Designing a Cost-Effective Variable Frequency Generator for Microwave Cooking

[mearvk]

What I am looking for is a cheap, lightweight way to generate a range of specific frequencies.

To expand a little, a microwave uses a constant frequency but may vary its power to cook the food. I've heard it tries to match the resonant frequency of the water molecule in order to agitate it which in turn creates heat. What I'd like is a way to input the desired frequency and amplitude and if possible have as many emitters as possible (think max. potential emission) per unit of surface area. I was thinking for starters just a flat board with unidirectional emitters hooked up to a power source with some kind of frequency control piece in the middle. I'm new to electronics but am reasonably intelligent.

What I'm looking for specifically is if this kind of thing already exists and if not can it be built cheaply and if so what is the parts list looking like?

Thanks.

 

[DragonPetter]

What I am looking for is a cheap, lightweight way to generate a range of specific frequencies.

To expand a little, a microwave uses a constant frequency but may vary its power to cook the food. I've heard it tries to match the resonant frequency of the water molecule in order to agitate it which in turn creates heat. What I'd like is a way to input the desired frequency and amplitude and if possible have as many emitters as possible (think max. potential emission) per unit of surface area. I was thinking for starters just a flat board with unidirectional emitters hooked up to a power source with some kind of frequency control piece in the middle. I'm new to electronics but am reasonably intelligent.

What I'm looking for specifically is if this kind of thing already exists and if not can it be built cheaply and if so what is the parts list looking like?

Thanks.

Is your purpose to vary the power absorbed by the food in a microwave?

 

[berkeman]

What I am looking for is a cheap, lightweight way to generate a range of specific frequencies.

To expand a little, a microwave uses a constant frequency but may vary its power to cook the food. I've heard it tries to match the resonant frequency of the water molecule in order to agitate it which in turn creates heat. What I'd like is a way to input the desired frequency and amplitude and if possible have as many emitters as possible (think max. potential emission) per unit of surface area. I was thinking for starters just a flat board with unidirectional emitters hooked up to a power source with some kind of frequency control piece in the middle. I'm new to electronics but am reasonably intelligent.

What I'm looking for specifically is if this kind of thing already exists and if not can it be built cheaply and if so what is the parts list looking like?

Thanks.

Welcome to the PF.

You have no experience with electronics, but want to build your own microwave oven? We do not discuss dangerous activities here. Sorry, thread is closed.

 

[berkeman]

After a PM discussion, I'll re-open the thread. Please keep in mind that there are regulations regarding transmitting radio waves. Anything you build will have to abide by these regulations.

 

[mearvk]

Thanks Berkeman. My interest is in resonant frequencies I was just using a microwave as a common example of such a thing. I do see how one might be confused. And I do see the need for the board to keep things safe.

 

[berkeman]

Thanks Berkeman. My interest is in resonant frequencies I was just using a microwave as a common example of such a thing. I do see how one might be confused. And I do see the need for the board to keep things safe.

Can you say more about the application? There are a number of ways to generate oscillations at different frequencies (both analog and digital methods).

 

[sas3]

If you describe more about the application "frequency range, power and what do you hope to accomplish " that would be a big help.

"berkeman" I see you are a ham, do we have many hams on this forum? just curious?
KB9JNA

 

[berkeman]

If you describe more about the application that would be a big help.

"berkeman" I see you are a ham, do we have many hams on this forum? just curious?
KB9JNA

[hijack] There are a few HAMs here on the PF, especially among the regulars in the EE forum. Always good to have folks with a range of practical RF knowledge! [/hijack]

 

[sas3]

Sorry about the (HiJack) but I certainly would not want the guy next door running a 5 watt signal generator that is spitting out signals all over the place, a call to the FCC would not be fun for anyone.

 

[berkeman]

Exactly!

 

[mearvk]

My interest in EE is pretty broad. I have a strong computer science background and am looking for fun projects in which to invest my time. For now I would be quite happy understanding how we generate frequencies and then transmit (send/receive).

