# Tag Info

## New answers tagged buck

0

OK, so you summed all your 5V loads and get 10A, so you decided on a 5V 10A converter... But then, you have to distribute +5V everywhere, at pretty high current, which needs fat wires. And you have 8V also, which means three power distribution networks all over the vehicle. I'm going to suggest another option: distribute 24V only, and use point-of-load ...

3

Using an adjustable LM317? No. Just no. There's a lot of reasons, but the top two are power consumption and it wouldn't work anyway. First, the LM317 is a linear regulator, and linear regulators work by burning up power in the output transistor (dig up a schematic and look -- the output current is always a bit less than the input current, and has to be). ...

2

The most important thing in a DC-DC layout is to minimize the area of the hot loop. This minimizes both its inductance (which causes L.di/dt spikes) and emissions. The hot loop is the loop with highest di/dt. In the case of a buck, input current is a square wave, output current is a triangle, so the highest di/dt is on the input side. For a boost, it's the ...

1

3.3 μH is too low for your application (You probably copied the typical application circuit from the datasheet). The calculated value is 5.8 μH. I don't know which Pi you're using, but depending on the model, a RaspPi may draw too much current at start up. So you should use a higher inductance with higher saturation current. Also you should consider the SRF ...

1

First, as @Kartman pointed out in the comments, replace electrolytic capacitors with ceramic. Both of them, C2 and C3. Second, I strongly recommend you to redo the PCB routing. First rule of these converters is that SW trace should be as small as possible. You have those huge copper pours sticking out beyond L3 and C3. Get rid of them, you already have ...

35

They key property of the LDO type linear regulators is the P-type pass transistor. This allows controlling the pass transistor with a voltage below the input voltage $V_{in}$ for all possible output voltages. Therefore, the output voltage can in principle be set arbitrarily close to the input voltage, especially for MOSFETs. Non-LDO type regulators that ...

10

A low drop out regulator (LDO) is a linear regulator. Linear regulators control the output voltage by adjusting the current through an output transistor. In all linear regulators, there is a minimum difference between the input voltage and the output voltage that you have to maintain. Take the ancient 7805 linear regulator. It needs the input voltage to be ...

31

An LDO is a type of linear regulator. All linear regulators have what's called a dropout voltage, a minimum input-to-output differential that they can't work without. The original linear regulators (including the still-popular 78xx and 79xx series) had dropout voltages on the order of 2~3 volts. Since the input-to-output differential is directly related to ...

4

A low dropout regulator is a type of linear regulator where the drop out voltage is relatively low. For example, an LDO might require an input voltage only a few hundred millivolts above the output voltage, while a classic linear regulator like the 7805 requires at least 2 volts. For more information, see: https://en.wikipedia.org/wiki/Low-...

0

The entire theory of switching power supplies is based on switching and its frequency. So no, there's no way to calculate any component, anything about the circuit, without knowing the switching frequency. And before you ask, no, there's no practical ways either, except trial-error.

2

But whether I can arrive at the formula of the output capacitor without any formula involving the switching frequency? The minimum value of output capacitance is based on the amount of ripple voltage; more capacitance means lower ripple voltage; higher switching frequency (same capacitance) also means lower ripple voltage. The ripple voltage is directly ...

1

Yes, this works and is normal. Note that there are some things to be aware of. If the controller is a current-mode one, it may not be able to handle negative current flow. While the circuit is boosting the V applied on the right to the load on the left; the regulation is on the right-hand side -- it doesn't regulate the output - it regulates the input. This ...

2

Yes, it’s possible to pass energy in either direction, but with a caveat: the left-side voltage must be higher than the right side. Have a look at this Q about boost converters: Difference between synchronous and conventional dc-dc boost converter? Notice something? The synchronous boost looks a lot like a synchronous buck, turned the other way. So the ‘buck’...

5

Yes, and boy, didn't it come as a surprise to me the first time I did it! I saw what was happening, said "T'oh!", and had to spin the board. If you put a capacitor across $V_{in}$ and look at the power flow from right to left, you'll see that you have a boost converter. To a first order (i.e., ignoring components going up in puffs of smoke, ...

0

I've designed buck converts with this exact configuration and it works very well. The main difference, application wise, is that you have to run two wires to your load. Aside from that, it's easier to control, calculations and performance are the same - even better - because it's easier / cheaper to find low resistance N-channel MOSFETs.

2

I'd say don't ignore N MOSFET based DC-DCs just because most can't run at 100% duty cycle. Anything that runs on 12V +/-5% should run fine on 11.8V too. So why not use a chip like LTC3851 that can run at 99% duty cycle? There are other such chips, and of course the PMOS based ones which can run at 100% duty cycle. Note that using a NMOS does not mean it can'...

0

I have had to do this long ago. This was for a 12/24 volt input with no gear change. This will give you a clue in how to proceed. I used a p channel scheme which made drive easy without any fancy highside chips. I could get RDs on of 27 milliohm which was not too expensive and adequate for from memory 6.3 A DC. Using discretes to drive the device was easy ...

1

You may consider what (almost) everyone else prefers, that is include high side differential amplifier for an external sense R. This must also consider the DCR of the L chosen and $I_{rms}^2R$. This R may be placed before or after the switch depending on optimal phase margin for damping low and high loads. Practical recommendation I would consider R=1mohm/...

0

I ended up replacing C2 with a schottky, L1 with a real inductor (SMD 0805) of an unknown value (too small for my meter to measure, less than 10uH or too low Q factor). Drove the circuit at 100 Khz with a PWM resolution of 80 bits. Works like a charm!

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