New answers tagged switch-mode-power-supply
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Turns out, the problem was unrelated to the design of the circuit but rather the startup of the input signal. The input signal defaulted to ground, which resulted in the mosfet shorting the power supply across the inductor. Connecting the ground wire of the Arduino last when powering up the circuit has fixed this issue.
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The body diode of the p-channel mosfet is forward biased, so it will always conduct through L1 which makes it fry: swap the drain and source of the PMOS.
Figure 12 of https://www.allaboutcircuits.com/technical-articles/analysis-of-four-dc-dc-converters-in-equilibrium/ and some other pictures on the internet incorrect as well.
The mosfet driver is missing its ...
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What’s the FB pin doing?
Anyway, looks like it’s current limiting. Try disconnecting it from the load. Another option is to use a bench supply to add current to the 5V (use current-limit mode). This would be supplementing the DC-DC output, so you could see what the load looks like.
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Some ideas...
600W is a lot. So some thoughts on how to manage that power.
A common technique with high-power drivers is to connect multiple drivers in parallel. You can do that with the 2n3055. This will reduce heating in each one.
Heatsinks that take multiple TO-3s are easy to find, either as new or salvaged from disused gear like linear power supplies (...
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There’s nothing inherently wrong with using an n-FET on the high side (and it's the preferred solution since n-FETS typically have lower Rds(on) for a given size), but watch the body diode (you have it backwards in your design). Also, to drive the high-side FET you need a bootstrapped high-side gate driver to make a high enough Vgs to turn on FET M2. (Huh, I ...
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You can't use M2 like that. MOSFETs have an anti-parallel body diode and yours is upside down freely conducting. You also need a high side gate driver.
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All half bridges using dual NFETs must use PWM only needed on low side with series cap to Vdd diode clamp to create a boost voltage >2Vgs(th) above Vdd to drive the high side gate voltage.
The high side then can be controlled independently for say a Boost / Buck conversion with an inductive Load by storing + or -ve inductive Current energy.
the clamp diode ...
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You're trying to use an N-channel MOSFET as a high-side switch. That isn't going to work without a special gate driver that includes a boost circuit.
Besides which, you've got it the wrong way around — think about which way the body diode is pointing!
You should probably be using a P-channel device at M2, but even then, you have to make its gate ...
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A switch mode regulator by its nature of 'switching' generates noise. You can kind of think of it like this: An LDO produces waste heat to drop the voltage while a SMPS produces waste noise.
Regardless of which you use, you generally want to do something about minimising the noise into them. While your LDO will produce less noise for a given input, it may ...
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You can use both. Use the DC-DC to drop the rail down to a voltage the LDO can use. You get the best of both worlds: DC-DC efficiency, and low noise for your RF stuff.
This is a really common design approach, that is, sub-regulating a local supply from a bulk DC-DC supply to create a rail with special requirements. The TI NMOS ‘cap-free’ LDOs are ...
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When comparing LDO's and Switching Voltage Regulators, which are generally the better performing option in terms of output voltage noise when considering an already noisy power supply.
As you have probably found searching internet, a LDO has generally less ripple/noise/EMI than a SMPS.
heat is not my only concern when considering which power supply
You ...
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The micro will surely be destroyed. You can prevent it by either preventing the short in the first place or by adding series or shunt protection. For example, you could add an LDO regulator on each board that is capable of withstanding 24VDC on the input (series protection).
Most RS-485 protection schemes are designed to handle transients, not a direct ...
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Your DC-DC won't work at all without at least one of the feedback transistors being turned on.
When they are turned on the VCE(sat) is directly in series with your feedback voltage. While VCE(sat) will likely only be a few mV in this case it still has an impact on the accuracy of you V(out)
You should be extra careful that the transistors used in the ...
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I would add a comment instead of an answer but my "reputation" is not high enough.
Check the following:
(1) Vin, Run and PWM pins. (Start with PWM of 0% duty)
(2) Check the FB pin is connected to the voltage divider [This needs feed back to work]
(3) Check if SW1 and SW2, are doing anything.
Never mind. Check the PWM pin; it should be <0.5V.
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You are asking for trouble; the drive frequency is 140 kHz and your primary series resonant tuning is precisely 140 kHz therefore, the L and C act as a short circuit at the switching frequency. You need to run the primary circuit either from an output stage designed to handle the series resonance or use a different approach - maybe add a current limit ...
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Faraday cage ideally do not have any holes or very less number of holes. So, the above might be probably used for cooling and safety.
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My method of calculating C bulk for AC rectifiers is based on energy storage required to prevent dropout if AC input cuts out for 1 cycle at max load.
The typical design on page 1 of your IC spec shows a Cin=680uF for a 12V to 5V @ 5A.
Let’s how close I get to their recommended design.
also Figure 23 shows the ripple current rating for the typical bulk e-...
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Since the LM2596 is a switching regulator, the input current will be much lower than the output current. Assuming a minimum of 24VDC input and 83% efficiency the input current for 2A output at 5V would be 2/(24/5)*1/0.83 = 0.5A. Using your formula the required capacitance is then 0.5*8.3*1000 = 4,150uF.
If more ripple is acceptable then a smaller ...
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For anyone interested, the issue turned out to be layout related.
It seems the feedback signal was such a high impedance, with the most sensitive part running near the transformer, noise was coupling into the system and being amplified.
There is an effective ~100x amplification of noise after the high voltage divider created by R9,10 & TM1 on the output....
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ATX Power Supply: is a minimum Load required on each supply?
I would suggest that the answer is still ...YES.
However the answer is more complex if you delve into the individual supplies.
Most supplies will meet the minimum Form Factors spec that @Jim quoted in the comments. However many of the top tier ATX supplies far exceed this spec.
In particular ...
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Over 20yrs ago a PC PSU would not perform as well as today and required some preload up to 10 % when good performance was expected with no load.
Today I have not had this experience and can easily operate all outputs with no load. So,there is no minimum load spec. The Intel spec is for no damage to the unit but shutdown is permitted.
Since the most ...
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How does the snubber capacitor acts as a short circuit during the
ringing frequency ? After all it is the one which decides the new
ringing frequency.
Texas Instruments are somewhat polarizing what happens - it doesn't act as a complete short but it does reduce the ringing frequency AND it also reduces the ringing amplitude. The amplitude reduces ...
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Does it really protect my digital circuit in case of Hazardous
situations like lightening and others?
The data sheet says this about surges: -
EN55024, heavy industry level (surge L-N : 1KV), criteria A
And all that means is that the unit will survive a 1 kV surge when applied to the AC lines. It doesn't mean that a surge will not produce a knock-on ...
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If higher resistance is not an issue, I'd still look into the fairly reduced current limit. At 3V, the FET will enter the linear region at about 1A of drain current. At this point the "RdsON" if the term is still valid will sky rocket, with potentially destructive power losses. Mind that your FET is 0.9 x 0.9 mm^2 !!!
Moreover, appart the 3.3V ON voltage I'...
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The voltage drop in a segment the supply wiring depends on the current in that segment.
The current in the segment from the supply to the top LED strip will carry the full 8.4 Amp, so you would use that current to calculate the voltage drop in that segment.
The 0.3 m to the second strip will only carry 7.8 Amp, and so on down the system.
The voltage drops ...
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This answer does not attempt to address your real question but is only to clarify your understanding.
This is my problem, I bought this circuit to receive caller ID over protocol DTMF.
As JRE has pointed out, you are confusing DTMF (dual-tone, multi-frequency) with Caller ID.
Figure 1. DTMF frequency matrix. Pressing any button transmits the row and ...
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