charging a capacitor through a transformer

[daviddoria]

so i know i need to simulate AC to make the transformer work... but if i do this with a 555 is there any particular frequency i need to achieve?

also, after i get through the transformer, it will still be pulsing right? can i connect that directly to the capacitor, or do i need to convert it back to "DC" first?

 

[Jerran]

Keep in mind when designing your circuit that the primary and secondary voltages are 90 degrees out of phase.

And yes if you input a sinewave or square wave you will get the same on the output.

 

[Optikon]

so i know i need to simulate AC to make the transformer work... but if i do this with a 555 is there any particular frequency i need to achieve?

also, after i get through the transformer, it will still be pulsing right? can i connect that directly to the capacitor, or do i need to convert it back to "DC" first?

What is your application? Be sure to run your frequency at a level that the transfromer can handle. If this is all ideal simulation, then you shouldn't have a problem. If you are actually building something, then you'll need to worry about core saturation and so on..

 

[Jerran]

Yes you will probably have to make it a DC current to use it with your 555, caps in an AC system act as a frequency controlled pot. They add a resistance to the circuit and bring the current 90 degrees out of phase(current leads the voltage).

Capacitive Reactance (Ohms)
XC = 1/(2pfC)
p = pie
f = frequency
C = Capacitance (Farads)

So the lower the frequency the greater the resistance the cap produces.

 

[daviddoria]

yes i'm actually building this.

http://bandtank.com/charge.jpg

will that work? all i want to do is put that cap C2 in there and charge it so it can be removed and discharged.

 

[Jerran]

if your wanting C2 to hold a charge I believe you are going to need a diode between it and the coil or its going to discharge threw the coil and R4 when the inducted voltage drops to 0. C2 will act as a battery, as you charge one side you are producing a voltage difference between it and the other cap plate.

 

[daviddoria]

this better?

http://bandtank.com/charge2.jpg

if that is ok... what diodes do i need to use? is the direction of the diodes correct?

 

[jem]

Hi:

I see several inconsistencies with your schematic. First of all, with the two diodes the way they are connected one of them will always be reverse biased at any time, and the capacitor will never charge. All you need is one diode in series with the capacitor (anode connected to top of transformer secondary, and cathode feeding the 1K resistor, which then feeds the +ve of the cap. The -ve of the cap connects to the bottom of the transformer winding).

Another thing is that I believe 1uH is quite low for the transformer windings. I am assuming you mean that these are the values of the primary and secondary inductances. With that kind of values, you will need frequencies well into the RF range in order to get any transformer action. Then, of course, the capacitor will be way too large, and the diodes will not rectify if they are ordinary Si rectifiers. You should have magnetizing inductances of a few mH's at a min.

Finally, can I suggest getting rid of the emitter follower stage, and driving the transformer directly from pin 3 of the 555 through a coupling capacitor (to remove all DC from the transformer primary). The 555 (do not use the CMOS variety though) is capable of sourcing and sinking current. With a 1:1 transformer, and 1K in the secondary, the 555 should be able to handle the current. That way you drive the transformer in a bipolar fashion. Otherwise, you would probably need a diode across the primary instead of letting the flux decay on its own. In any case, if you use an emitter follower with a 1K resistor as shown, there will be barely any voltage across the primary of the transformer. Also, you should aim for approx. 50% duty cycle in your 555 output (R2 should be much higher than R1 to approach 50%).

Are you winding your own transformer? or can you get the specs of the transformer you are using? That will determine the frequency of operation you should select.

Hope that helps,

Jem

 

[daviddoria]

wow, sounds like you are quite the expert on this stuff...

heres my updated one:
**broken link removed**

my 1uH markings were just the default from the program. i labeled them based on the resistances that i measured, dont know the actual specs on this transformer.

i changed R2 to 100K ohm and R1 to 1K ohm

this should give 7.164 Hz (is this an appropriate frequency? i couldnt see how you would keep a 50 percent duty cycle and have the frequency any higher than this??)

does everything else look ok?

thanks alot

 

[jem]

Looks good! If that transformer in question is similar to that of your other posting, may I suggest a frequency around 15 - 20 KHz. You might also want to use a schottky diode (faster) instead of a general purpose diode like 1N4006. Reverse blocking voltage of the diode should be around 300-400 V, or else the diode will break down. You might also want to play with the 1K resistor in the primary side. Lower it to get to 100 Ohms to start (with 10 V supply). Finally, don't forget the voltage rating and polarity of the capacitor!

Jem

 

[jem]

Forgot this! You have to add a capacitor between pin3 of the 555 and the top winding of the transformer. Try something around 0.22uF or so (a quick guess) if you are using around 20 KHz.

Jem

 

[daviddoria]

**broken link removed**

3 questions:

1)
what does the cap between pin 3 and the coil do?

2)
what does the value of the resistor in series with the large capacitor do?

3)
in order to get 15KHz, i had to use a smaller capacitor. is there anything wrong with using this small of a value in a timing circuit (1nF)?

if everything else looks good, i'm gona order that diode and get to work!

also...
once everything is in place to charge this puppy, i also need some way of discharging it (a spark). i am in the process of designing an adjustable spark gap thingy (for lack of another name), but i dont know how to dump this capacitor that we now have charged across my spark gap. i would just do it with a switch, but then i can't implement my timing (delay) circuit that i've been working on. any suggestions?

