Power transformer questions

Thread Starter

ronph

Joined Feb 4, 2013
25
Newbie here so please bear with me as I have only some knowledge on electronics and am replicating an INVERTER CIRCUIT.

Using a low voltage power supply to produce 115~120 volts just to light an LED bulb or CFL ungutted lamp here. I have found a suitable circuit. Having said that I am using a 24v/110 center tap (12-0-12) transformer and in my experience using 8AA (12v) batteries as source and with the circuit I am only getting around 80v output. I know there are some losses whenever transistors and resistors are involved hence I may not be getting the ideal 110~115 volts required although I am able to light up a 18watt (50W equivalent and getting about 75% light output) CFL.

Given that, scenario and barring any heat problems with the transistors (IRF540), IF I INSTEAD GET A TRANSFORMER 9-0-9v CT (18V) and powering it up with 12 volts, would the resulting HV output be more than 80v as earlier mentioned? Crude calculations (newbie) inputting 12volts into a 9-0-9 transformer would realistically increase the HV voltage to about 33% [12v/9v=1.33].

Any help on this would be welcome.

Audioguru

Joined Dec 20, 2007
11,248
When a transformer has a load then its voltage drops. So maybe your transformer is made to produce 15V-0-15V with no load then it drops to 12V-0-12V with its rated load. Then when it is connected backwards in an inverter the output voltage is too low.

Yes, if you use a 9V-0-9V transformer with the circuit powered from 12V the output voltage will be higher if it is not overloaded.

Your inverter circuit is simple so it produces a square-wave output that a common multimeter cannot measure accurately.

An AA alkaline cell produces 1.5V only when it is brand new and is not overloaded.
You have 8 of them and if your load is 18W then the power from each cell is 18/8= 2.25W which is very high. The current is 2.25W/1.5V= 1.5A which is extremely high. The battery cells will drop to 1.2V each in about 15 minutes and will be dead in half an hour.

Here is a voltage graph with a 1A load:

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Thread Starter

ronph

Joined Feb 4, 2013
25
Thanks Audioguru. Very informative indeed. Well, since I have not done burn tests on how long the light would last I can not comment on the circuit until I get the consumption in maH. With a 12-0-12 withou load, my output was only 81V so considering that a transformer would hardly twich at a few volts, I went and used 9-0-9. 12v divide by 9v I get 1.33 and its the. 33 that made a huge difference (note source voltage is still 12v). So 81v * 1.33 = 107v. Actual test came out at 107.2v which is not bad! I was getting less consumption using LED bulb than a CFL bulb. I do like the fact that this circuit 4047 in combination with IRF540 I could light both types. Based on my calculations its only rated 23watts output with 2amp transformer. The biggest cfl wattage I have available is 18 watts so I have yet to push the circuit to its limits and expect some heat from the transistors. The schematics did specify 5amp but its either too much to spend and am reasonably happy at 23 watt limitation. Still have to conduct burn tests if this circuit is practical. If its runs for at least 1 hour at USABLE brightness on 8x AA then am good. Of course should I decide to extend run time I will have to rely an 7AH 12v battery.

Ron

Audioguru

Joined Dec 20, 2007
11,248
The output is a square-wave. What are you using to measure the output voltage?

A multimeter usually detects the much higher peak voltage of the sine-wave from the mains then shows 70.7% of it (RMS) so it reads the voltage of a square-wave much too low.

When the square-wave has a peak voltage the same as the RMS voltage from the mains electricity then a heater or an incandescent light bulb (also a heater) are fed the same amount of power. But electronic items (a compact fluorescent light bulb, maybe an LED bulb and most electronic products including motor speed controls) operate poorly at the much lower peak voltage of the square-wave.

Thread Starter

ronph

Joined Feb 4, 2013
25
Hey Audioguru, well, the circuit is pretty much simple (see below) with a 9-0-9/110V transformer (not 220). As mentioned I am testing it on 8AA batteries and your right, 30 minutes is the max I got out of the batteries running 18W CFL. I have to do a test on a 7.5W LED bulb (power consumption is about 350-400 milliamps. What I see on CFL's is that on the circuit, it initially draws high amperage to "start up" then settles down but I have not yet measured the amp draw.

My goal at this time is to find a circuit using
a. 12 volts
b. small battery either a 4.5AH SLA @12V or AA x 8 = 12v
c. lighting CFL up to 20 watts
d. run time >1 hour

With LED bulbs the battery lasts longer so am focused now on longer run time using CFL.

Perhaps someone out there can suggest a circuit simple to construct yet lights up both LED & CFL's with longer run time.

Ron

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Audioguru

Joined Dec 20, 2007
11,248
Your CD4047 is missing a supply bypass capacitor from its pin 14 to its pin 7.

