18
If proper engineering is followed, the switching noise shouldn't impact the readings and the oscilloscope weighs less, takes up less space, and draws less power. Thus, many benefits to a single increase in engineering difficulty.
13
The SMPS advantages over linear supplies, of compactness, high efficiency so low heating, light weight and wide input voltage range, are very valuable in portable equipment like an oscilloscope.
The principle disadvantages of SMPS over linear supplies are switching noise on the output and radiated EMI.
(Circuit complexity used to be considered another but ...
4
The LNK302 datasheet that you linked recommends that you change R4 to 2.4Kohms or less if the circuit has zero load.
3
The high frequency switching used in the SMPS can inflict some quite considerable high frequency common-mode noise on the DC output. This is due to the interwinding capacitance of the internal isolation transformer (aka flyback transformer).
To reduce this noise (and obtain EMC approval) requires that the DC output is either grounded or capacitively ...
3
I'd recommend trying to describe a simpler circuit than the one you posted in your question.
A buck converter is a much simpler SMPS. It is also DC-DC, so you don't need to worry about AC rectification:
(source: Wikipedia)
In a buck converter, the output voltage is less than the input voltage. I might describe it this way:
This type of SMPS works by ...
3
I'd suggest you use a 3.3V LDO with a series SCHOTTKY diode such as the SDM40E20LS or SDM40E20LA dual diodes. These diodes have very low Vf values of around 310mV @100mA.
Connecting two in parallel will get you about 1A capability at 410mV drop.
The LDO may have some minor supply current (about 10mA) if your SMPS drops by 0.1V during normal operation, but ...
2
In the technical literature, the parameter \$M\$ refers to the dc transfer function of the considered converter: \$M=\frac{V_{out}}{V_{in}}\$. For a boost converter operated in the continuous conduction mode or CCM, it can be shown that \$M=\frac{1}{1-D}\$ with \$D\$ the duty ratio.
2
it's called "burp mode" and it's normal.
the normally ultrasonic noise of the power-supply becomes audible when the power-supply stops an restarts several times per second because the load is too light to have it run continuously,
Usually it's one of the magnetic parts making the noise, but some capacitors can also act as transducers.
1
it's effectively a comparator relative to 1.25V.
It's not a comparator, it's an error amplifier ("EA" on the block diagram). The 33nF cap and the internal series RC are part of the frequency compensation.
It's unclear what kind of voltage range you're trying to achieve. Maybe you want something like Vfb = +1.25V + a*Vout- Vpwm, which would ideally yield ...
1
You have already described linear power supplies transformers rectifiers and smoothing capacitors, then you are half way there.
A switch mode power rectifies the mains to give a high DC voltage. This voltage is then "switched" or converted back to AC at a high frequency. The high frequency AC is reduced in voltage using a transformer and this low voltage AC ...
1
Short answer: if you can't explain inductors to your grandmother, then you cannot explain switching power supplies. Can you explain resistors? Capacitors? Yes? Well, first you have to "own inductors," and understand them as totally as you understand resistors, but without math and without lingo.
Most people understand resistors, but less about ...
1
The basic-basics: input -> magic -> output, except the magic part doesn't violate the laws of thermodynamics, so power out = power in * efficiency. Meaning you can't get 10 watts out with one watt in, for example. It explains why a boost converter generates high voltage/low current on an output with low voltage/high current on the input, it explains why a ...
1
Yes the supply current will be about double on 12V as compared to 24V.
The average current through the pair of active MOSFETs is the coil current, which is fixed. The power dissipation moves from one MOSFET to the other as the supply voltage changes. For example, one might conduct for 10% of the time with 24V and the other conduct 90%, but that would change ...
1
The answer of Verbal Kint makes most sense.
If you multiply the DC gain on slide 35 by \$ \frac{R_L}{R_L} \$
$$H_0 = \frac{k_0-k_i}{g_f - (g_0+g_i) - g_r - \frac{1}{R_L} } \frac{R_L}{R_L} $$
and substitute the equations given in slide 27 into the term \$k_0-k_i\$,
then \$H_0\$ evaluates to
$$H_0 = R_L \frac{\frac{1-D}{R_i}}{g_f R_L - (g_0+g_i) R_L ...
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