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With a lot of advancement in solid state electronics and signal manipulation, isn't it easier to simply take high amplitude signals with frequencies near 1 MHz and multiply the signals using diodes and frequency filters(LC/RLC) than to use a magnetron?

Since in frequency multiplication the amplitude is halved, we can take much higher amplitudes for the low original frequencies which is easier to do than amplifying a very high frequency signals.

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    \$\begingroup\$ there are simpler ways yo make such frequencies than synthesising them \$\endgroup\$
    – Jasen Слава Україні
    Oct 19, 2019 at 7:07
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    \$\begingroup\$ high power gigahertz transistors are a fairly new thing. mwrf.com/industrial/… \$\endgroup\$
    – Jasen Слава Україні
    Oct 19, 2019 at 7:11
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    \$\begingroup\$ robustness; a magnetron is a milled-out cavity in a copper block \$\endgroup\$
    – analogsystemsrf
    Oct 19, 2019 at 17:04
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    \$\begingroup\$ You are underestimating a lot of things. Microwaves need high power. What's cheaper for high power? A hunk of metal or a bunch large semiconductor crystals that have to be grown? Have you thought about how big these expensive transistors, caps, and diodes have to be to handle 1500W? You're not talking about a little radio here. It's like saying it's easier to make really loud sounds with a speaker than it is to just banging something together really hard. \$\endgroup\$
    – DKNguyen
    Oct 19, 2019 at 19:45
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    \$\begingroup\$ This reminds me of a question "why do we bother using mechanical relays when there are SSRs". "Obsolete technology" doesn't mean there's something newer. It means there's something both cheaper and with better performance. \$\endgroup\$
    – Dmitry Grigoryev
    Oct 21, 2019 at 9:01

3 Answers 3

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Magnetrons are cheap, reliable, pretty efficient (65% or so- and they tolerate high temperatures so heat sinking is easy) and made with mature technology. They are also reasonably tolerant of VSWR issues (if the user does not put a proper load in the oven, for example). They don't really allow the frequency to change much without expensive mechanical tuning which is not available on consumer ovens- so standing waves tend to appear in the oven.

To get 1000W-ish of microwave power any other way would be more expensive and possibly more fragile. It's possible today, but too expensive. Of course the semiconductor makers are always looking for the next big market, but the oven market is going to have to wait more years I think. One of the few advantages they might have is to allow the frequency to be modulated which could reduce or eliminate the need for turntables and stirrers. However that could have implications in other areas of the oven design such as the door, which is designed to attenuate one particular frequency.

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    \$\begingroup\$ That covers it well. +1 I developed a low cost "in situ" fiber probe to measure temperatures in a microwave oven, after developing such for commercial companies (who could afford to pay more for anything that would work within an operation chamber.) I was able to get it down to about $70 in small qty parts. But even with the possibility of low cost to the manufacturer at high volume, I wasn't able to interest Whirlpool/Amana, etc. I thought it would be nice to set a temp and have it shut off when reached. They did too. But they felt the market wouldn't care enough, in the end. \$\endgroup\$
    – jonk
    Oct 19, 2019 at 21:42
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    \$\begingroup\$ However that could have implications in other areas of the oven design such as the door, which is designed to attenuate one particular frequency. The diameter $d$ of the holes is not tuned for one particular frequency. Waves of wavelength $\lambda$ are attenuated by $T=(d/\lambda)^4$. So if you want a given $T$ for safety, and $d$ is fixed, you can just calculate a minimum $\lambda$ and not go under that. See physics.stackexchange.com/questions/141562/… \$\endgroup\$
    – user46399
    Oct 20, 2019 at 15:02
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    \$\begingroup\$ @BenCrowell I’m talking about the door seals (not the door window) which are often tuned. See, for example, US 4523069. \$\endgroup\$
    – Spehro Pefhany
    Oct 20, 2019 at 15:08
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    \$\begingroup\$ Also inverter-driven magnetrons (e.g. many Panasonic models) give some of the benefits of solid-state drive, like variable power without resorting to very slow PWM. I can't find hard numbers on the efficiency compared to conventional drive circuits, but I assume inverters are a little better \$\endgroup\$
    – Chris H
    Oct 21, 2019 at 9:38
  • \$\begingroup\$ I wouldn't consider losing a third of my power "pretty efficient" \$\endgroup\$
    – Michael
    Oct 21, 2019 at 16:25
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The domestic microwave oven needs high power to cook the meal and high frequency to excite the water molecules. What is not needed is high stability because the water energy absorption spectrum is broad. (1, 2) The magnetron does this cheaply. The low price and low duty cycle of the domestic microwave means that they should last for many years despite falling magnetron output.

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For a linear circuit, In the best-case, you can transfer 50% of your input power to the wave and the other 50% energy heat up your circuits. "https://www.microwaves101.com/encyclopedias/maximum-power-transfer-theorem"

For high power amplifiers (with some tricks), the power efficiency is about 70-80% for example in class B amplifier. It varies with an impedance of the load. for this example, changing the food condition generate a new impedance for the circuit and this changes the efficiency.

So a hard mechanical body can withstand temperature when you transfer high energy.

It is a cheap technology and It has a longer history than many known circuits.

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