Designing a Cost-Effective Variable Frequency Generator for Microwave Cooking

[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.

 

[mearvk]

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.

I don't see how a resistor 'drops voltage'. A resistor drops current by increasing the resistance; the voltage remains constant.

From what I understand the voltage, minus any voltage regulators, is constant in this circuit.

 

[vk6kro]

You might have the wrong idea here.

Suppose you had 10 resistors in series across a 10 volt battery. ("In series" means they are connected end to end, so the same current goes through all of them.)

Let all the resistors be 470 ohms.

So, the total resistance is 4700 ohms. (when resistors are in series, their resistances add up like that).

The current will be ^E / _R so the current will be 10 / 4700 or 0.0021 amps.

Now, this flows in all the resistors so the voltage across each resistor is I * R or 0.0021 * 470 or 1 volt.

So, you can see that the resistors share the voltage according to their resistance.

Since you are obviously interested in Electrical things, could you get into a class and have someone teach you this stuff properly? Some of the concepts look simple, but they are full of traps and you could easily get the wrong ideas if you try to teach it to yourself.

 

[mearvk]

I don't understand the 'voltage across each resistor' as a concept.

Are you saying that if I test for voltage after the 10th resistor I will get 1v or 10v?

If I test for voltage after the first resistor do I get 9v or 10v?

Looking at I = V / R we'd expect that adding resistors to a 10v circuit would lower the current. It is not intuitive why the voltage would be shared or divided or anything else. The equation simply states that, assuming V is constant, that increasing R decreases I. Says nothing about V changing across changes in resistance and in fact says otherwise.

Intuitively:

After R1 you'd have I = V / R or 10v / 470 ohm or 0.02 amps.

After R2 you'd have I = V / R or 10v / 940 ohm or 0.01 amps.

But what you seem to be saying is that the equations should look like this:

After R1 you have 9v / 470 ohm worth of current.

After R2 you'd have 8v / 940 ohm worth of current.

This seems massively flawed since we'd have to constantly add voltage back into the circuits so that every time we encountered a resistor we didn't change the voltage from 3.3v or 5v to something unexpected.

 

[mearvk]

I think I see what you're saying now.

The more resistance, relative, to the sum of total resistance, the greater the voltage across that particular resistor. But what does this mean?

I struggle with the meaning of 'splitting' a voltage across several resistors. Can you explain that a bit?

 

[vk6kro]

If you measured the voltage with a voltmeter, there would be 1 volt across each resistor.

The voltages across all the resistors adds up to the supply voltage.

They don't have to all be the same resistance, so the voltage will divide itself in proportion to the resistors present.

If you measured the voltage across, say, 3 of the resistors then there would be 3 volts showing on the voltmeter. If you measured across all of them the voltage would be 10 volts.

The current in all the resistors is the same if they are in series.

 

[mearvk]

So I could take a 9v battery and via a small circuit, connect equal 9 resistors and splice into the portion between R5 and R6 to create a 5v supply?

 

[vk6kro]

Yes. You could do that.

You would take two wires out. As long as there were 5 resistors in between the two wires, there would be 5 volts out between the two wires.

It is usual to have one of the wires connected to the negative side of the supply..

 

[mearvk]

Cool.