If you all would be more comfortable about this if we stuck to the reception side that would be OK. But essentially I would be quite happy making a homemade radio with some intelligent, programmable controls from scratch. I have maybe $200 I could dump into this.

Blah blah blah and thanks.

 

[berkeman]

My interest in EE is pretty broad. I have a strong computer science background and am looking for fun projects in which to invest my time.

A common way to make frequencies is with a frequency synthesizer IC:

http://en.wikipedia.org/wiki/Frequency_synthesizer

Although those are square wave frequencies used for clock signals, typically. You can also feed a higher speed clock into a memory circuit to read out codes that you convert to analog waveforms (like how a CD player works, or the sound card in your laptop).

You can also use analog oscillators to make sine waves, although you generally have less control over those in terms of changing their frequency by much.

If you all would be more comfortable about this if we stuck to the reception side that would be OK. But essentially I would be quite happy making a homemade radio with some intelligent, programmable controls from scratch.

The problem with making a radio transmitter is that it is easy to cause electromagnetic interference (EMI), which in many cases is a violation of federal law. With enough knowledge and equipment, it is very practical to build radio transceivers that comply with the law and do not cause EMI with other RF devices (like police and Fire radio systems...). HAM radio folks do it all the time, for example, and I work on such devices in my job as an EE.

You can start with experiments with wired communication with analog waveforms -- kind of like the radios without the antennas. Using wired connections will teach you a lot about the circuits, without the concern over causing RF EMI.

 

[mearvk]

I'd like to understand how signal is actually generated. Supposed I have a simple constant frequency that a quartz can provide. How do I sample the crystal at a given time, t, to derive the corresponding point in the output wave?

What is the actual physical process that is happening at the nuts and bolts level? Am I feeding constant current into the crystal and listening as it oscillates naturally? From here I would need to map the quartz' frequency onto a real space/time map from which presumably I could do something useful (reception, generation, etc). How does this mapping occur?

At this point I'm trying to understand the theory so that I have a more solid foundation from which to operate.

Edit:

Couple more questions. If I want to receive radio (for example) frequencies of say 105.1 MHz I would need to receive the signal at that frequency and then subtract the carrier signal then do something with the remnants (output to speakers). What parts would I need to only receive a range of frequencies? The antenna receives a range of frequencies all the time, so from this how do I isolate only the desired frequency? Is it as simple as sampling the antenna data at the desired frequency, i.e. 105.1 million times per second? Wouldn't you also get harmonic frequencies (2x, 1x, .5x, etc) using this thinking? Wouldn't you also get some non-harmonic data? Would I then need a 'CPU' that operates at least as fast as the signal I'm trying to sample?

Thanks.

 

[mearvk]

www.allelectronics.com/make-a-store/item/CRX-503/50.000-Mhz-Crystal-Oscillator/1.html

I found this among many others. My question is how does one know what the pins do and what kind of currency they may expect, which one is ground, which one is feedback, output etc.

The only thing I see is a black circle, which I would guess is the ground terminal.

 

[sas3]

Here is a link to the pin outs for an other 4 pin DIP package CRYSTAL OSCILLATOR I believe they are all the same.

http://www.vectron.com/products/xo/xo400.pdf

 

[berkeman]

www.allelectronics.com/make-a-store/item/CRX-503/50.000-Mhz-Crystal-Oscillator/1.html

I found this among many others. My question is how does one know what the pins do and what kind of currency they may expect, which one is ground, which one is feedback, output etc.

The only thing I see is a black circle, which I would guess is the ground terminal.

Here's an intro to crystal oscillators:

http://en.wikipedia.org/wiki/Crystal_oscillator

.

 

[mearvk]

I've been looking around and I have a few simple questions.

I was thinking of using a 9v battery to power the breadboard for now. The oscillator mentioned by sas3 works on either 5v or 3.3v and supply current says <80mA. I take this to mean that I should feed its input no more than 80 mA. How do I know what the minimum mA is? Anything from 1 to 80 mA is OK?

