Questions on DC buck converter components - Page 1
Forum section: electronics / beginners

author: joeyjoejoe
author_rank: Frequent Contributor
timestamp: September 29, 2018, 01:44:53 pm
content: I've been playing with a simple homemade DC-buck. (Sure makes me glad I picked up a 4 channel scope instead of a 2 channel!) This is a continuation of my project to build a battery charger - the driving goal: to learn various things. China offers many battery chargers for much cheaper, but with no learning involved either!I'm curious how each component affects the buck converter. I've read some literature by TI, and I guess there are tradeoffs to be made by choosing different components.My understanding so far:- The higher the PWM frequency, the smaller the coils can be for the same set of parameters- The output capacitor is mainly chosen for output current ripple and affects the response time to large changes in current- The Diode can be replaced with another MOSFET, which primarily (solely?) improves the efficiencyQuestions1. What generally dictates the voltage range of a DC buck converter? Right now, with a 1% duty cycle, I can't get the voltage very low. (The input voltage could be adjusted of course, but I picked 12V as a common adapter size.)2. What does the size of the coil affect? My suspicion is bigger coil = more energy stored = more available current?3. I have seen some circuits with a capacitor on the input side of the coil as well. I'm not sure where this fits in, and if I can ignore it for now while figuring out the basic circuit.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 29, 2018, 02:49:59 pm
content: Well, what's an inductor?The law of inductance is:V = L * dI/dtSo, you apply a voltage across an inductor, and the current ramps up (dI/dt is a change in current over time).Say you apply a voltage for a brief moment, then turn it off and wait. The inductor starts at zero, charges up to some dI = V * dt / L, then discharges through the diode. (If the output voltage is also about zero, then the inductor discharges into the diode, with a forward drop Vf ~ 0.7V say.)That current flows into the filter cap, which has a similar law:I = C * dV/dtIf we've applied an average (dI/2) to the capacitor in this time, then its voltage will have changed by the amount dV = I * dt / C.Applying a little bit of calculus, we can get more precise. The ripple voltage waveform on the capacitor will be a parabolic section, because it's the second integral of the input, which is a switched square waveform. The integral of a square is a triangle wave, and the integral of a triangle is a parabolic section (and so on to higher order curves). So rather than applying a constant current and getting a ramped output, we will see the ripple curving upwards, but changing by the same total amount. Nothing tricky about that.When the switch turns off, inductor current ramps down, and the capacitor stops charging (again, along a downward curved parabolic section). When the inductor is finished discharging, capacitor voltage either sits still (no load current), or drops at a rate proportional to the load current. If the load is resistive, the current depends on voltage drop, and the voltage discharges along a decaying exponential curve.The waveform marches on in this way, every time it repeats. If the switch turns on while the inductor is still discharging, then the output ripple doesn't see a decaying exponential section, the inductor ripple current dominates instead.What does this do for output voltage?If the inductor is fully discharged between pulses, then inductor current is discontinuous mode (DCM). We can use an energy argument here: E = V*I*t, but in particular, the integral of V*I (power) with respect to time. When the switch is on and the filter capacitor is relatively large, the applied voltage on the inductor is constant, while the current is a linear ramp. The area under the curve V*I is therefore the area of the triangle: (Vin - Vout) * Ipk * t_on / 2. Which is also the inductor energy E = L * I^2 / 2 (homework: show that this is the case). The output power is the average energy, E * F_sw (F_sw = repeat rate of the switch). V and I are then given by Pout and R_L.In CCM (continuous current mode), the output voltage is simply the average of the switch voltage waveform, but be careful that we're not simply averaging PWM. I mean, that happens, but it shouldn't be your focus. We are charging and discharging an inductor. The inductor is key, and we must always control its current first and foremost. If we drive PWM directly, we're doing what's called voltage mode control, and it's very easy to blow up the switch (which has a hard current limit -- you aren't going to protect a MOSFET with a fuse!). With current mode control (average current mode control is easiest for CCM, peak current mode for DCM), you canguaranteethe switch will never blow up, and as a side effect, the input and output currents are limited, too!Tim

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 29, 2018, 02:52:46 pm
content: Also, you've drawn the MOSFET backwards... the body diode is pointing into the load, so for that exact case, Vout = Vin - 0.7V or so.If you use a N-ch FET the right way up, note you need to apply Vgs(on) to gate-source, which since source is moving with respect to ground, Vg(on) has to be Vin + Vgs(on). This is usually done with a bootstrap gate driver (you can roll your own, or get an IC). Otherwise, there's a P-ch FET where Vgs(on) is negative (i.e., "hanging down" from Vin), but performance is about 2.5 times worse (i.e., more Qg and Rds(on)).Tim

author: joeyjoejoe
author_rank: Frequent Contributor
timestamp: September 29, 2018, 03:14:54 pm
content: Woops, that was before the 2nd coffee, I just whipped it up in KiCad to get a schematicI'm indeed using a P-channel mosfet (IRF5305Pbf), with a BC547 to trigger the gate.So how do some of these PMIC's (ie. LM2596) get such a wide range of voltage (1.2V - 37V on the adjustable version) with fixed components? Or can they?I notice in their data sheet, they do indeed specify maximum input voltages to give a certain output voltage at a certain load. Presumably, a similar problem to what I'm having.I suspect this is one of the trade-offs for the efficiency? Whereas a linear regular might easily work from 10mA - 5A, with a much finer tolerance, at the expense of wasted energy.