Hi,

I am trying to build this circuit. The problem is, LTSpice gives me 17V on the DC rail but in real life I get 22V. I tried putting lowering the voltage with diodes in series but I got instead 16V on output of the rectifier and 7V after the serie of diode (wich was supposed to give me a drop of 7V and give me 15V at the other end of those diodes). So I taught my problem was else where. I tried testing from DC ground to the emitter of the 2n2222, and a spark came off and I burned the BJT... anyone can help me?

 

You are using the default diode in LTSpice.

Change them to the diodes that you are actually using.

I don't know what it is that you are trying to simulate, but at this moment it does not look very promising.

 

Thanks for the reply. I tried with me diodes and I get the same thing. Why are you saying it doesn't look promising? I'm trying to get a regulated 12V rail on the last rail. R 3.04k is in fact a potentiometer.

 

Here are some possibilities...

• I don't like your transformer. It's way too lossy, you're losing power in it because of the 10Ω(R2) in the primary input.
• To simulate a small 5 VA transformer for 60Hz, 120VAC input, use an 11 Henry inductor on the primary, and try 0.196H for secondary for a nominal 16VAC RMS, 22V peak, on the sec..
• A turns ratio of 7.5:1 needs an inductance ratio of 7.5^2:1 (56.25:1). Therefore the secondary inductance is 11H/56.25=0.1955H.
• Set V1 to the peak line voltage i.e. to get 120VAC RMS: 120 X 1.414=170V.
• Winding resistance(series R) estimate: 300Ω primary, 10Ω secondary.
• If you need more power, just reduce the Inductors internal resistance. You can leave the inductance as is, but in the real world inductance goes down as VA goes up.

Good Luck,
Ifixit

 

Hi,
I'm not sure what you're trying to do, but I have several comments and suggestions for you. First, based on the LTSPICE circuit, I calculate that your DC Output should be just about 20V; recall, the 120V supply is the PEAK voltage and not the RMS value. The turns ratio is the square root of the ratio of the output to input inductances, into which the RMS value is: 120/sqrt(2) = 84.86 VRMS and the turns ratio is: 4.256:1, but since the load on the secondary DC output is quite low, I expect the DC to be peak-charged, so actually, we do use the 120 V Peak value * 1/4.256 = 28V. Since it didn't make sense, I ran the same file in LTSpiceIV and got the same result, then changed the 10 ohm resistor in series with the transformer to 1mOHM and then, I got the 28VDC for the Supply value. Now, if you want to limit the supply to 15VDC, why not use a resistor-zener shunt regulator. Your load is about 33mA, so you can use a 1 watt zener at 15V, and a series 260 ohm, 1 watt resistor. This will provide 50mA thru the resistor, and give you 33mA into your circuit and provide a 17mA buffer against drops in line voltage. You can use a 1/2watt zener at 15v and a 350 Ohm, 1 watt resistor if you drop the surplus current to 5mA, which is likely ok for your circuit. Your simulation problem was the series resistor in the transformer primary and if you're trying to model an actual primary resistance, you need to model it differently, or in fact, leave it out. Transformer primary resistance doesn't cause voltage drops, but secondary DCR does. Attached is my LTSPICE schematic with just a few changes. Try it, it solves your 15VDC issue. See the attached image file in png format.
Good luck.
Regards,
Kamran Kazem

 

Thanks to all of you. I'll try de zener method. I can't get why the BJT burns down. I tried another MOS I had (rated 100V 800mA) and it returned to dust also. If I have 22V and want to regulate the emitter to 12V, I need to set R7 but in no way I should be over 1A (max current for a 2N2222)? Almost no current in the MOS, so you have 4K||40k||40k so an approximate of 4K load. 12V/4K = 3mA.

 

A 2N2222 transistor is rated for 800mA current, but 500mA is a realistic limit when Ib=Ic/10.

However, that's when it's being used as a saturated switch. When used as a saturated switch, Vce is quite low (say, around 0.2v or less) so power dissipation is also low.
P=EI, so P = 0.2v*.5A = 100mW; well within the 625mW power rating of a TO-92 package.

You're not using it as a saturated switch. Vce is much higher.

 

So I should use a NPN which is (22V-12V)*0.5 = 5W in the current case (80% of the max rating) or 1.5W in the case I use the 15V Zener. Am I right? Do you know of any BJT which would fit in there?

 

You might use a 2N3055. Unless I'm mistaken, Radio Shack still carries these.

Some of them are in the steel case TO-3 package. These have a pretty low thermal resistance from the junction to the package at 1.52°C/1W.

A TIP3055 is a similar type in a TO-218 package (similar to TO-220) with a thermal resistance of 1.4°C/1W. You might be better off using one of these, unless you have a suitable TO-3 heat sink available.

With either type, you'll need to use a good-sized heat sink, or one cooled by a fan. For a TO-218 or TO-220 package, you can make a pretty decent heat sink from an older copper Pentium heat sink that has a fan; just drill a hole for a self-tapping screw for the tab. Make certain to use heat sink compound (Radio Shack stocks it in a small tube). Don't forget to power the fan with 10v-12v DC.