So my first step was to try and change the 9v to a 5v. Radioshack has a voltage regulator that takes up to 35dcv and moves it down to 5v. So my next question was how to know how much amperage a 9v battery can push so that could be regulated as well. 9v batteries are rated, evidently, in terms of amphours and not just amps. So how do I have to worry about regulating a 9v battery from its normal output of 100 to 200 mA back down to under 80mA or do I just trust that the oscillator won't pull more than 80?

Final question, does regulating the voltage have an effect on the current? If I lower the voltage from 9v to 5v would I also lower the current from 9x to 5x? Or would I merely be lowering the wattage from 9x to 5x?

Thanks.

 

[vk6kro]

The oscillator works on 5 volts and that is all you have to supply. As a matter of information, the data supplied says that the oscillator will draw less than 80 mA if you supply it with 5 volts.

You don't have to control the 80 mA. This is just the current the device will draw if you supply 5 volts.

You can find the ground terminal of a packaged crystal oscillator by measuring the resistance from each of the pins to the metal case. The ground terminal will be connected to the case and the +5 volts input will usually be diagonally opposite this terminal.

The other pins are used for the output and for a voltage control input. This voltage control gives you a little bit of control over the frequency so that it can be set to an exact value. Computer oscillators are usually cheap ones that are not very accurate and do not have this voltage control.

If you buy a crystal oscillator, get one with a frequency you can listen to on a radio or observe on an oscilloscope. 50 MHz is too high a frequency for common short wave receivers and it is also too high for many older oscilloscopes.

Computer cards often have these crystal oscillators in them. A very common frequency is 14.318 MHz but other lower frequencies would be interesting to collect.

Forget about microwave signals. They are very hard to generate with simple devices and test equipment.

9 volt batteries are a very expensive way of getting power. You can get holders that will accept 6 alkaline 1.5 volt batteries and this would be a much better way of providing power.
A discarded computer will often have a fully functioning power supply in it and this already has 5 volt outputs.

A linear regulator (like the 7805) will drop the voltage from, say, 9 volts to 5 volts so the load (if it is resistive) will draw current according to the formula I=^E/_R (Ohm's Law).
So, for example, a 100 ohm resistor will draw 9 volts / 100 ohms or 90 mA from a 9 volt supply, but it will draw 5 volts / 100 ohms or 50 mA from a 5 volt supply.

 

[mearvk]

Lots of good ideas in that response; thanks!

I guess my next question is about providing current at a given voltage. Say I have a 10 amp source at 5v and a 1 amp source at 5v. Would it hurt the crystal oscillator to hook it up to the 10 amp supply at 5v? I know you said you could hook up a computer PSU so I'm guessing you can do it without liability but I'd like to be sure.

How many amps at 5v should I start to be concerned about electrocution? Or does electrocution only occur above some voltage threshold?

Also, I'll probably be looking for an oscilloscope as a confirmatory tool. Don't really want to spend more than $200 to $300. From what I can see they get expensive quickly. Since radios operate from 88 to 108 how would I verify what's going on with a fast crystal or a signal generator? Can't see paying $1000+ for that ability. Suggestions very welcome.

 

[sas3]

A great tool for entry electronics is one of these>>>> https://www.amazon.com/Vintage-Sports-Cards-MX-906-Electronic/dp/B00005K2SY/ref=sr_1_9?ie=UTF8&qid=1339296968&sr=8-9&keywords=200+in+1+electronic+project+lab


that is what started me out.. it will teach you all about frequency generators power supplies and transmitters and a lot more.

 

[vk6kro]

Lots of good ideas in that response; thanks!

I guess my next question is about providing current at a given voltage. Say I have a 10 amp source at 5v and a 1 amp source at 5v. Would it hurt the crystal oscillator to hook it up to the 10 amp supply at 5v? I know you said you could hook up a computer PSU so I'm guessing you can do it without liability but I'd like to be sure.

How many amps at 5v should I start to be concerned about electrocution? Or does electrocution only occur above some voltage threshold?

Also, I'll probably be looking for an oscilloscope as a confirmatory tool. Don't really want to spend more than $200 to $300. From what I can see they get expensive quickly. Since radios operate from 88 to 108 how would I verify what's going on with a fast crystal or a signal generator? Can't see paying $1000+ for that ability. Suggestions very welcome.

