New answers tagged switch-mode-power-supply
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Choose a ceramic capacitor with an X7R temperature coefficient and a voltage rating of at least 25 volts, and an inductor with at least a 1 amp rating at 125 Celsius. The schematic is
simulate this circuit – Schematic created using CircuitLab
Keep the traces as short and wide as possible, and run the ground trace from C7-ground directly to the ...
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I would imagine with 15% loss max at 2x160W Class D = 48W , it will need forced air with a tiny case fan with a plenum to maximize the air velocity over the fins.
Here the coils are twinned to support the current needed with 8 coils. Basically they need 2 per channel for differential Class D switching to double the voltage and power.
https://www.st.com/...
answered 18 hours ago
Tony Stewart Sunnyskyguy EE75
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0
As long as you follow the layout guidelines, there should be no concern over not passing EMC.
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For this power class, it would still be work to use a linear regulator. Indeed, it can get a bit warm, but as long as you have some proper grounding for thermal dissipation, it should be alright. Furthermore, you would not have to worry too much about EMC.
A quick search at TI website gives the following LDO LM317, which delivers at least 500mA if the ...
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Never heard about "burst-mode". Pulse skipping is a simple method to limit the output voltage / current within the maximum permitted margins. The simplest analogy is a children swing. You push every cycle in resonance as long you have enough amplitude, then you skip pushing until the swing goes lower. In a resonance circuit there is a similar behaviour, if ...
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root of mean of squared current, thus:
$$RMS= \sqrt{ 0.38 I_x^2 + 0.62 I_y^2 } $$
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Your power supply will still work. However doing so means that you no longer have isolation between the primary (input) and secondary (output) sides of the power supply.
Whether or not this is a problem or a violation of some requirement depends on the particulars of your system. Many systems require that the primary and secondary returns (grounds) of ...
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When you short the output ground to input ground, there is no isolation anymore. However, the outputs still give you the specified voltages referred to input ground.
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It is not a mistake, the author is correct.
..at \$\ f_c\$ (where \$\ T=1\$, with associated \$ 90^{\circ}\$. phase lag), the closed loop gain \$\ G(s)\$ is 3dB down..
Let \$\ T=1\angle90^{\circ}=\cos(90^{\circ})+j\sin(90^{\circ})=j\$
The closed-loop transfer function is:
\$\ G(s) = \frac{1}{K_{FB}} \cdot \frac{T}{1+T}\$
Substituting \$\ T=j\$ and ...
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My question is where should the V_sense resistors connect, before or
after the secondary LC filter?
It should almost certainly NOT go after the 2nd LC filter because you'll turn the circuit into a power oscillator.
With feedback systems you have to take care of not introducing too much phase shift or delay or negative feedback (desirable) becomes ...
-1
You are absolutely right. The loop gain (open loop gain times the feedback fraction) is unity at the frequency where the closed loop response is down 3dB at -45 degrees. It is only for unity closed loop gain (feedback fraction = 1) where the closed loop gain is down 3dB when the open loop gain is unity. For a closed loop gain of 2 (feedback fraction = 1/2), ...
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You need to specify the conducted emissions, on both input and output voltages. You also need significant input L+C filters and output filters. You need a Ground Plane under this region. Without these methods, you can pick "the best switchReg" and still fail.
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My answer is concise in comparison to others:
It's best to know the switching frequency of your power-supply, measure or read from the datasheet the spectrum that it outputs and then tune filter networks for those particular frequencies and tones you're up against. Just bulk filtering can be difficult and often times not so effective.
It's relatively easy ...
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The leading-edge blanking circuitry or LEB is a common circuit found in modern switching power supplies. It is usually built as shown in the below circuit:
When the drive goes high, a short pulse of \$t_{LEB}\$ duration blinds the IC for a small period of time. We are talking about 250-350 ns for ac-dc controllers operating below 100 kHz and around 100-150 ...
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This is a classic in switching power supplies. The auxiliary \$V_{cc}\$ is an image of \$V_{out}\$ and is dependent upon the transformer turns ratio linking the auxiliary and power windings. That is to say, if you have a 12-V output and a turns ratio of 1:1 on the auxiliary winding, then the auxiliary will be 12 V also. This is a theoretical approach because ...
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In a word: don't
The best way to drop 10 volts at 0.25A is already answered by other people, so I will not cover that.
Your question, however, aims at powering an ESP32 module specifically, and you don't disclose more details about the intended usage. However, if you are targeting an IoT scenario with WiFi communication, you need to consider that the ESP32 ...
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I'd use a circuit similar to one given in Tony's answer.
simulate this circuit – Schematic created using CircuitLab
\$D1\$ (ES2J - but can be anything suitable) protects the circuit against reverse polarity. Power rating of both \$R2\$ (470R) and \$D1\$ (3V9 zener) should be at least 1W because their power dissipations increase under no- or light-...
