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PICList Project The Burdette switcher:

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. teknoloji , basın , yayın , tatil , örgü , dantel ,
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).

Possible modifications:

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.

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LED 12 Volt Lead Acid Battery Meter

In the circuit below, a quad voltage comparator (LM339) is used as a simple bar graph meter to indicate the charge condition of a 12 volt, lead acid battery. A 5 volt reference voltage is connected to each of the (+) inputs of the four comparators and the (-) inputs are connected to successive points along a voltage divider. The LEDs will illuminate when the voltage at the negative (-) input exceeds the reference voltage. Calibration can be done by adjusting the 2K potentiometer so that all four LEDs illuminate when the battery voltage is 12.7 volts, indicating full charge with no load on the battery. At 11.7 volts, the LEDs should be off indicating a dead battery. Each LED represents an approximate 25% change in charge condition or 300 millivolts, so that 3 LEDs indicate 75%, 2 LEDs indicate 50%, etc. The actual voltages will depend on temperature conditions and battery type, wet cell, gel cell etc. Additional information on battery maintenance can be found at:

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Telephone Ring Generator Using Switching Supply

The telephone ring generator shown below generates the needed high voltage from a simple switching mode power supply (SMPS) which employs a CMOS Schmitt Trigger square wave oscillator, 10 mH inductor, high voltage switching transistor (TIP47 or other high voltage, 1 amp transistor) and a driver transistor (2N3053). The inductor should have a low DC resistance of 1.5 ohms or less. The switching supply must have a load connected to prevent the voltage from rising too high, so a 22K resistor is used across the output which limits the voltage to about 120 DC with the phone ringer disconnected and about 90 volts DC connected. The output voltage can be adjusted by changing the value of the 150K resistor between pins 10 and 11 which will alter the oscillator frequency (frequency is around 800 Hz as shown). The supply is gated on and off by a second Schmitt Trigger oscillator (pins 12/13) so that the phone rings for about 2 seconds and then the circuit idles for about a minute between rings. These times can be adjusted with the 10K and 300K resistors connected to pin 12. The push button shown is used to manually ring the phone. The 25Hz ringing frequency is generated by another Schmitt Trigger oscillator (pins 1/2) which controls the H bridge transistor output circuit. The 6 transistors in the output stage (4 NPN, 2 PNP) should be high voltage types rated at 200 volts collector to emitter or more. The ringer will only draw around 10 mA, so the output transistors can have a low current rating but must have a high voltage rating. I used TIP47s and small signal PNPs of unknown numbers that I had on hand, but other types such as NTE287 (NPN) and NTE288 (PNP) should work. Both have a 300 volt C-E rating and cost about $0.95 from mail order houses.

The two 470 ohm resistors connected to the output serve to limit the current in case the output is shorted. I never tried shorting the output to see how effective the resistors are, but I did lose a couple transistors and then decided to add the resistors. They should limit the surge to around 120 mA which should be low enough to prevent damage. The circuit draws around 250 mA when the ring signal is present so if you want to operate it from batteries, six 'D' type alkaline cells are recommended. It probably won't work with a small 9 volt battery.

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This is the circuit which outputs 100 V of the alternating current from the input of 12 V of the direct current.
It is convenient to use the equipment which works in the alternating current using the battery (DC12V) of the car.

It is using the IC-type multi-vibrator for the oscillator of the alternating current. The frequency is about 60 Hz. I used 7400 as the IC for the oscillator but 7404 is OK.
The signal of the oscillator has the switching operation with TR1-TR4. TR2 and TR4 are the transistor for the main switching. Because these transistors are difficult to drive directly from the IC, they make amplify in the electric current using TR1 and TR3.

The connection between TR1 and TR2, and TR3 and TR4 connects in the way of being called "the Darlington connection".
The transformer that the input is 100 V and the output is 24 V in the one with the 12 V center tap makes the input and the output opposite and uses.

Because the comparatively big electric current (about 3 A) flows through the part of the line that the circuit diagram is bold, the thick wiring materials are used.

The output voltage is the square wave ( ).
Depending on the equipment, there is one which can not be used.
When the load is added, the wave form of the output voltage changes by the inductance of the transformer.

I was asked about 220V output from some readers. The output voltage of the inverter is decided only in the transformer. You can use the transformer with 220V as for primary(input) and 12V as for secondary(output). At my circuit, primary and secondary should be used oppositely. Then, you will be able to get AC220V from DC12V.

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Battery Charger Ideas

Here is the schematic for the automatic charger I have been using for my kids' battery cars. The charger is a small molded unit that probably doesn't supply more than an amp and this circuit would have trouble with much more. No current limit is provided by this circuit - it relies on the charger for that. The circuit could be modified to provide more current by lowering the 470 and 330 ohm resistors in the 5195's base circuit and the 10k in the collector of the 4401. A relay could also be used in place of the pass transistor.

