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1

Pls reduce the resistance in Vin to maintain 15V across the IC UC38xx . The MOSFET minimum required gate drive voltage is 10v other wise it will be in ohmic region and generate heat as well as reduce power due to low contuctance


0

Designed output power = 12V x 10A = 120W High efficiency say 95%, heat dissipate = 120W x 5% = 6W You have to use a large heat sink with fan.


9

If it is confirmed the MOSFET needs a better drive then the best is to advise a suitable circuit. A simple bootstrap circuit described by Monsieur Balogh in a TI application note can do the job nicely in a cost-sensitive application. As noted in some of the comments, the UC384x was not really meant for hi-side drive - unless you make it entirely float and ...


6

I’m guessing the N-FET isn’t being turned on all the way. It's being biased in linear mode and so has a substantial IR drop across it, which is being shed as heat. You don't want that. How bad is it? Let's assume the FET Vgs threshold is about 4.5V. Then when it's on: Vd = 60V Vg = 36V - 2V = 34V (limited UC3845B output swing) Vs = (Vg - Vth) = (34 - 4.5V) ...


6

MOSFET ON time and OFF time is decided by a UC3845B based circuit The UC3845B has a push-pull type output but, unless this chip is powered from a supply that is several volts higher than the 60 volt rail (as shown in your circuit) you won't efficiently drive the MOSFET in your buck regulator. Given that the UC3845 is only rated at a maximum of 36 volts, you ...


0

You can check if a heat sink is needed by calculating (at least theoretically), by checking the numbers in the datasheet. Normally these are called Tj (Junction temperature) and all related heat dissipation figures are shown together. Also read https://en.wikipedia.org/wiki/Junction_temperature.


1

Between the +5V and the -12V outputs, you have a total of 17V, definitely not what you need. Connecting the two together will damage the power supply. Not good either. The proper way to get +7V is to use a buck regulator to regulate the +12V down to +7V with good efficiency. This is usually easier than to convert from a negative voltage. The Alibaba web ...


1

Nicely done photo - I wish tehy were all that good. That's not "missing a leg" - it has been blown apart - presumably by electrical failure. There is a reasonable prospect that there is other damage. Datasheet here. Look for damage on lines connecting to pins 2 (input) and 3 (output) BUT could be anywhere. The capacitor equivalent to C2 in datasheet ...


-1

Type conversion is possibly a easier and more efficient way to do this. For hex to std_logic use binario <= to_stdlogicvector(hexadecimal); Make sure you include the ieee std_logic_arith package.


2

Shortly, it's the effect of \$V_L=L\ di/dt\$. Here's the equivalent diagram of your setup (neglecting the real model of the inductor): simulate this circuit – Schematic created using CircuitLab The inductor sees a positive DC voltage for the half of the period and a negative DC voltage for the rest of the period. So, from \$i=\frac{1}{L}\int{V_L\ ...


6

No. Modulation (and demodulation) imply that the information is being imposed on a carrier of some sort, such as the audio tones used in voiceband telephone modems, or the RF carriers used in cable TV and wireless systems. Ethernet (except for its wireless variations) uses baseband signaling, in which the information symbols are transmitted directly over ...


8

All modems are PHYs (physical layers) in the OSI network model, but only some PHYs are complex enough to be considered modems. Why? I think it depends on four things: simple or complex modulation carrier or baseband WAN or local historical use For example, an Ethernet PHY uses simple encoding depending on the speed in use (MLT, PAM). These schemes have a ...


2

In a flyback transformer, a gap is mandatory for energy-storing reasons but also for stabilizing the inductance in production. We can show that the magnetizing inductance of an iron-core inductor featuring an air gap is expressed by: \$L_{mag}=\frac{N^2A_eµ_e}{l_m}\$ with \$µ_e=\frac{µ_rµ_0}{1+µ_r\frac{l_g}{l_m}}\$ in which \$µ_e\$ represents the ...


1

I'm inferring that LED stability is important for your application. The ripple will definitely affect the output with a noise spectrum of harmonics. My suggestion? Use both. Use DC-DC to make the primary voltage, followed by an LDO (perhaps wired as a current regulator?) to post-regulate the ripple.


0

You could add a power lowpass filter to remove the fundamental and higher frequencies. There are a ton of examples and calculators if googled. The simplest may be a series resistor of a low value (maybe 1 Ohm) and a parallel capacitor of say, 1000µF. The resistor drops a little power while operating, but the cap conducts AC, causing the AC component (ripple) ...


2

The ripple on a buck converter will have a negligible effect on LED intensity, and will be far to high a frequency to be visible in any case. Note that such a converter could also be set up to operate as a constant current source (also with negligible, high frequency ripple), providing greater control of the intensity over all conditions.


3

As the name implies, a SMPS relies on the principle of transferring energy from input to the output cyclically, and this is accomplished via a switching device and a corresponding driver. Thus, there will always be high frequency components (main switching frequency + harmonics + higher order frequency components) at the output of your power supply, but it ...


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