Donald L Burdette [dlburdette@JUNO.COM] Wondered:
...could a switcher be designed that would use parts available from ANY Radio Shack or well-stocked junk box? I'm happy to report the answer is YES!
I went into my workshop yesterday afternoon, and after a few hours had cobbed together a design that works pretty well. For the inductor it uses the secondary of a 12V step-down transformer. The control element is a '555 timer and the power transistor (yes, only one) is an old PNP thing I had (2N5981). A 1N4001 diode, a 12V zener, some 2N3904's, a few resistors and caps finished the parts list. The output cap I used is 6000uF, but your output ripple voltage requirements may change that.
I get about 11.25 V because my zener is 10.7V instead of 12V. I get about 0.7A before the thing comes out of regulation. The power is limited mostly by the gain of the power transistor and the rating of the transformer. It would probably work better with a FET or newer high-gain bipolar. I get about 100 mV peak to peak ripple, most of which is due to the ESR of the capacitor, which must be about 0.05 ohms.
Mine runs at 2000 Hz, but you could increase that to maybe 5K or even 10K if you have a good FET. At 5K my transistor gets hot (that would be ok if I put a heatsink on it). Unfortunately, the transformer gets less and less efficient (hotter) as the frequency goes up.
The advantages and disadvantages I see (compared to "simple" designs) are:
- Parts are easily available.
- Layout and construction methods are not critical.
- Design can easily be tinkered with to scale up or down, or adjust to your requirements.
- You can get any voltage or any group of voltages you need. If you have the right transformer, you can get +5V, -5V, and -12V from one inverter.
- Parts are larger. My transformer is about 1.5 x 1.5 x 2 inches. Caps are bigger too.
- Parts count is higher, though maybe not compared to 3 separate "simple" designs.
- Parts cost may be higher unless you have a good transformer on hand. But again, maybe not compared to 3 separate "simple" designs.
- Regulation is not as good.
The core of the 12V inverter consists of Q1, T1, D2, and C3. When Q1 is on, D2 will be reverse biased and current will begin to build in T1. When Q1 turns off, that current will be forced to flow through C3 and D2, which will now be forward biased. Due to the high inductance of T1, the circuit operates in continuous mode, which means that the current through T1 never drops to zero, but ramps up and down about an average. I haven't actually measured it, but I suspect in my prototype the variation is less than 5% of the average.
Q2 is used to drive Q1 and invert the output of the '555. Note that since the base drive current of Q1 is 100 mA, R2 will dissipate significant power. The calculation is: 0.1 x 0.1 x 120 = 1.2 watts. Multiply by the duty cycle which may be as high as 60 or 65%, and you get about 0.78 watts. R1 ensures that Q1 turns off quickly, minimizing power dissipation and improving efficiency.
U1 is the oscillator which provides the PWM signal to Q1. R4, R5, and C5 create a 2000 Hz signal of approximately 60% duty cycle. As the output voltage increases, the feedback reduces the duty cycle and therefore the power supplied to the output. The frequency is also somewhat increased.
D3, Q3 and Q4 provide the output voltage sense and feedback to alter the PWM duty cycle of U1. As the output voltage increases, D3 conducts and turns on Q4. Q3 mirrors the collector current of Q4 to the control pin of U1 and protects that pin from being pulled below ground, possibly damaging the IC.
The -5V output is derived from the tap on T1, and is regulated by the autotransformer nature of T1. This means that if pin 3 is at -12V, pin 2 will be at -5V. In other words, the voltage ratio of the outputs is the same as the turns ratio of T1. Other voltages could be derived similarly by using other taps on the transformer. This is the way most PC power supplies are built - one transformer, many taps, only one of which is explicitly regulated.
It should be noted that if the -12V supply is unloaded, the -5V supply will not be well regulated. A certain minimum load must be placed on the -12V supply. I don't know what this is, but it should not be very high (10-20% of maximum load should be plenty).
- Use a FET in place of Q1. This could provide much higher output currents. My prototype runs out of steam at 0.7A because I don't drive Q1 hard enough to get more current through it. However, this provides one advantage. When first powering up, the current in T1 can build fairly high before C3 is fully charged. If Q1 were able to provide unlimited current, T1 might saturate, and the current would suddenly rise very high, possibly destroying Q1 or blowing the fuse. Since Q1 is current limited by virtue of its low gain at high current, we have somewhat of a soft startup.
- Increase the frequency. I measured the switching time on Q1, and it is around 400nS. This suggests that frequencies of 50-100 kHz are not unreasonable. An iron core transformer would have very high losses at this frequency, so you'd need ferrite. Be warned that a small ferrite would be much more likely to saturate on powerup. If you are using a FET, beware! You might also want to replace D1 and D2 with high speed diodes. You can use smaller output caps, but beware that that often means lower ripple current ratings and higher ESR!
- Change the transformer. The large core of a 12VA iron transformer is way overkill, even at 2000 Hz. Unfortunately, when you go to a smaller core, you get smaller wire and lower current ratings. What would be ideal is a smaller core with fewer turns of bigger wire. What that means is a lower voltage output rating with the same current rating. Good luck. If I really wanted to reduce the transformer size, I'd get a Miller high current torroidal inductor from Digi-key and boost the frequency to around 50 kHz. If I wanted more than one output voltage, I'd wind more turns and taps on the thing. That's not likely to be hard, since they don't have lots of turns on them.
- Reduce output ripple. My prototype has about 100 mV peak to peak ripple. About 30 mV of this is due to the current, capacitance, frequency combination, while about 70 mV is due to current and ESR (1.4 A x 0.05 ohms = 0.07 V). Note that the peak to peak current in C3 is around twice the output current. Since it's nearly a square wave, the RMS current is about half the peak to peak value. You could also set the output voltage higher and use a linear regulator to get really clean output.
- Improve the voltage detector. What's here is kind of crude. Though it works reasonably well, a reference diode and op-amp might improve it a little. Beware of making the gain too high though. I initially had a 1K resistor in R8, and that caused oscillation when Q1 turns off - when Q1 turns off, output voltage increases suddenly due to current suddenly flowing through the ESR of C3, which turns on Q4, which pulls down U1 pin 5, which turns Q1 back on, which causes the output voltage to decrease. Reducing the ESR of C3 should minimize this problem.
- Overload protection. If you overload this thing, you are likely to destroy Q1.
|file: /Techref/piclist/burdetteswitcher.htm, 8KB, , updated: 2013/6/27 15:13, local time: 2018/9/26 00:08,
|©2018 These pages are served without commercial sponsorship. (No popup ads, etc...).Bandwidth abuse increases hosting cost forcing sponsorship or shutdown. This server aggressively defends against automated copying for any reason including offline viewing, duplication, etc... Please respect this requirement and DO NOT RIP THIS SITE. Questions?|
<A HREF="http://www.sxlist.com/techref/piclist/burdetteswitcher.htm"> the Burdette Switcher</A>
|Did you find what you needed?|
Welcome to sxlist.com!
& kind contributors
just like you!
Please don't rip/copy
Copies of the site on CD
are available at minimal cost.
Welcome to www.sxlist.com!