 

[Jerran]

Looks to me that C3 is being used as a AC coupling. It will block a steady DC current but allow AC/pulsed DC to flow thru it. R4 limits the current into the capacitor, without it Id imagine the transformer would draw more current then the primary could provide(current threw the primary is determined by the load on the secondary).

Another diode in parellel to the primary would also probably be advised. Without it reverse EMF could cause problems in your supply.

more on this in the following link.
https://www.electro-tech-online.com/threads/12v-sealed-lead-acid-battery-charger.8546/

 

[daviddoria]

jerran:

http://bandtank.com/charge5.jpg

is that what you mean by another diode in parallel to the primary?

do you have a recommendation for the model # of both of these diodes?

also, did anyone see the error in the calculation giving 3000v on the secondary?

also, C3 is unncecessary if i know that only pulsed DC will be coming out of pin 3 correct? is it just good practice to put there?

 

[Jerran]

Almost. D2 should face the otherway, it will ground out any excessive voltage spikes. The anode should be attached to the top of the coil ( side your current is entering), and cathode attached to the bottom. I typically dont have a resistor in series with my coils/relays so Im not sure off hand if the cathode should be between the coil and the resistor or attached directly to ground. So id suggest directly to ground unless someone else objects The diode on the secondary side needs to have a high enough voltage rating for your application, Ill do a little browsing threw some catalogs and see whats available. Ive never had a project that needed over 20 volts so Im not sure atm.

Leaving C3 in is good practice, it helps ensure a clean pulse to your coil. DC causes an offset in an AC system. Simple example is if you have a 20 volt peek to peek sine wave and intraduce a +4 volt dc voltage into the system your wave offsets by that value. It will swing from +14 volts to -6 volts.

For the 10volt supply side you could probably get by with a switching diode, Ive used 1N4001's in small h-bridges and alot of 1N4148's aswell.

 

[Jerran]

Since you dont know the amount of turns on either side of the transformer you will have to find the ratio of turns in the primary to the secondary.

A simple test would be to put a decent sized resistor across the secondary and measure the current and voltage of both coils. You need to know the number of turns or ratio of primary turns to secondary turns before you can figure the voltage on the secondary(typically the ratio is determined by dividing the higher voltage by the lower one). Example a primary of 240 with a secondary producing 96 volts under a load would be 240/96 = 2.5 : 1 (left side is primary right side is secondary). This instance would be a step down transformer, its primary will be 2.5 times larger then its secondary voltage. A step up of the same number of turns would be 1 : 2.5, meaning the secondary will be 2.5 times greater then the primaries voltage. Also note power in will always equal power out (out will always be slightly less in the real world but for figure purposes we assume 0 loses for friction and outside forces). Primary voltage x primary current = secondary voltage x secondary current.

<Edit>
Also note the direction the current flows from the secondary coil, you will have to know which lead has the + voltage to charge the cap.

 

[jem]

Re. your latest schematic. First of all, it is R2 which I suggested you change to 100 Ohms. This is just to limit the current drawn from the 555 primary when the capacitor in the secondary is fully discharged. In such a case, for a brief instant, the capacitor will appear to be a short circuit, and secondary current will be limited only by the series resistor, and the DC resistance of the secondary winding. Because it is a high turns ratio transformer, the reflected impedance (on the primary side) might be too low for the 555.

Also, I think the diode is unnecessary. The 555 drive will source as well as sink current. So the primary will be driven in a bipolar fashion. Such diodes are typically needed when the primary is driven single ended (for example when the top of the primary winding is connected to Vcc, and the bottom of the primary is connected to the collector of an output transistor -assuming an NPN output stage).

Finally, I do not think you will get much of a spark with the 200 - 300 V which you will get from this circuit. Typically, the output feeds a Xenon flashbulb which is triggered using a separate trigger pulse transformer. If drawing sparks is your goal (as opposed to charging a capacitor), maybe you should consider using a LOPT (Line Output Transformer) which you can scavenge from an old color TV set (maybe even include the voltage multiplier stage). You can also pick up the driver transistor from the set as well. You will however need a fairly high B+ voltage. Alternatively, you could consider a car ignition coil arrangement.

Jem

 

[daviddoria]

no no, the spark is plenty big. we have built some prototype smaller models (the same fuel air ratio) and it ignites fine.

you are saying the diode on the primary side is unnecessary? the diode on the secondary is a half wave rectifier correct? so really i need to divide by 2 the output voltage, yes?

 

[Jerran]

Aye Jem's talking about the diode on the primary. I wasnt aware a 555 could sink current so you should be ok without it.

A half wave rectifier chops half of a AC wave off(example at http://www.geocities.com/SiliconValley/Peaks/6319/halfwave.html ). So no you dont have to divide your voltage by 2. In your circuit its just used to keep the cap from discharging when the inducted voltage from the coil is 0 volts.

 

[daviddoria]

but i thought you had to charge the cap with DC.... doesn't it come out of the secondary AC??

dont i need some sort of "convert the voltage back to DC" part of the circuit?