The circuit is fairly efficient. Then a longer run time must use a larger battery or a dimmer light.

Thread Starter

ronph

Joined Feb 4, 2013
25
Thanks Audioguru, yup a larger battery as you said. As mentioned, I ran it on 7aH battery and transistors did not heat up. Transformer warmed a bit to 75 degrees which is unnoticeable running a 7.5watt cfl. Question now is improving the circuit. You mentioned a bypass cap between pins 7 & 14? As a newbie am not sure what would be the benefits and what type of capacitor is required. I did installa a 2200uF 25V cap between the + & - of the line going into the circuit if that is what you meant.

My next question is, will this circuit work using a car battery. Considering this to be sort of an emergency/camping light. I know that car batteries have in excess of 12 volts and would limit the voltage to just that (scenario here is that the engine is not running) and the circuit will be connected via the cigarette lighter socket of the vehicle. Reason for limiting the 12volts is also to limit the HV output between 110V~115V. That given, would like to integrate into the circuit an LDO regulator such as the LM2940-12 (given that it does have a voltage drop of 0.5v) knowing that a stand alone car battery's current would be between 25~90AMPS. Will the regulator get fried (even with a heat sink?).

If anyone has a simplistic add-on circuit I would really be interested.

Sorry, need to pick your brain or someone else's brain but am on a learning curve at the moment on these circuits.

The actual circuit I have made is on http://www.youtube.com/watch?v=cB1UmypEdC8

Ron

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Audioguru

Joined Dec 20, 2007
11,248
Are you going to use this circuit to power a 110VAC cfl? Why not use a 12VDC cfl instead? They are made for campers and recreational vehicles (motor homes).

The vehicle's battery has many voltage spikes caused by an inductive load switched off. Therefore the CD4047 needs a filter capacitor and a zener diode.
Also the capacitor keeps the voltage constant for the CD4047 because each time its output switches it draws much more current than when it sits with its output high or low.
Without the capacitor the voltage drops each time the output switches.

There is no way a little 1A voltage regulator can power this 4A circuit.
Does a light bulb need voltage regulation? In some countries their AC voltage varies all over the place and their light bulbs are fine.

I can buy a very nice 75W inverter for \$9.99 on sale frequently. It is guaranteed for a year or two and if it fails then the store will give me another one.

Thread Starter

ronph

Joined Feb 4, 2013
25
Hey Audioguru, my circuit did include a bypass cap intended for those battery spikes. Yes, to power a 110AC CFL.

As far as my question on using a 2940-12 regulator is if and when say my battery pack's power goes below the required voltage to power the circuit, I could utilize the 12 volt socket to power up the circuit. More research on the 2940-12, some opined and state in the transistor's data sheet that it was intended for use in the automotive industry hence I was leaning toward incorporating a 2940-12 into the circuit.

Yes, it would be easier to just buy an off the shelf 75W inverter. Having said that though, I have also read that such inverters in the market do consume over 1 amp even when no load is connected while the circuit which I refer to based on limited instruments show only a consumption of between 375~500 milliwatt "loaded" which means less car battery drain and longer run time on the bulbs whether CFL or LED's.

I have yet to get the 12volt LED's to test out and see how much power consumption on each bulb. Again, googling through a plethora of information on bulbs reveal that the internal circuits vary to a degree so some may/may not work on DIY type circuits.

I guess the only way to find out for me is to incorporate a 2940-12 and see if it can handle the car battery power or simply fry the darn thing.

I guess the fun in replicating circuits is 1. does it actually work, and 2. modifying/improving the circuit to suit my requirements.

Nothing ventured, nothing gained - perhaps some burned circuits in the process

Audioguru

Joined Dec 20, 2007
11,248
Hey Audioguru, my circuit did include a bypass cap intended for those battery spikes.
Your circuit had a resistor feeding spikes to the CD4047. The capacitor you had across the battery was huge so it had a high inductance which makes it useless as a spike filter. If you add a 0.1uF ceramic capacitor across the supply pins of the CD4047 then the 100 ohm resistor and the ceramic capacitor make an excellent spike filter.

As far as my question on using a 2940-12 regulator is if and when say my battery pack's power goes below the required voltage to power the circuit, I could utilize the 12 volt socket to power up the circuit. More research on the 2940-12, some opined and state in the transistor's data sheet that it was intended for use in the automotive industry hence I was leaning toward incorporating a 2940-12 into the circuit.
The LM2940-12 has a maximum output current of 1A. Its maximum dropout voltage at 1A is 1V.
So if the current exceeds 1A like your circuit or if the input voltage drops below 13V then it will stop regulating and it will reduce its output voltage.

