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I think it acts as a switch turning on to increase the voltage on the 10000u cap.... however I am probably wrong... I was wondering if instead of a darlington you could use a mosfet there...
Well it doesnt make a whole lot of sense to use two switching regulators. Why two switchers instead of just one?
Also, note how large that inductor is, 1mH.
I wanted to use a switching system to obtain the drop from 240 V to ~30 V. I did want to have adjustable voltage and current limiting. I wanted to create a hybrid power supply using an switching pre reg and then use a linear reg to create a low noise output... I guess thats my goal... 10Amps and of course learn on the job...
I built the transformer a couple of days ago (primary is 10.4mH each secondary is 272uH) so it would be good if I could use that... I wanted to learn about the TL494 too. I would really like to use the TL494 to do the tracking too but it seems difficult to make it track the output of the Linear reg...
I am of course open to suggestions here as this is a discovery process for me... A buck inverter was mentioned but I am not sure if I can use that to track the output of the linear reg... I am not sure how to configure such a system too..
Your gate drive transformer L6,7,8: Usually a transformer "rings" some and you may need to add a gate to source resistor on each MOSFET. You can simulate this by adding some leakage inductance to the transformer model. (make it more real world) It is common to have a 100 or 220 ohm resistor there. OR Add a RC there where the RC time is related to the ring frequency and is quite fast compared to the switching frequency. When you make one watching the ringing on the Gate. For more information look up gate drive transformers. coilcraft.com
You have a oscillation. It might be in the time constant in the 494 error amp. You might need to have at least 0.5A load to stop the oscillation.
With a little work you will be able to remove Q1. Keep U2 and R8 and feed back that point to the TL494. You will need to change the ground and power for U2/R8 to the ground and power that the TL494 lives on.
At the end the ground for the TL494 is connected to the power line. It is not the same ground as the output side ground! It is OK for simulation.
The idea to eliminate the Q1 circuitry would be to feed the output of U2 to the TL494. That gets rid of that big 1mH inductor too, and if your TL494 switches fast enough you can use a much smaller inductor to aid filtering of the dual diode rectifier.
The DTC pin of the chip is the dead time control pin, and it is shown as open (no connection) on his schematic, so i thought i would inquire. It should go to ground.
Yes I have fixed that and put in a soft start option. I am working back on a way of making the TL494 track the LT1083 and am running sims at the moment to see if this works.. will post the outcome although it is taking a while to sim.
If you post the .asc file and all the models used i'll give it a shot myself too.
Usually it's not too hard to get this to work.
BTW, the circuit i posted a while back was meant for you to study a bit and get familiar with because that is your basic feedback circuit so if you understand that you'll understand this more complex circuit too.
Ok so I have bundled the files into a zip folder.. I had trouble finding a model of the TL494 and did eventually find one on the LT user group site that worked, however its been modeled using a hierarchical page structure. All you need is to have the sub pages in the same folder and when you add the symbol it should/will work.
There are two asc files. The first is the last one I posted which uses a pre-regulator after the TL494 and before the LT1083 as per the datasheet. The other is one I hacked up a couple of days ago and revisited. (when you indicated that two pre-regulators was an issue - which I agree) In this case the error amp uses two inputs - the inverting input measures a scaled down version of the output and is the reference. It Changes via the voltage control of the LT1083. The non inverting input has a scaled down value of the input voltage to the LT1083. In this way if the output changes - that is, increased - then the difference between the two is decreased and the TL494 will adjust to increase the voltage... if the voltage on the LT1083 is decreased then the input will initially be a lot higher than the output as the filter cap needs to drain but eventually matches up... that's the basic idea of this.
In this circuit I am having trouble with the voltage control of the LT1083 (it could be spice or just the way the circuit works) however I am unable to set a resistance (on the LT1083) where it reaches ~30 volts - in this case with the two LT1083s being hooked up to the TL494 circuit. However it did work perfectly with the single LT1083 circuit which is still on the sheet but not hooked up to anything.
All this aside, if there is a better way of hooking up the TL494/half bridge/step-down transformer to match the TL1083 +2~3 volts I am always interested.. in this I am learning so willing to try anything
I think the half bridge is a good idea. You get a 1/2 voltage drop by using this method. This way you are not switching the power line voltage but 1/2 line voltage appears across the transformer.
