested, since Q_G is 146 nC, I_G is 100 nA (not sure I'm reading the right cell for this one value, I took the value of I_GSS, "gate to source forward leakage"), and I'm aiming for a 100 kHz switching frequency, I compute the following:

$$C_B \ge \frac{146~\mathrm{nC} + 100~\mathrm{nA} \cdot \frac{1}{100~\mathrm{kHz}}s \cdot B}{\Delta_{V_B}}$$

where \$B\$ is the duty cycle and \$\Delta_{V_B}\$ is the acceptable voltage drop at the gate of the capacitor, which I will assume to be 0.1 as in the answer I am taking the inspiration from, which sounds reasonable because my micro-controller will output 3.3 V and the IRF3205 starts to turn on at V_GS(th) = 2 V and the 3.3 V will only be the voltage applied to the gate for the first cycle if I understand correctly, before the bootstrap capacitor has had a chance to charge from the high voltage source of the load.

So the "worst case scenario" for this equation would be with \$B\$ = 1 so I should be able to determine C_B with:

$$C_B \ge \frac{146~\mathrm{nC} + 100~\mathrm{nA} \cdot \frac{1}{100~\mathrm{kHz}}s}{0.1~\mathrm{V}}$$

The units in the equation are homogeneous, but the equation will give me C_B in \$\frac{~\mathrm{nA} \cdot ~\mathrm{s}}{~\mathrm{V}}\$ which are just nF, so I get:

$$C_B \ge \frac{146 + \frac{100}{100000}}{0.1} = 1460.01~\mathrm{nF} \approx 1460~\mathrm{nF} = 1.46~\mathrm{\mu F}$$

Thus I am planning to use a 2.2 μF capacitor for B because that is the nearest capacitor value I have that is greater than the computed result.

Does that seem reasonable to you?

Answer 1

^{simulate this circuit – Schematic created using CircuitLab}

It would be good to incorporate a form of current regulation as shown above, although you might want to add a base resistor for Q1. Easiest way to drive the LEDs is to put them from the supply to an NMOS device.

The simulation shows a current limit of about 7A, equally shared between D1 and D2. This may not be so with real components.

I realize this does not answer the OP's question, but it does provide what is IMHO a much better means of achieving their goals.

(edit) Upon reading the specs, it appears that these LED modules have their own driver modules, so external regulation may not be needed, and may interfere with PWM. And the 50W version appears to be for 36VDC.

If they do not have internal driver, then the linear current regulation proposed here will dissipate a good deal of power. With a 16V supply and 12V on the LEDs, 3A will be 12W AT 100% pwm. That can be corrected with an inductor in series with the LEDs, and a commutating diode to carry current during PWM OFF. This results in a buck converter with triangular wave in continuous conduction mode.

However, further discussion of the actual implementation of the LED driver should be done in a separate question to be posted by the OP.

^{simulate this circuit}

That was my idea for making this a simple switching regulator, but something doesn't seem to work as I think it should. I'm dealing with some brain fog right now so I'll have to revisit this. Maybe the OP should start a new question specifically on ways to implement an efficient way to drive these LEDs.

The problem is that the square wave 50% PWM causes runaway continuous conduction mode. By using a smaller duty cycle, I was able to get good results. But further discussion and design details should be moved to a separate question, if the OP is interested in following this approach. My LTspice simulation: