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What are those parallel soldered-traces on the PCB traces?
They are traces where some of the soldermask has been removed in order to allow solder to build up to improve the current capacity of the trace.
The reason they left the narrow parallel lines of soldermask is so that the solder distributes itself more evenly across the width of the trace. ...
9
What are those parallel soldered-traces on the PCB traces?
When you are designing a PCB with high current traces, you want to keep the trace resistance as low as possible. You can do this by multiple methods:
(a) Use a thicker copper layer eg 70 microns instead of 35 microns.
(b) Use bus bars to carry the current instead the PCB traces. These bars are ...
5
In case of high power adapters, high current has to be supplied. Higher current requires either a wider copper trace and/or a thicker copper trace.
Widening the trace
Widening a trace reduces the trace resistance and allows a bigger area for the electrons to flow across without much resistance. While this method works in most of the cases, it fails when ...
5
What are these soldered/un-soldered dotted-shapes?
These act as test points. As pointed out by @oldfart, these could help facilitate bed-of-nails test. It looks similar to how it sounds:
This is important in case of mass manufactured items. In this case, before you label it as working, you might want to run some tests. For ex - is the output voltage ...
3
Because lab supplies will actually be tested by people with unpredictable loads and high-quality measurement equipment.
Thus, the quality of voltage stability and accuracy needs to be significantly higher for lab supplies.
Do a test: get a 2 Ω power resistor and attach it to your notebook supply to draw ca. 10 A; measure the voltage with that load, and ...
3
No I do not think you can.
The Hybrid Pi models are small signal models meaning the model is a linearization of the behavior of the transistor when biased in a certain state.
The Hybrid Pi models are useful for analyzing and designing circuits where the transistors are used as more or less linear amplifiers.
For circuits where the transistors are ...
3
In addition to the other good info in answers;
Heavy parts soldered to large circular pads need more mechanical strain relief than electrical conductivity to avoid a microcrack annular ring , which is quite common due to shipping vibration and handling. The heavy track void lines also provide additional surface area for bulged solder to conduct some more ...
2
It is certainly possible.
The size of capacitor needed would likely be in the super-capacitor class, with a capacitance over 1 F. However, you would now need to adapt your power electronics to handle this new load. When the super-cap is discharged it almost looks like a short-circuit from the perspective of a low power linear regulator.
It would be much ...
2
You are not back powering the SMPS regulator you are back powering the regulator load. Providing the load current on the switching regulator (Buck or Boost) is much higher than any current from a clamp there should be no problems at all.
However if the current drawn from the regulator is less than the current from the clamp (or many clamps) then the VCC ...
2
It is pretty standard practice to clamp a signal to Vcc by using a diode and resistor. However, when this happens, some small amount if current (maybe ~50mA) is flowing back into the Vcc line. When using a switching regulator to supply the Vcc line, what implications does introducing this current source have on the system? Is this safe?
Brief: Energy ...
2
There may be problems if you have unpowered peripheral devices connected to a powered microcontroller. You need to make sure that all microcontroller pins that are connected to these peripherals are set to a low voltage (logic 0) before removing power to the peripherals. Otherwise, significant current will flow through the input protection circuits of the ...
2
This is what I get out of the setup you envision:
SMPS -> VReg* -> SMPS
And your asking about the rail with the * on it.
In this case, the last SMPS will affect the regulation of the Vreg because it is switching current and because the Vreg has source impedance, and cannot respond infinitely fast to regulate the voltage.
The question then becomes, how ...
2
L1 reduces the noise generated by the converter. If too much electrical noise gets into the mains cord the product will fail EMC.
L1 is far too small to reduce mains frequency artifacts. If L1 had enough inductance to attenuate mains frequency harmonic noise, the small cheap SMPS could double in size. Passive mains harmonic filtering is not cheap and when ...
1
You need to be looking at the trip curves for MCBs when planning for inrush:
https://studyelectrical.com/2014/07/miniature-circuit-breakers-mcb-types-characteristic-curves.html#Selection_of_Right_MCB
You can see that 10ms and below is not on the graph. Inrushes below 10ms will not cause a trip, by design.
From having designed similar equipment I would ...
1
0.3-0.8 is really low voltage but, on the other hand, no other info are provided therefore i assume that we can draw as many current as we wish.
Forget to do something at 0.3V but starting from 0.5V we can harvest something.
The U1V10F5 can operate from 0.5V, with a max current of 1.2A. The output voltage is 5V, and the efficiency is between 70%-90%, we can ...
1
Your high side fet is not driven with high enough voltage. You can't connect the gates of the FETs in parallel like you have done. TC4427 is intended to have a P-channel FET on the high side. And actually the high side FET is not even needed here, you can use only the low side.
So change to P-channel FET on high side or remove it completely
1
The inverting converter, your first figure, operates this way.
When S is closed, current builds up in the inductor. The node joining S and L is positive. The output node is negative, which means D isolates the input and output. Current builds up in L with no reference to the output conditions.
When switch S opens, we have a current in L, and a stored ...
1
The output voltage on the MAX757 is set by a voltage divider between ground and the output which is connected to the feedback input (pin 2), the formula to calculate the required resistors is:
VOUT = (1.25) [(R2 + R1) / R2]
To get 5V I have used 30K for R1 and 10K for R2, for 3.3V you could use 18K for R1 and 11K for R2.
If you can find it an alternative ...
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