Here is how it works: When the battery voltage is low, the voltage at the base of the first 2N4401 (on the right) is not sufficient to turn it on and the second 2N4401 is biased on by the 10k resistor. The power transistor is turned on and the LED lights. When the battery is fully charged the voltage will exceed a somewhat arbitrary "over-voltage" value slightly below 14 volts and the regulator will switch off. The 470k feedback resistor gives the circuit some hysteresis so that it will not turn back on until the battery voltage drops below about 13.5 volts. When the battery is nearing full charge the light will begin to flash on and off and after a few hours the light will only come on occasionally. This occasional over-voltage jolt sure seems to keep the batteries in great shape.

Here is an experimental (and simple!) regulator for alternator chargers.

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Gelled Lead Acid Batterry Charger

Part List:
All resistors are carbon, 1/4 watt, 5% tolerance, unless otherwise indicated.
R1 = 500 ohm              C1 = 0.1uF (100nF), ceramic       U1 = LM350T
R2 = 3K                      C2 = 1uF/40 volt                          U2 = LM301A
R3 = 820 ohm              C3 = 1000pF (1nF), ceramic        S1 = Pushbutton switc(normallypen)
R4 = 15 ohm               C4 = 0.1uF, ceramic (see text)
R5 = 230 ohm              D1 = 1N457 (or equiv.)
R6 = 15K                   Led1 = Red, 5mm, ultra-bright
R7 = 0.2 ohm, 5W, WW      Q1 = 2N2905, PNP, TO-39 case

This high-performance circuit first quickly starts (and holds) the charge at 2 amp, but as the voltage
rises the current will consequently decrease.
When the current falls below 150mA, the charger automatically switches to a lower 'Float' voltage to prevent
overcharging.
At the point that a full charge is reached, Q1 will bias and the LED will light
The is a 8-pin OpAmp. Transistor Q1 is a PNP, Silicon, AF-Out type with
a TO-39 metal case and can be substituted for a NTE or ECG129. Diode D1, a Si, GP Det. type, can be substituted
with a NTE177 or ECG177. Theneeds to be cooled.
The input voltage should equal or about 18volts
R1's ******** is to bleed some of the input voltage to the output and vice-versa. A 1N4002 or similar diode can be
used also.
R2 and R5 are actually metal-film type resistors. To get the 3K for R2 use two 1K5 (1500 ohm)
resistors in series. For R5 use two 470 ohm resistors in parallel. Or whatever combination to get to these values.
For R1, 500 ohm, you can use two 1K in parallel or 470 + 33 ohms in series.
R7, the 0.2 ohm resistor, is a 5 watt wire-wound type. Do notuse the standard carbon type.

C4:This (optional) 0.1uF (100nF) Ceramic capacitor needs to be mounted over the power lines and as close to the LM301 (U2)
as possible. It will filter off any possible residue hf ripple, which otherwise may prevent this op-amp from working
properly. Use only if you have problems with the LM301 not switching off
S1 is a subminiature pushbutton switch, normally open. I received a couple emails in regards to this switch.
I thought the diagram was pretty clear. It shows an 'open' switch with the arrow indicating a 'momentary'
connection when pushed. Nothing out of the ordinary here folks and a standard symbol for electronics...
When the start switch is pushed, the output of the charger goes to 14.5 V. As the batttery approaches full
charge, the charging current decreases and the output voltage is reduced form 14.5V to about 12.5V, terminating
the charging process. Transistor Q1 then lights the led as a visual indication of a full charge.

Note: This circuit will both work for Gel cell and regular L.A. batteries, and is from an application note found
in the National Linear 1 Databook as part of their regulator spec datasheets (#400041, Rev.1, Page 1-147).

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DC/AC inverter (1)

This is the circuit which outputs 100 V of the alternating current from the input of 12 V of the direct current.
It is convenient to use the equipment which works in the alternating current using the battery (DC12V) of the car.

It is using the IC-type multi-vibrator for the oscillator of the alternating current. The frequency is about 60 Hz. I used 7400 as the IC for the oscillator but 7404 is OK.
The signal of the oscillator has the switching operation with TR1-TR4. TR2 and TR4 are the transistor for the main switching. Because these transistors are difficult to drive directly from the IC, they make amplify in the electric current using TR1 and TR3.

The connection between TR1 and TR2, and TR3 and TR4 connects in the way of being called "the Darlington connection".
The transformer that the input is 100 V and the output is 24 V in the one with the 12 V center tap makes the input and the output opposite and uses.

