New answers tagged

1

Here are a couple of inductors used often as radio-frequency chokes. They are wound so that capacitance is minimized: Both inductors have no ferrite cores, but this kind of winding is appropriate on ferrite. These windings are divided into segments (4-segments for the smaller one, 5-segments for the larger one) - the space between segments reduces ...


1

does segmenting the secondary into 10 or so segments of a couple of hundred turns each, separated by 1 or 2 millimeters of plastic help to reduce the capacitance? Yes In a transformer winding, inter-turn capacitance is generally negligible compared to inter-layer capacitance. We can approximate the winding self-capacitance within a small factor by ...


1

This is a bit late, but... For this frequency range proximity effect losses will be much higher than skin effect losses. Proximity effect loss is reduced by using Litz (also good for reducing skin effect losses) or bunched conductors (less effective at reducing skin effect losses). Loss resistance due to proximity effect is proportional to \$ number\_of\...


2

Measure the dynamic resistance as per the instructions in the datasheet. Plot a I-V curve of your LED around the bias point and draw a tangent at the bias point to have a line with dV/dI. Then you can calculate a value for the capacitor.


0

Ideally the first step would be to determine if this is even a problem. Maybe the 1080 Hz signal will not cause any problems for you. But if it does, one simple idea is to put a 100 uF or 470 uF capacitor on the output of your SMPS, if possible (if it will remain stable with this much capacitance). At 1080 Hz, the 100 uF capacitor is a load of 1.4 Ohms or so....


1

For reference about mains signals injected ... https://goughlui.com/2014/02/02/project-analyze-mains-power-flicker-issue-ripple-signalling/ Here is an example of removing 1050 Hz, 10 V peak ... with coupled inductors (soft coupling K=0.5). Without filter With filter, AC analysis ... And transient analysis results Another topology that seems more efficient ...


3

It looks like this paper uses fractals to prevent current crowding in a high power MOSFET. In high power transistor design, you want to maximize current by minimizing resistance. If you start with a single basic transistor (fist order) with two metalization finger contacts, the wider the transistor is, the lower the transistor active region resistance is. ...


1

Solutions to problems: Agreed. C1 is there to mitigate this. If the external trimmer introduces too much noise, you can increase C1, at the expense of step-response performance. Perhaps more importantly you need to consider your external trimmer part of the signal integrity loop. I think by the "one trim terminal" remark you're indicating the rail ...


0

I thought I had stumbled on a way to use Fibonacci to filter the input/output of a power converter, by using LC filters according to a Fibonacci progression of values, but as was pointed out to me, my example did not necessarily prove anything. Therefore, in place of my original answer, I have improved my answer by merely providing one good example of ...


6

R4 and R5 apply a limited supply to the chip during the power-up-from-cold scenario. Once operating, the chip receives sufficient power from the secondary winding of the main inductor (named T1). So, now R4 and R5 can be forgotten about. What you are left with is the secondary winding of T1 producing an AC voltage (due to MOSFET Q1 switching). That secondary ...


2

The data sheet is says this: - Can anyone say, what is the exact difference between these? I think you'll need to speak with the supplier or try and work it out yourself by programming a few settings.


2

The common grounds need to be connected, but you should avoid sharing the wires to the power supplies. The common connection should be close to the destination, not the source. You don't want current spikes in the 24V common wire to affect the 5V power.


0

You most likely need a power supply rated for "inductive loads" or "constant current limiting".


3

As with most things when it comes to switching converters, performance under a wide range of loads or even no load is highly dependent on the topology. Resonant converters are an entire class of topologies, rather than a single topology, so there is only so much that can be generalized about them. What you describe is true for series resonant converters (...


2

If the output ripple is 'small', then the voltage across the inductor is approximately constant, and its current has a linear ramp. This linear-ramping current will generate a parabolic ripple on the capacitor (the net capacitor current is the inductors ramp minus the (constant) load current). If the ripple is very large (equivalent to C being small, and ...


3

This circuit behave as an Underdamped second order system if we consider voltage source as an input and inductor current as an output. And we know that for an Underdamped second order circuit, initially output is proportional to input (you can verify by looking a graph of step response of 2nd order Underdamped system) but after some time output start ...


4

in DC transient response of an LC circuit we get sinusoidal waveforms, so why not here as well? Short answer is "we do get a sinusoidal waveform" but it's only for a short period of time. Quite literally, the boost converter switching speed is so much higher than the natural resonant frequency of the inductor (and output capacitor), that you never ...


0

Real engineers do things like this all the time. The key is to ask yourself what could go wrong, and what would be the consequences? Then you ask yourself what can you do to make sure the worst consequences don't take place? If it seems like I am equivocating that is because I don't know you and what experiences you have had. Every time you energize a device ...


0

The solution for First issue is found! The reference design and MAX17690EVKIT evalkit has an unnecessary load circuit, D303 + R323, on the secondary side. And the heat source is that circuit. Resoldering the unnecessary circuit solves the heat issue. Here are some pics before and after the modification: However the primary side still seems hot. I will try ...


1

The fusable resistor differs from a normal resistor in that the fusing characteristics are specified and civilised .On high energy stuff like your single phase mains or forklift batteries random resistors can do bad things like Explode or go on fire and not self extinglush or release foul gas .Remember that you are poking Kw into your 1 watt resistor if ...


0

Technically, the equations remain accurate, but if the values change in operation, so do the equation variables. This is why we use power-factor-correction capacitors on motors - it helps to ensure the load maintains its 'proper' reactance.


0

PROBLEM in the initial picture schematic : short between pins C1 and C2. Replaced 2 secondaries by one. Bridge rectifier used. Same behavior. Q1 - This note , figure 12 tell you the input voltage you can use ("input" 0 -> 20 mV) for amplifier (so around Vref = 5 V). See figure 11 for properly biasing techniques ... Q2 - You should use simulator ...


4

You're probably right about the inductor being the problem. According to the datasheet 33uH might not do it -- 22uH or 15uH is called for. Try this inductor, Chilisin Electronics BWVS00808040150M00 (3.6A Isat, 15 µH Shielded Inductor 3.2A 50mOhm SMD) from Digikey for 48 cents. EDIT: I apologize -- I read the datasheet wrong because you said that you had a ...


3

For a linear PSU I would really recommend not using the negative-bias trick to try to get 0 V from a 3-terminal regulator. As per this question stability of the negative rail directly affects output stability. Linear regulators are simple enough to make building them with an op-amp no problem (in which case as @Michal observes you can get as close as the op-...


6

Without using a negative voltage the bottom of regulation range is always dependent of what minimum voltage you are able to amplify (for closed loop control). For single transistor without biasing it is about 0.7v A single transistor with biasing about 0.2v A transistor differential pair about 100mV A rail-to-rail opamp 50mV Considering the basic follower ...


7

For switching converters that use a 1.25V reference, you need to bias the bottom end of the feedback chain to be at the reference voltage. You can do this by using a second supply (eg 5V from a phone charger), this 5v can the be used to power the front panel instruments also. simulate this circuit – Schematic created using CircuitLab


1

Here's a CircuitLab model of the output stage of the MC34063, driving the push-pull pair of transistors you have implemented. The 1.6nF capacitor represents (somewhat simplistically, I admit) the input capacitance of the MOSFET. The simulation shows that the output cannot drop below 0.6V or so, which is not surprising for an emitter follower configuration: ...


7

Here's a reference for a 0-3V adjustable LM317 power supply without the minimum 1.2 - 1.3V output voltage limitation. https://www.edn.com/use-an-lm317-as-0-to-3v-adjustable-regulator/#id2782607-48-a


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