# Tag Info

21

Switching regulators and switching controllers are very similar, and essentially perform the same function. Both of them are DC to DC converters. Both switching regulators and switching controllers can be obtained/configured in either buck (output voltage < input voltage), boost (output voltage > input voltage), or both topologies. The buck/boost ...

19

"Is there any benefit to using a voltage booster on batteries that are below a device's cutoff/operating voltage?" Of course there are benefits in that situation: a battery that would otherwise be dead can still be used for some time. But probably not for long, so it is debatable whether this is usefull. What DJ (IMO correctly) argues is that the Batteroo ...

17

Linear regulators work by effectively putting a controlled variable resistor between the source and load. All of the current for the load flows through this resistive element. And the voltage across it is equal to the difference between source voltage and load voltage. So the power dissipated is $P_{lin} = I_{load}\times{}(V_{src}-V_{load})$. Switching ...

13

Every voltage drop in a regulator is a source of inefficiency. Synchronous regulators replace the normal schottky diode with a MOSFET that is open when the primary MOSFET is closed and vice versa to reduce or eliminate even the small inefficiency caused by the diode. This is true in both buck and boost regulators.

11

Usually switching regulators are more efficient, but not always. An ideal linear regulator has a voltage drop $V_{IN} - V_{OUT}$ and there is a linear pass element such as a transistor that acts like a resistor, so the power loss in the ideal case is P = $I \cdot (V_{IN} - V_{OUT})$, as you say. That's the ideal case, in reality the regulator needs a ...

10

simulate this circuit – Schematic created using CircuitLab Figure 1. Opto-isolated signal detector. An opto-isolator electrically isolates and protects your micro from the AC and eliminates any ground-loop issues. R1 limits the current to about 10 mA rms. D2 provides reverse voltage protection for the opto-LED, D1. D1 will illuminate on positive ...

10

Yes, that works. The advantage is that the low side switch is easier to control, since its input is ground-referenced. The downside is that the load is not ground referenced. If you are sure you have a floating load, then this is a very valid thing to do. I usually drive solenoids with a similar circuit, for example. This topology also works if the load ...

10

10 bits at 200kHz requires a 200MHz peripheral clock, and 12 bits at 0.5MHz requires a 2GHz peripheral clock unless you have a fancy enhanced resolution PWM peripheral. So lower PWM frequency means more expensive, heavier larger inductor. It’s difficult to get the MCU to regulate its own power. “Soft” firmware disruptions or bugs can cause physical damage. ...

8

Using a switcher instead of a linear regulator is a no-brainer without even doing the math. Also a LDO specifically is silly, since your problem is that you have a very large drop range. Any linear regulator would do, not just low-dropout ones. In case some still need convincing, let's do the math. A linear regulator will drop 19 V, which times 100 mA is ...

8

Our goal is to keep the load on the batteries running as long as possible. In general, these loads are either fixed resistance (like a basic flashlight) or fixed power (like almost anything electronic beyond a certain complexity). A fixed power load is generally a switching regulator, which has a minimum dropout voltage. A fixed resistance load doesn't much ...

8

I think you're a bit confused about what these parts do ! The THD 12-2412WI DC-DC converter is an isolated power converter. It is like a mini-mains adapter but with a low voltage input. It takes power from the input and using switching and a transformer makes the power available at the output. But as it is isolated, there is no electrical connection between ...

7

It is well-know that switching regulators are more efficient then linear regulators. To a point. Putting 3.5V into a LDO 3.3V linear regulator gives an efficiency of 94%. You'd be hard-pressed to find a switching regulator that can do that. I also know that linear regulator have to dissipate the different between the input voltage and output voltage ...

7

Quite likely, the choke isn't broken. It is normal for chokes to act as speakers, especially under high momentary current (think of transformer hum on loaded transformers). So your choke got noisy, because there are irregular current peaks across it. You observed the irregularities on your scope. The most likely reason for erratic voltage at the choke (...

7

Ignacio has covered the main point. A lesser point but still useful to know is that a synchronous buck regulator (for instance) will be more stable than a non-sync buck generally. This is because the output voltage equals the input voltage x duty cycle and, if you have very good switches (low on resistance) then all you have to do to get decent output ...

7

Look at the datasheet, section 7.2, the functional block diagram: Note how the Feedback pin connects to a resistor voltage divider (R1, R2) which generates an error signal which goes into the Error Amplifier. This error signal is compared against a 1.23 V reference voltage. If you add a resistor in series with pin 1 as shown in your schematic then this ...

