feedbackpower supplyvoltage-regulator
I'm designing a simple 15V to 5V DC voltage feadback mode non SMPS regulator.
I'm using E13007E2 power NPN switch mode transistor. And to drive that transistor
I'm using C945 another transistor.
This is my schematic diagram of the circuit.
It gives 4.5V output when it's not loaded. But when it's loaded and after some
time the C945 had gone. It's 2 junctions are open now. c945 is to shunt just only
6mA. Why it can't handle 6mA?
Any idea how could I calculate the operating points precisely?
High side switching is always tricky. There are no easy and simple ways, only various tradeoffs.
PMOS transistors are nice in that they can work within the existing voltage. The gate voltage needs to be pulled below the input voltage by 12-15 V to turn them fully on. The downside is that P channel MOSFETS usually have a little worse characteristics than the equivalent N channel.
N channel may have a better combination of Rdson, voltage tolerance, and cost, but require you to somehow make a voltage higher than the input to drive them. Some high side FET driver chips include a charge pump or other trick for this purpose. Another downside of a N channel high side switch is that the gate must swing a much larger amount, from zero to 12-15 volts above the input. This is because the gate voltage is relative to the source, which is now riding up and down with the voltage being switched. This requires high slew rates to stay out of the partially on region as much as possible, and provides more opportunity for noise pickup elsewhere.
There is no easy solution.
However in your particular case you may not need a high side switch at all. As W5VO mentioned in a comment, a flyback topology only requires a low side switch on the primary. The high side can stay connected to the input voltage.
A center tapped primary with the transformer run in forward mode is another possibility. The center tap goes to the input voltage with a low side switch pulling each end alternately to ground. Again there is no free lunch, which in this case is exhibited by the low side switches now having to withstand twice the input voltage. This is why the center tapped topology is more used for lower input voltages and usually not for worldwide "universal" power, which needs to handle up to 260 V AC or so. That would mean 368 V peaks, and 735 V stress on the low side switches. Transistors with that kind of voltage capability give up other parameters, like gain in bipolars and Rdson in FETs.
There is no free lunch.
I meant to say this earlier but somehow it slipped thru the cracks. You will most likely need a transformer anyway to get isolation. Unless you really really know what you're doing, you want the resulting supply to be isolated from the power line. The main exception is if the power stays completely inside a sealed box and there is not even a ground connection to the outside world. Otherwise, you run the risk of a user getting connected to the hot side of the AC line should even a few simple things go wrong. There is good reason commercial power supplies are mostly isolated.
Given that you probably want isolation, the problem becomes how to drive a transformer as apposed to how to make a buck switcher directly.
The problem is your current measurement method and assumptions for load current.
I get I(R1) = 177 mV dc /150 Ω = 1.18mA with a 1K load at 9V out on 470uF cap and 9.7V Zener with Iz=17mA pk using Si diodes.
Since nominal Zener Vz= 9.4~10.6 @ 5mA and Zt is around 15 Ω at 17mA pk
I used 1k load for 1.18mA average. Assumptions for load on mine were different than yours.
Your measurements were:
Your assumption on how multimeters work is wrong for AC current.
To make accurate complex current measurements with cheap multimeters, use a shunt resistor and 10K R with 100uF cap and measure DC voltage.
Best Answer
Circuit Lab has simulations
Link to the above circuit
They did not have your parts number in their library.