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Best? no , Simple Yes
Get a junkyard Festoon lamp holder but instead of a 12V bulb, buy a 24V 10W bulb designed for 250mA constant current with just enough voltage drop to run any 3.3V LDO and only dump < 0.5W in the LDO and 2.2W out of 10W in the Lamp, so it will be dim.
If the bulb is 10W rated at 24V it will be 3 Ohms cold and if rated at 28V, it ...
answered Jan 17 at 5:12
Tony Stewart Sunnyskyguy EE75
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4
you're planning to burn up 2.5 watts a TO220 part is going to need a heatsink at that power level, but you can put a resistor in series with the input and make the heat there where it's not damanging anything
simulate this circuit – Schematic created using CircuitLab
at low current the reguilator sees near full voltage, but the current is low so ...
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How do you know the noise is 13KHz?
On the compensation pin, there is a 0.1uF capacitor in each circuit (C59 and C67). Will it cause the control loop unstable?
If the beat frequency is the cause, shutting off the one will eliminate the noise totally. Or, changing the load from very light to heavy should change the noise level, even frequency, significantly.
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Often it's a good idea to read the connector drawings from the connector manufacturer -- I used to keep the Pin-1 markers of all kinds of connectors in my notebook.
However, very surprisingly, Molex doesn't mark the Pin 1 on the KK-5.08 connectors, though they do for the KK-3.96!
(Source: Molex drawing for KK-3.96 connector)
(Source Molex drawing for KK-5....
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The switching power supplies don't like to work without any load - in most cases the output voltage will go too high (as you see). The current range for CH1 and CH2, shown in datasheet of yours SPS, starts from 0.2...0.4A instead of 0. Try to load your SPS (with resistors or lamps), then do the measures.
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Pin 1 is on the right side of the connector which is a bit unusual, and from the pictures I guess you thought pin 1 was on the other side, so that explains the wrong polarity.
The power supply has a minimum load on two channels, notice it says "0.4-5A" on channel 1, not "0-5A". This is common on this type of switching supplies.
It only has feedback (and ...
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try this circuit. It worked for me after 3 weeks of testing and burning out components. It works fine from 4.1V to 23.5V. You need a 10K trimpot (DS1804) and an op amp (LM311).
I tried with other op amps, but without success. Good luck.
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What you are hearing is most likely the piezo-electric effect from the MLCC capacitors - as various other comments and answers stated.
Your oscilloscope shot shows a 3.81 kHz oscillation with 300 mV on the 5 V line. This is very much audible. As your switching frequencies are all outside the audible range, the source of the oscillation is very likely from ...
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What it means is that the neutral output of the inverter is not connected to ground. On a car, ground would be the chassis.
The ground pin might be connected to the car chassis, or it might be totally disconnected.
If you are just plugging one appliance in, it isn't necessarily unsafe. If the appliance is double insulated, with a 2-pin plug, it's largely ...
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Many smaller inverters (and my Honda EU2000 generator) have "floating neutrals".
I don't think "floating" is quite the right term here, as the Neutral will alternately be connected to +170 V or Ground. What happens is that the inverter has a single high voltage supply (+170 V or so), and connects the Line to that supply on one half-cycle, and Neutral to ...
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It means that the output 115VAC is not connected to the input power in any way (not common with the minus, which is connected to the car chassis, nor to the +12V input). Similar to adding an isolation transformer to your mains that comes out of the wall (well, without the pesky design, safety testing and certification ... ).
That's a fairly subtle ...
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According to Onsemi conductors PDF manual for this IC, page 5, this IC has a ripple at the output of 400 mV. Their designs often suggest use of a low value inductor (LC network) to remove most of such ripple. However inductors and capacitors before any extra filter are subject to whine and chatter. This should only affect a few parts, and you CANNOT make ...
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1) All your noise problems come from two issues:
Timing cap* inductor pair are too big. Reduce L 20uH to 33uH low DCR. Move up to 30 kHz. I suspect you have resonance with your 4.7uF load and 100uH.
The e-caps are STD rated for low ripple current and must be low ESR. Thus the current is limited and the piezo ceramic caps are audible with too much current ...
answered Jan 12 at 4:44
Tony Stewart Sunnyskyguy EE75
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5
Whine or high-pitched whistling usually comes from coils and ceramic caps.
Coils whine due to magnetostriction: core material expands and contracts according to the magnetic field, so they convert current ripple into sound.
High-K ceramic caps (ie, not C0G) are piezoelectric: they expand, contract and flex depending on the voltage applied to them, which ...
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The feedback loop around the TL431 is a classic. It works by pulling the opto LED cathode to ground which, in turns, modulates the peak current setpoint in the primary side via the optocoupler emitter or collector current. This works well for a constant-voltage output. Then, when the output current \$I_{out}\$ increases, the voltage starts going down and, ...
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Basic external
Add a current sense (small resistor and comparator and reference).
When I is high switch in a resistor Rdrop using a FET.
You'd need to juggle the sense limit or add hysteresis so it limits initially and stays that way until I falls to a value where it will be safe when Rdrop is removed.
Rdrop can be eg Nichrome wire - high wattage at a low ...
