Experimental Circuit DC - AC

Thread Starter

Standisher

Joined Jan 16, 2015
129
Please bear with me. I am relatively inexperienced in electronics and am therefore experimenting with battery operated circuits before moving on to AC powered circuits. Having read the theory, one of the things I want to master, and create with confidence, is a full wave bridge rectifier I tend to learn best by hands on and therefore wanted to create AC voltage from a DC Battery source so that I can then go on the build a bridge rectifier to convert back to DC (albeit at a lower voltage because of the diode voltage drop). I found the following circuit which, I thought would allow me to create 9V AC from a 9V DC battery:

http://www.sentex.ca/~mec1995/circ/555dcac.html

Built exactly as in the schematic , it does work, in the sense that I do indeed get an AC output BUT the multimeter AC reading is telling me that I am getting only 4.5V AC (half the DC input voltage). If I then rectify this with a Schottky diode bridge, I end up with a DC reading of approx 3.6V DC.

Am I missing something here? Clearly I still have much to learn before progressing.
 

Thread Starter

Standisher

Joined Jan 16, 2015
129
Hi,
How you made L1 (1uH) and what type of transformer you used as output transformer?
Hi,
I didn't 'make' the inductor, I bought it. Perhaps I should have also made clear that I was not intersted in putting the output into a transformer. I simply wanted to convert, as near as possible, 9V DC to 9V AC (I did not want to increase the voltage any further). I took AC positive reading directly after the inductor (and the negative from ground). In fact, you have just made me wonder if the type of inductor I am using has any bearing on the issue.
 

tcmtech

Joined Nov 4, 2013
2,867
Yes, the inductors ratings have a major effect on the workings of the circuit.

If you want to simply turn DC into AC read up on how an H-bridge circuit works. You can get AC out without using any inductor.

Also, some reading about the different types of self-oscillating push-pull converters would be of interest to you.
 

Thread Starter

Standisher

Joined Jan 16, 2015
129
Yes, the inductors ratings have a major effect on the workings of the circuit.

If you want to simply turn DC into AC read up on how an H-bridge circuit works. You can get AC out without using any inductor.

Also, some reading about the different types of self-oscillating push-pull converters would be of interest to you.
Thanks for the reply; I will certainly read up on 'H-Bridge' circuits and "self-oscillating push-pull converters". I appreciate that the inductor 'value' in this circuit is important and I did purchase the correct value http://www.ebay.co.uk/itm/111417809595?_trksid=p2060353.m2749.l2649&ssPageName=STRK:MEBIDX:IT. I was
 

OBW0549

Joined Mar 2, 2015
3,566
Your circuit (minus the transformer, as you indicated) converts 9V DC into 9V peak-to-peak AC. For a square wave, the RMS value is half the peak-to-peak value, or 4.5V in this case-- exactly what your meter is indicating. Nothing wrong happening here; all is well.

BTW: the inductor L1 serves no useful purpose in this circuit. It is completely extraneous. Remove it, and you'll see no significant change.
 

ian field

Joined Oct 27, 2012
6,539
Please bear with me. I am relatively inexperienced in electronics and am therefore experimenting with battery operated circuits before moving on to AC powered circuits. Having read the theory, one of the things I want to master, and create with confidence, is a full wave bridge rectifier I tend to learn best by hands on and therefore wanted to create AC voltage from a DC Battery source so that I can then go on the build a bridge rectifier to convert back to DC (albeit at a lower voltage because of the diode voltage drop). I found the following circuit which, I thought would allow me to create 9V AC from a 9V DC battery:

http://www.sentex.ca/~mec1995/circ/555dcac.html

Built exactly as in the schematic , it does work, in the sense that I do indeed get an AC output BUT the multimeter AC reading is telling me that I am getting only 4.5V AC (half the DC input voltage). If I then rectify this with a Schottky diode bridge, I end up with a DC reading of approx 3.6V DC.

Am I missing something here? Clearly I still have much to learn before progressing.
The simplest is a mains transformer with centre tapped secondary - the drive circuit is more or less an astable multivibrator. Its a bit tricky since what used to be the primary now produces high voltage that is possibly dangerous. The transformer from an old midi-HiFi probably has pretty hefty tapped secondary and at least 1 other low voltage winding - just tape up the high voltage one so you don't get electrocuted.

Ferrite cored transformers are much smaller for a given power rating, but they only work at higher frequency - at least 10kHz. And you'll have to use fast recovery diodes in the rectifier. But its much easier for DIY - you'd be winding a few hundred turns (at most) instead of a few thousand.

You can often find DC to AC converters in flatbed scanners that are old enough to have a fluorescent tube. Also AT era ISA card network adaptors - but those are usually potted.
 

Thread Starter

Standisher

Joined Jan 16, 2015
129
Your circuit (minus the transformer, as you indicated) converts 9V DC into 9V peak-to-peak AC. For a square wave, the RMS value is half the peak-to-peak value, or 4.5V in this case-- exactly what your meter is indicating. Nothing wrong happening here; all is well.