Because the comparatively big electric current (about 3 A) flows through the part of the line that the circuit diagram is bold, the thick wiring materials are used.

The output voltage is the square wave ().
Depending on the equipment, there is one which can not be used.
When the load is added, the wave form of the output voltage changes by the inductance of the transformer.

I was asked about 220V output from some readers. The output voltage of the inverter is decided only in the transformer. You can use the transformer with 220V as for primary(input) and 12V as for secondary(output). At my circuit, primary and secondary should be used oppositely. Then, you will be able to get AC220V from DC12V.

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Unregulated Power Supply

A basic full wave rectified power supply is shown below. The transformer is chosen according to the desired load. For example, if the load requires 12V at 1amp current, then a 12V, 1 amp rated transformer would do. However, when designing power supplies or most electronic circuits, you should always plan for a worst case scenario. With this in mind, for a load current of 1 amp a wise choice would be a transformer with a secondary current rating of 1.5 amp or even 2 amps. Allowing for a load of 50% higher than the needed value is a good rule of thumb. The primary winding is always matched to the value of the local electricity supply.

Notes:
An approximate formula for determining the amount of ripple on an unregulated supply is:

Vrip = Iload * 0.007 / C

where I load is the DC current measured through the load in amps and C is the value of the capacitor in uF.The diagram below shows an example with a load current of 0.1 amp and a smoothing capacitor value of 1000uF.

The calculated value of ripple is (0.1 * 0.007) / 1000e-6 = 0.7 volts or 700mV. The value of peak-peak ripple measured from the graph is 628mV. Therefor, the equation is a good rule of thumb guide for choosing the correct value for a smoothing capacitor in a power supply.

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DO NOT TRY TO DO THIS MOD !!! IF YOU ARE NOT AWARE OF THE SHOCK

DO NOT TRY TO DO THIS MOD !!! IF YOU ARE NOT AWARE OF THE SHOCK RISK !!!

I got two old PC powersupplies for free, to be used for this test..
They are named DTK Computer model PTP-2008. 200 Watt Output.
Original outputs are:
+5V 20A
+12V 8A
-5V 300mA
-12V 300mA

After modification:
+13.5V 14Amp cont. 20 A for 20 sec.

The external 230 Volt AC power ON/OFF switch is removed and bypassed.
Old unused outputs are removed. Over voltage protection changed to only protect one output at 16 V,
Voltage regulating resistor net changed to only monitor a single output,

Do it like this:
Cut: white, orange, blue, and yellow wires as close as possible to the pcb.
Cut: all plugs away in the other end of the black and red wires, parallel all red and black wires..
Desolder: Fan wires, L1, L3, L4, R25, R26, R27, R29, R50, R51, R52, R61, R66, D10, D16, D17, C29, C28, ZD1
Mount a 680 Ohm 1/4 Watt at R50 location.
Mount solderpins in the holes for R26, R61 and Fan connection.

This is a fast part-drawn schematic that only covers what I wanted to know.
Mount 13.5 K Ohm at the solderpins at R26. (13.5 Volt output adjust point)

Mount 15 V Zener and 100 Ohm in series in the ZD1 holes. (Over voltage protection)
If two or more powersupplies needs to be paralled, then cut R30,
now it is possible to enter constant current mode opperation without shutting down.
This is also needed if your load (or radio) has big capasitors in parallel with the power supply line.

Orange wire connects unused cap to the new 13.5 Volt output (old +5 Output).

Transformer low voltage outputs are cutted,
and 12 volt output vindings are connected to the high current double diode.

The fan is reverse mounted so that it will blow cold air into the heatsinks and transformer.
The NTC is glued with epoxy to the heatsink with the powerdiode,
The fan controller is changed so that the fan starts to rotate at +40 C on the heatsink,
If the temperature goes further up, the fan will rotate faster.

Mount potentiometer 47 K at R61 solderpoints. (adjust to fan start at 40 C then change to normal resistor)

Output ripple is under 5mV pp at 20 Amp. (0 - 100 Mhz)
I have tested it with my HF VHF and UHF rig, and could not hear any more noise than usual.
Output power has been tested with 14 Amps cont. for one hour, no problem at all !!
Efficiency at full max load is 60 %

This mod done April 2002 by me OZ2CPU.
Dont ask for schematics or any more info. All I have is here on this page..

Idea to this mod was found here

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PICList Project The Burdette switcher:

LED 12 Volt Lead Acid Battery Meter

Telephone Ring Generator Using Switching Supply

Telephone Ring Generator Using Switching Supply

Battery Charger Ideas

Gelled Lead Acid Batterry Charger

DC/AC inverter (1)

Unregulated Power Supply

DO NOT TRY TO DO THIS MOD !!! IF YOU ARE NOT AWARE OF THE SHOCK