A 5 volt supply which can deliver 10 amps would just run cooler at a lower current. It will only deliver the full 10 amps into a 0.5 ohm load (5 volts / 0.5 ohms = 10 amps). So, the lower current won't do it any harm at all.

There are a lot of very good digital oscilloscopes coming from China now so the price of older analog ones, second hand, is dropping.
There are some good analog oscilloscopes which would suit your expeiments perfectly.
You could be looking for a 20 MHz dual trace oscilloscope and you may find a good one for around $100.

You can test at higher frequencies by making an adaptor which rectifies some of the signal on a tuned circuit and produces a DC signal out. So, you can monitor what is happening to a tuned circuit by watching the rectified signal on an oscilloscope or even a multimeter.
This rectifying device is called an "RF probe".
Like this:

 

[mearvk]

So the 0.5 ohm resistor would make the PSU's current work that much harder to power the oscillator? That sort of makes sense.

Where my brain struggles to keep up is where are we keeping up with the amount of current necessary to power the crystal? If we have a 80 mA oscillator then we need to make the resistor a function of that. Wouldn't we need a 0.125 ohm resistor to make it draw the 10 amps?

80 mA * 0.125 ohm = 10amps.

I may have slaughtered the units but this is where my thinking takes me. Is this close to being real? Anyone? Anyone? Bueller?

Thanks sas3 for the link. I may end up getting one but I'm afraid I won't learn anything from them!

I noticed that some of the people were complaining that certain oscilloscopes only had 1 input and that this seemed to be a big deal. And vk6kro noted that I might check out a dual trace oscillo. What's the big deal with dual inputs or dual traces?

I'm nowhere near being able to read that electronics diagram. Where can I get versed on all the symbols because I really want to learn all this stuff.

Thanks!

 

[vk6kro]

No, the rating of a power supply only refers to the MAXIMUM current the supply can deliver. The actual current will depend on the load. (The "load" refers to the size of the current being drawn by the device the supply is powering. A large current is referred to as a heavy load.)

If there is no load, then there will be no current.

In fact, 80 mA seems a bit high for a crystal oscillator and many oscillators would actually use about 20 mA or ¹/₅₀th of an amp.

A dual trace oscilloscope is useful for comparing two signals, especially when one may be causing the other.
However, it is also an indication of a better quality oscilloscope. These tend to be more stable and give a more detailed, thinner trace than single trace ones. Single trace oscilloscopes may also be older than dual trace ones.

That circuit was not very complex (if you ignore the Norwegian writing) and it was just a type of rectifier intended for high frequencies.

 

[mearvk]

I was watching a Youtube video about electrical generators. The example had a small fan blade powered by wind spinning a coil inside magnet. The proof was that a light in the circuit illuminated a bit.

My question is since current is drawn (not pushed) how is it possible to burn out a light bulb with too much current? You take the crystal oscillator and it won't draw more than 80 mA but the bulb can evidently draw too much and burn out the filament. I get that a bulb is a sort of a special resistor but how does the oscillator differ logically and know not to burn itself out when presented with 5v and 10 amps?

Thanks.

 

[vk6kro]

An oscillator or any circuit can burn itself out if the voltage is too high.

A resistor is fairly predictable and you can calculate how much current will flow if a given voltage is put on it.
If this exceeds the power rating of the resistor, it may burn out by getting too hot.
Resistors also have a maximum voltage rating where they may arc internally and get damaged like that.

Lamps are like resistors except they get very hot in normal operation, so their resistance increases with temperature. However, they can burn out if the temperature gets hot enough to melt the filament and this happens if the voltage is too high.

Transistors have maximum voltages and they can be damaged if these voltages are exceeded.
They can also get too hot and the soldered joints inside them can melt, and then the transistor will not work.

The crystal oscillator will work OK on 5 volts but it may burn out if you put 6 volts on it. The current doesn't matter as long as there is enough available to supply 80 mA.
It is a bit like putting fuel in your car. Adding fuel won't make your car go faster unless you have an empty tank.