BTW: the inductor L1 serves no useful purpose in this circuit. It is completely extraneous. Remove it, and you'll see no significant change.
Thanks for that OBW0549, I did wonder if it was half peak-to-peak I was seeing at the AC output, but was surprised that the full wave rectified DC output was only around 3.6V and this led me to believe there was something wrong with the AC output I was applying to the full wave rectifier (I did expect there to be a voltage drop across the Schottky diodes but thought this should be not more than 2V overall so expected an output of circa 7V minimum). Could this be due, I wonder to the very low current at the rectifier? Thanks again for your helpful reply.....oh, and BTW, I did try removing the inductor and you're quite right, it adds no value in this circuit :)
 

Thread Starter

Standisher

Joined Jan 16, 2015
129
Thanks all for your replies and advice. This experimenting, has really all been about building a full wave rectifier and, to do so, I needed to create an AC voltage in a safe environment without the possibility of damaging a transformer. I'm still not happy that I have mastered, in practice, the theoretically simple bridge rectifier circuit (although it certainly looks correct) and want to be certain that I fully understand this aspect before applying to (and potentially damaging) the transformer I have.
 

OBW0549

Joined Mar 2, 2015
3,566
Thanks for that OBW0549, I did wonder if it was half peak-to-peak I was seeing at the AC output, but was surprised that the full wave rectified DC output was only around 3.6V and this led me to believe there was something wrong with the AC output I was applying to the full wave rectifier (I did expect there to be a voltage drop across the Schottky diodes but thought this should be not more than 2V overall so expected an output of circa 7V minimum). Could this be due, I wonder to the very low current at the rectifier? Thanks again for your helpful reply.....
If you are connecting the AC terminals of the bridge rectifier in place of the transformer shown in the diagram you linked to, then 3.6 volts out of the bridge rectifier is about what I'd expect. The bridge is being driven with a +/- 4.5 volt peak square wave (9V p-p), and each half-cycle of the driving waveform is rectified to 4.5 volts DC, minus two Schottky diode drops. So 3.6 VDC out of the rectifier would be quite reasonable.

oh, and BTW, I did try removing the inductor and you're quite right, it adds no value in this circuit :)
The writeup for the circuit you linked to stated that the inductor is there to filter the waveform driving the transformer, which I assume the author means to attenuate some of the high-frequency components of the square wave coming from the 555. However, 1 microhenry is way too small to have any significant filtering effect in this circuit, which is why I said it was extraneous. Had it been a hundred microhenries, or a millihenry, that might be a different story. Might.
 

MrAl

Joined Jun 17, 2014
8,248
Please bear with me. I am relatively inexperienced in electronics and am therefore experimenting with battery operated circuits before moving on to AC powered circuits. Having read the theory, one of the things I want to master, and create with confidence, is a full wave bridge rectifier I tend to learn best by hands on and therefore wanted to create AC voltage from a DC Battery source so that I can then go on the build a bridge rectifier to convert back to DC (albeit at a lower voltage because of the diode voltage drop). I found the following circuit which, I thought would allow me to create 9V AC from a 9V DC battery:

http://www.sentex.ca/~mec1995/circ/555dcac.html

Built exactly as in the schematic , it does work, in the sense that I do indeed get an AC output BUT the multimeter AC reading is telling me that I am getting only 4.5V AC (half the DC input voltage). If I then rectify this with a Schottky diode bridge, I end up with a DC reading of approx 3.6V DC.

Am I missing something here? Clearly I still have much to learn before progressing.

Hello there,

Cute little circuit.
Yes, you only get 4.5v out because when the top transistor is on the output is at about 9v and when the bottom transistor is on you get about 0v output, so that is 9v at the positive peak and 0v at the negative peak and the average is 4.5v.
Your meter may be reading the average voltage so try using a small cap in series with it on the AC scale. It wont be AC really though.

You wont get any higher than that without a transformer or a different circuit.

If you dont want to use a transformer, then you can build a transistor "H" bridge. That will give you what looks like +9v at the positive peak and -9v at the negative peak. That will be 18v peak to peak or 9v peak, and if that was a sine wave it would be about 6.4vac (RMS), but since it is a square wave you'll get a higher output if you rectify it back into DC.

So go check out what a transistor H bridge is, then try to build one. It wont be too much harder than what you have done already but will require a few more parts. Also note that then the output will not be ground referenced anymore.
Of course you can ask about that here too :)
 

Thread Starter

Standisher

Joined Jan 16, 2015
129
This conversation really has helped me a lot. Thanks to all contributors. Mind you, I'm now trying to get my head around H-Bridges but just about everything I'm finding so far via Google talks about motor direction control :confused: Never mind, I'm sure I'll get there.
 

MrAl

Joined Jun 17, 2014
8,248
This conversation really has helped me a lot. Thanks to all contributors. Mind you, I'm now trying to get my head around H-Bridges but just about everything I'm finding so far via Google talks about motor direction control :confused: Never mind, I'm sure I'll get there.
Hello again,

When you see an H bridge with a motor just imagine that the motor terminals make up your OUTPUT, not really a motor. So if you had a resistor as a load the resistor would be connected where the motor is instead of the motor. Also, the switching frequency may be higher too.

BTW in your original drawing that was what is often called a "half bridge" and it is being driven with a 555 timer. The two transistors make up the half bridge.
 
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