It is hard to see your schematic but it looks like your input voltage is 680 volts. Where did you get this from? 440vac * 1.414=
Hi Ron, I could of course be wrong but I live in Australia and the mains here is 240v RMS measured 320 on a multimedia so peak to peak 640... I have this correct or not? something basic but so basic its not mentioned much.
Try it. Make a AC voltage source with 240volts 60hz. Check to see if it is 240 RMS, 320pk and 640p-p.
Add 4 diodes and a cap and load resistor. Look at he DC voltage.
OR
The 320 volts is pk through the diodes, to the cap, charges to 320 volts. The -320 volts pk through the diodes, it gets inverted, and charges the caps to 320 volts. The caps get charged to 320 twice in 60hz. (not 680V)
Crap... yep I have this wrong.. I did actually use a bridge rectifier and it did come out as 320v DC however I got confused with the way the half bridge worked.. I have just rerun my spice SIM with the value of 320V not 640V.. and it works as I had hoped...
Yes that voltage is was way too high.
I just noticed that in the schematic L3 is flipped. The dot for L3 and L4 should be pointing in the same direction. I flipped L3 before i ran the simulation. With L3 the way it is now you dont get push pull operation you get push push, which isnt right.
Also, how did you decide on the inductance values for L3 and L4 and L2, and the driver transformer inductances too?
What target switching frequency were you after?
One of the first things you should do when you run a sim like this is to check that all your signals are appropriate. Check both outputs of the TL494, then both outputs of the driver transistors, make sure you have a clean rectangular wave of the right height, make sure you have a rectangular wave out of the driver transformer, etc. You might need a small resistive load on the output of the power transformer secondary to eat up and large spikes just to be able to monitor the output there a little better.
Well I have to say I am a bit surprised it actually worked when I got some pretty major things wrong... Making those changes it now works really well.
I also added in a ripple reducing capacitor at the base of the LT1083 and in parallel with the current sense transistor as per the datasheet's specifications although it says a 22μF capacitor I only have access to 10μF capacitors.. it still works pretty well.
I also bumped up the Pre_reg output filter cap to 10,000μF and the output filter capacitor on the LT1083s to 100μF - more in line with a real world design although with these very small you can see the response a lot better.
One thing I might ask is what value should the resistive load be? I worked out that at 2 watts a resistor value should be around 450Ω however I placed a 1k resistor as the load... Is this the kind of value you were thinking of?
Here is a picture of the transformer. I measured the inductances of the primary and each of the secondaries. Its those values I am using in the sim. I designed this using a spreadsheet I built which is also attached.
1k might do it. When you run this up you could also look for high voltage spikes across the MOSFETs as you apply more and more input.
The 15v power supply has to be able to handle the current that all the devices take. You have to look at the data sheets and find the current your devices will draw, then add that all up. If the driver only has to put out a peak current but lower after that, you can reduce the current requirement for the power supply as long as there is enough capacitance across the driver chip to provide the required peak current without drooping the supply voltage too much. You can look for low going spikes on the 15v power supply near the driver chip to see if there is enough bypass capacitance.
I had some trouble identifying what the current draw of the components I am using... Looking on the datasheets for the following components - LM324, IR2183 and TL494 it was not obvious which value should be taken..
For the LM324 there are several values listed for current. Should I use the supply current which says only 1.5mA or take the value from the output current which on the datasheet says 40mA source. Looking at maximum power disipation it says 1130mW for the DIP package so if I am using this with say 15 volts that 75mA.
For the IR2183 this appears to deiliver a large but very narrow spike to the gate of the mosfet.. however it says "Output high short circuit current" - Typical - 2.3 amps.
For the TL494 it says the recommended current for each transistor is 200mA.. so should this be say 400mA as there are two outputs...
If I assume that I only require the 2.3 amps to charge the Mosfet capacitor which would be 150nC / 2.3A = 65.2ns where the 150nC is the required gate capacitance of the IRFP450 - that is ~170th of a Hz cycle so an average of ~13mA over the course of a full cycle then...
I need a pretty small power supply for this... 75mA + 200mA + 13mA so around 300mA minimum and thats at 15 volts I am assuming so a transformer that transforms this down to 15 Volts would be Vs/Vp=Ip/Is so 15/240 * 0.3A =18mA so a transformer of say P=V* I => 240 *18mA = 4.5VA....
When I run an average on the current for the component IR2183 - Ix(U2:1) in Spice it comes to around 5.7mA.
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