 

[mearvk]

An oscillator or any circuit can burn itself out if the voltage is too high.

A resistor is fairly predictable and you can calculate how much current will flow if a given voltage is put on it.
If this exceeds the power rating of the resistor, it may burn out by getting too hot.
Resistors also have a maximum voltage rating where they may arc internally and get damaged like that.

Lamps are like resistors except they get very hot in normal operation, so their resistance increases with temperature. However, they can burn out if the temperature gets hot enough to melt the filament and this happens if the voltage is too high.

Transistors have maximum voltages and they can be damaged if these voltages are exceeded.
They can also get too hot and the soldered joints inside them can melt, and then the transistor will not work.

I asked this question elsewhere but let me ask it within the context of this thread. If I raise the voltage but lower the amperage equally then would I burn out the bulb? 100 watt bulb expects 120v and 0.83 amps. Well if I reverse that and put 8,3v and 120 amps would I nuke the bulb or would it be business as usual?

 

[vk6kro]

You can't lower the current. The current just depends on the voltage and the resistance of the load.

If the supply can't deliver the current necessary, then it won't be able to maintain the voltage when the load is connected.

A 100 watt 120 volt lamp will have a current of 0.833 Amps ( 120 volts times 0.8333 amps = 100 watts)
and it will have a resistance of 120 volts / 0.8333 amps or 144 ohms

Now if you put 8 volts across 144 ohms, the current will be 0.0555 amps or 55.5 mA. (8 volts / 144 ohms = 55.5 mA). (So, no, you wouldn't "nuke" the lamp. It probably wouldn't even light up at all.)

Even if the supply COULD deliver 1000 amps to a different load, it can only deliver 55.5 mA to this load.

Still puzzled? There is "internal resistance" in all power sources which limit the current they can supply.
This is why you can't start a car with flashlight cells, even though the voltages of the cells may add up to 12 volts.
If you try to start a car with them, the voltage will be lost in the internal resistance of the batteries and the car will only get less than 1 volt

 

[mearvk]

You can't lower the current. The current just depends on the voltage and the resistance of the load.

If the supply can't deliver the current necessary, then it won't be able to maintain the voltage when the load is connected.

A 100 watt 120 volt lamp will have a current of 0.833 Amps ( 120 volts times 0.8333 amps = 100 watts)
and it will have a resistance of 120 volts / 0.8333 amps or 144 ohms

Now if you put 8 volts across 144 ohms, the current will be 0.0555 amps or 55.5 mA. (8 volts / 144 ohms = 55.5 mA). (So, no, you wouldn't "nuke" the lamp. It probably wouldn't even light up at all.)

Even if the supply COULD deliver 1000 amps to a different load, it can only deliver 55.5 mA to this load.

Still puzzled? There is "internal resistance" in all power sources which limit the current they can supply.
This is why you can't start a car with flashlight cells, even though the voltages of the cells may add up to 12 volts.
If you try to start a car with them, the voltage will be lost in the internal resistance of the batteries and the car will only get less than 1 volt

So theoretically if the bulb's resistance were 0.64 ohms you could have a 100 watt, 8v bulb powered by 12.5 amps?

How does one know the internal resistance of a given power source? 9v batteries, car batteries, etc?

 

[Bobbywhy]

mearvk, from your Opening Post:
“What I'd like is a way to input the desired frequency and amplitude and if possible have as many emitters as possible (think max. potential emission) per unit of surface area. I was thinking for starters just a flat board with unidirectional emitters hooked up to a power source with some kind of frequency control piece in the middle.”
Is this still what you want to do? One crystal oscillator will not give this configuration. And, although some have a frequency control, as vk6kro has pointed out, this is a “trim” control…it only gives you a slight change in output frequency. It is certainly not a variable frequency generator, as your post title says.
Will you please explain your goal/objective in more detail? Then you can expect members here to contribute useful ideas and suggestions.

Cheers,
Bobbywhy

 

[vk6kro]

So theoretically if the bulb's resistance were 0.64 ohms you could have a 100 watt, 8v bulb powered by 12.5 amps?