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You are running class D, so to hit 250W BTL you will want somewhere in the region of about 65-80V or so assuming a single rail design and an 8 ohm load, dual rail would be +-35V to +-40V or so.
Now the 50W number tells you how to size heatsinks, but 5 seconds is forever in semiconductor junction terms so there is not a lot to be saved in the sand, it still ...
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Peak power in a SMPS means peak Voltage and Peak current .Your semiconductors must be rated for the job .Your ferrite inductors must not saturate at these high peak currents .Your input and output caps must not give too much ripple at these high peak currents .So the design is much like a 250 watt SMPS but heatsinking needs are much less due to the average ...
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If I use a 19V 4.7A supply to a MAX 250W ATX power supply am I an idiot or should it be safe enough?
If I understand this correctly, you intend to use this "PeakPSU", apparently a cheap PicoPSU knockoff:
...which is a board with a bunch of DC-DC converters intended to generate the usual voltages to power a PC motherboard from a 19V laptop power supply.
The "specification":
Note it doesn't spec all the stuff you'd really like to know about a switching ...
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Most (all that I used) of the impulse power converters have VOUT connected to the coil. This one has VIN connected to the coil.
Apparently you are used to buck converterns, then. It might be worth a web search. A buck converter works by switching power on and off to the coil, so the coil input sees an average voltage less than the supply. Coils try to ...
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Here are the innards:
This is a classic boost regulator.
Figure 3. A classic boost regulator. Source.*
The transistor shorts the inductor to ground causing it to "charge" up magnetically. when the transistor is turned off the inductor voltage rises to feed the output.
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Everyone has their own preferred simulation tool depending on what their background is and what they are simulating most often. You can get Almost all simulation tools to simulate any circuit(within reason) some just are a little easier in some scenarios. I would suggest you look at what simulation software the people around you or the people that you are ...
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There are many parameters to take into account while designing a transformer, so it's difficult to explain in details from scratch how to design it, what i can suggest is to start learning through webinars and so on.... then once you are more familiar to come back with specific questions, it might be easier for every one here i believe.
As a start, you can ...
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You possibly won't find a standard pre wound ferrite transformer but you can obtain the ferrite core on which you can custom wind a transformer. For 200W 1sq.cm transformer core is more than enough.
Since IR2153 is used, only option is push pull operation and no automatic stabilisation of voltage. I suggest you use a proper smps ic like SG3525 or TL494 etc,....
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"How to determine parasitic capacitance of a diode according to the datasheet?"
There are two simple answers.
1) Read the data sheet and take the value you see . In the case of the data sheet you link, it is stated as being 12 pF.
2) Unless the data sheet provides a graph, you can't.
In general, reverse-biased diodes show a capacitance which decreases ...
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The Miller effect is due to having a capacitance across two points with a negative voltage gain between them.
[Intuitively you can think about one side of the cap getting pulled down by the gate-drain gain while the other side is pulled up by the gate driver, so you need to add the extra charge that the drain is trying to pull out.]
It multiplies the ...
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Functionally you can get a similar result in both your examples using a differential amplifier with some limited gain.
simulate this circuit – Schematic created using CircuitLab
answered Dec 31 '19 at 18:54
Tony Stewart Sunnyskyguy EE75
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6
To add another specific to what The Photon said, error amplifiers in switching regulators are often implemented as gm amps. So they take the difference in voltage on the input and convert it to a current with a gain of gm (A/V).
That's a really easy way for IC designers to save a pin as the compensation network goes between the output of the gm amp and ...
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Not every triangle you see in a schematic is an op-amp. All kinds of amplifiers are drawn on schematics as triangles.
In this case, the amplifier symbol likely represents the complete difference amplifier subcircuit, including both the op-amp and feedback resistors.
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Source Impedance as a function of frequency can be measured that is inverse to load regulation. The open loop impedance can be defined by RdsOn or Rb/hFE. Closed loop impedance reduces output impedance less with rising frequency due to a fixed GBW. This is reduced further by a shunt array of low ESR capacitors as long as it results in a stable "servo" ...
answered Dec 31 '19 at 16:24
Tony Stewart Sunnyskyguy EE75
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2
The TL431 is (probably) the lowest cost and best value for money IC in existence.
A TL431 in even 500 quantity costs under $US0.01 in China (and probably not vastly less in large quantities).
For that you get 0.5% accuracy voltage reference, 50 ppm/C variation including all parameters across whole temperature range, an internal comparator, useful ...
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My question is why the turn on speed is linked to the diode reverse recovery time ?
It isn't. What they are saying is: usually the FET can turn on faster than the diode can turn off, however that results in both the FET and the diode being on simultaneously, ie cross-conduction, and increased losses. So it is more efficient to switch the FET a bit slower to ...
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How does the LDO controls the output voltage of the buck converter as well as the switching frequency of the transistor?
Can't tell you – they're tying together input voltage and ground, so this circuit is operating the 78xx outside its operational boundaries; so, this definitely depends on the properties of one or two specific 78xx implementations, and ...
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