6

Wow, this is a tough one. First, there is clearly no component that would cover the whole range. Due to huge investment that is required for an IC development, any such component is designed for specific market. Although each manufacturer will try to support as many applications as possible, nobody will invest in such small niche. I can say it's small for ...

6

Yep, that voltage makes sense -- look at U1. It's a non-synchronous converter; even if it's "off", you will see a diode drop on 5V from VBATT because current can flow through the inductor and through the diode with nothing stopping it. In a synchronous converter, that diode may be a switch (MOSFET) which would prevent the issue you are seeing. EDIT: Here's ...

6

No, this sounds reasonable, if it is what you want to do. One problem is that you will drop 9-6=3 volt at 4 amperes, that's 12 watts. You need a hefty heat sink for that. A better option is perhaps to have a 6V output from a switching regulator, and a 5V low-dropout linear regulator, converting 6V to 5V. The 5 volt regulator would only have to drop one volt,...

6

simulate this circuit – Schematic created using CircuitLab Figure 1. This circuit does everything you asked for in your question. Label your switches as required.

6

Your inductor is connected wrong. You should have wired it up to pins 1 and 3 (diagonally opposite): As a result the inductor is basically on open circuit, so you will get no output. Given your VCC trace passes right next to pin 3, you should be able to simply scape off any soldermask from the trace where it passes the inductor pin, and make the connection ...

6

You don't do anything with them, they are fine if they are floating and left open. If you're worried about a charge developing on the coil (which probably won't happen), you could tie one end to ground. Tying both ends to ground would result in a short and a considerable loss in energy.

5

Switching regulator design is complicated indeed, and it even requires a proper PCB layout thanks to the high frequencies and EMI involved. But it's by no means impossible when building it around a controller IC and following its datasheet's recommendations. But consider that for about the same 3 euros, you could buy a ready-made switching regulator module ...

5

See the datasheet (page 4, note 2): The LTC3388E-1/LTC3388E-3 are guaranteed to meet specifications from 0°C to 85°C junction temperature. Specifications over the –40°C to 125°C operating junction temperature range are assured by design, characterization and correlation with statistical process controls. The LTC3388I-1/LTC3388I-3 are guaranteed ...

5

Here is a quite short list of TI's Buck/Boost converters which have PDIP packaging. There are only 2 of 33 DC-DC converters which are available with though-hole package , and the only difference of the listed items is the Operating Temperature Range (C°). So as you can see and as @Dave Tweed said, through-hole design is not popular. Let's have a look on the ...

5

I would take a close look at the National Semiconductor (now TI) LM2575 family. Very, Very easy to use. LM2575: 1 Amp buck converter LM2576: 3 Amp buck converter I would suggest a couple of things, though. Have a bunch of simple PC boards made up that the students can assemble. Although this SMPS family is very forgiving, grounding DOES matter if you ...

5

Although the absolute value of a digital potentiometer's resistance may vary 30%, the matching between the internal resistors is really good. This means that for a voltage divider (to wit, a potentiometer), the voltage division accuracy is quite good, since the voltage division factor relies entirely on the ratio of the resistors used, not their absolute ...

5

One possibility is to have a boost converter bypassed by a diode. When the output is 95 V or above, it does nothing (the raw voltage is just passed to the output thru the diode). When the output drops below 95 V, the boost converter runs and makes 95 V from the 12-95 V input. Now use that to run a regular "universal input" off the shelf 12 V power supply. ...

5

TI's got a tool, named WEBENCH which can make a lot of charts and calcualtions for you. Here is its output with your parameters in pdf. Let me highlight the one about the efficiency. The simulations shows that this IC has a better efficiency when Vin is 20V, but this difference is not that much. It is not just the Vin that matters, if you change the ...

5

Well, unless some kind of charge-balancer is used, most types of energy storage devices in series (batteries, capacitors, etc.) will charge unevenly. Since a supercap may be rated +/- 10% in capacity, imagine what happens when strung together and charged in series: the one with the -10% capacitance charges first, and the voltage across it goes higher than ...

5

To clarify, it seems you want to detect the presence of 12 V AC in a processor that apparently has 5 V digital inputs. Here is a simple circuit: The positive halves of the AC input signal will turn on the transistor, which drives the output line low. D1 prevents the transistor from getting damaged by excessive B-E reverse voltage during the negative half ...

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