How does one know the internal resistance of a given power source? 9v batteries, car batteries, etc?

Yes, you could have a bulb like that. Bulbs are usually made for 6 volts, 12 volts etc, though.

You can measure the internal resistance of a supply by putting loads on it and observing the voltage drop.

So, if you put a load of 5 ohms on a 12 volt supply and the voltage dropped to 6 volts, you could say the internal resistance of the supply was 5 ohms.
The current flowing would be 1.2 amps and there would be 6 volts dropped across the internal resistance and the other 6 volts dropped across the external resistor.

That would make the calculation very easy, but you don't have to match the resistors exactly like that.

Suppose you had a 10 volt supply and the voltage drops to 8.5 volts if you put a 4 ohm resistor across it?
Current in the 4 ohm resistor = 8.5 volts / 4 ohms or 2.125 amps
Current in the internal resistance = 2.125 amps
Voltage across the internal resistor = 10 - 8.5 volts or 1.5 volts
Internal resistance = 1.5 volts / 2.125 amps = 0.705 ohms.

Try not to skip over that last calculation as it is important to know how to do calculations like that.

 

[mearvk]

mearvk, from your Opening Post:
“What I'd like is a way to input the desired frequency and amplitude and if possible have as many emitters as possible (think max. potential emission) per unit of surface area. I was thinking for starters just a flat board with unidirectional emitters hooked up to a power source with some kind of frequency control piece in the middle.”
Is this still what you want to do? One crystal oscillator will not give this configuration. And, although some have a frequency control, as vk6kro has pointed out, this is a “trim” control…it only gives you a slight change in output frequency. It is certainly not a variable frequency generator, as your post title says.
Will you please explain your goal/objective in more detail? Then you can expect members here to contribute useful ideas and suggestions.

Cheers,
Bobbywhy

This thread got off track. Yeah I understand that the oscillator's only give trim control. I'd like very much to get it back on track. How does one typically multiply the oscillator's frequency?

 

[mearvk]

Yes, you could have a bulb like that. Bulbs are usually made for 6 volts, 12 volts etc, though.

You can measure the internal resistance of a supply by putting loads on it and observing the voltage drop.

So, if you put a load of 5 ohms on a 12 volt supply and the voltage dropped to 6 volts, you could say the internal resistance of the supply was 5 ohms.
The current flowing would be 1.2 amps and there would be 6 volts dropped across the internal resistance and the other 6 volts dropped across the external resistor.

That would make the calculation very easy, but you don't have to match the resistors exactly like that.

Suppose you had a 10 volt supply and the voltage drops to 8.5 volts if you put a 4 ohm resistor across it?
Current in the 4 ohm resistor = 8.5 volts / 4 ohms or 2.125 amps
Current in the internal resistance = 2.125 amps
Voltage across the internal resistor = 10 - 8.5 volts or 1.5 volts
Internal resistance = 1.5 volts / 2.125 amps = 0.705 ohms.

Try not to skip over that last calculation as it is important to know how to do calculations like that.

I see how 5 ohms and 5 ohms internal resistance would get you to 1.2 amps current but I don't see at all how we're getting the voltage lowered. I = V / R so increasing R decrease I, yeah I see that, but I don't see why it would alter V.

 

[vk6kro]

The two resistors (internal and external) are in series, but the output terminal of the supply is at the junction of the two resistors, so you get the result of the voltage drop across the external resistance.

 

[mearvk]

The two resistors (internal and external) are in series, but the output terminal of the supply is at the junction of the two resistors, so you get the result of the voltage drop across the external resistance.

Not sure I understand. Can you explain more generally?

 

[vk6kro]

Suppose R and r in the above diagram are equal. Can you see that the voltages across them will also be equal because they both have the same current flowing in them?

Now the output of the power source is at the terminals marked A and B, so the voltage there has to be the internal voltage minus the voltage drop across the internal resistor, r.

So, if the resistors are equal, then the voltage across the resistor R will be half the internal voltage so you know what the size of the internal resistor is.

And that is what you were trying to measure.