I guess the only way to find out for me is to incorporate a 2940-12 and see if it can handle the car battery power or simply fry the darn thing.
The battery power has nothing to do with connecting a 12V regulator.
The battery can supply 400A to start a cold engine. But the regulator takes 1A maximum.
Its 1A current rating is too small to regulate the 4A needed by your circuit.
The regulator is smart. It will not burn up. If the current is too high, if the input voltage is too low or if it gets hot then it simply stops regulating and it reduces the output voltage.

Thread Starter

ronph

Joined Feb 4, 2013
25
Hey Audioguru,

A lot makes sense to me as I have been reading up on the bypass cap.

As far as the 2940-12 I read up on that and....yeah 1amp is too small. Probably a LM338 in a TO220 package would work but could not find it in the local shops here and outrageous prices on ebay and elsewhere.

I asked an electronics guy and he says that the circuit will only take whatever current it needs whether its a small battery or a car battery. He cautions though about the battery spikes and same as your suggestion using a bypass cap. So I will put a 0.1uf cap on on the + & - sides just before the circuit. Where do I put the 100 ohm resistor...should it be right after the capacitor on the postive rail? Wondering if that would affect the rest of the circuit?

Ron

Audioguru

Joined Dec 20, 2007
11,248
So I will put a 0.1uf cap on on the + & - sides just before the circuit. Where do I put the 100 ohm resistor...should it be right after the capacitor on the postive rail? Wondering if that would affect the rest of the circuit?
Sorry, MY circuit uses 100 ohms but your circuit uses 330 ohms.
I forgot to post my circuit which is attached here.
I show the 0.1uF bypass capacitor connectred between pin 14 and pin 7 of the CD4047 which are its power supply pins. I also added a 16V zener diode between the pins to ensure that a voltage spike cannot get to the IC.
These parts do not affect the rest of the circuit.

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Thread Starter

ronph

Joined Feb 4, 2013
25
Hey Audioguru, got it! Yeah, I have seen that somewhere in another circuit out in the internet. I will incorporate that one instead. So 16v 1watt would be fine?

Thanks. Only thing now is to box it up and am good on this circuit.

Such a great FORUM!

Ron

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Audioguru

Joined Dec 20, 2007
11,248
Your circuit uses a 330 ohm resistor that feeds the CD4047 and feeds the 16V zener diode when the voltage is higher than 16V.
If the supply voltage goes to 20V continuously (very unlikely) then the power in the 330 ohm resistor and in the zener diode is:

20V - 16V= 4V. The current is 4V/330 ohms= 12mA.
Then the power in the resistor is 4V x 12mA= 48mW so it can be a 1/10W or 1/4W resistor.
The power in the zener diode is 16V x 12mA= 192mW so a 1/4W zener diode is fine.

If a voltage spike is 60V the resistor and zener diode will be fine since the spike is for a very short duration.

Thread Starter

ronph

Joined Feb 4, 2013
25
Very interesting indeed Audioguru. OK so one last modification. As you know the voltage into the transformer determines the HV output. Using a 9-0-9 ideally makes up for the loss. My battery pack 8AA voltage is not 12 volts since its rechargeable on the assumption that each only puts out 1.25 volts respectively total would be 10 volts. I tested out a 12-0-12 transformer and I get only around 81 volts (yes, it does light up the CFL but the lumens output is around 50%. With a 9-0-9 transformer @ 10 volts I get 107 volts (close enough) and light output is much better. However, if I use a small 4.5Ah battery which puts in a respectable 13.5 volts, the voltage would be around 124+/-. Knowing that most bulbs are only rated at 110 and even with a safety margin +/- 10% = 122 volts. Would you think the circuit in the bulb will not last long?

I was therefore thinking of using either a potentiometer or a wirewound varistor to: 1. adjust the voltage to 110~115 volts, 2. act also as a dimmer, and 3. using it to find the "sweet spot" where the transformer's hum (or ring) would not be noticeable. It will be in series with the + of the source. In your opinion which of the two - carbon type pot or wirewound varistor would be more uhm how do you put it.....suited for the purpose.

Ron

richard.cs

Joined Mar 3, 2012
162
Take a look at this post http://www.fieldlines.com/index.php?topic=129295.0 on fieldlines.com - it uses a SG3525 chip and two mosfets to drive a transformer from an old PC psu and gives a regulated 240 V dc output suitable for driving CFLs. The output voltage would be independant of the input voltage over a fairly wide range.

Changing a few components would allow a 120 V output if that's needed for your application. I'm not sure if 120 Volt CFLs have an internal doubler or not? I assume they don't as the tubes are very short they won't need the extra voltage.