[theory] FM Transmitter/ Oscillator Circuit Theory Explaination

CDD_CDD · 8 days

Wikipedia Colpitts oscillator. I think you'll find your answers there.

jubjub7 · 13 days

I think noise is going to be a real problem for you here. I'm not aware of any amplifier with low enough input noise to measure to a nanovolt signal over 500kHz of bandwidth. The lowest noise LT amplifier I could find has an integrated noise close to 1uV for a bandwidth of 500kHz. What exactly are you trying to measure? If you can narrow your bandwidth of interest, filtering out the noise will become much easier. Is signal averaging a possibility for you?

There is nothing wrong with cascading multiple CE stages, but for your application I think you're going to want as much gain as possible in your first stage.

Goobyalus · 1 month

Just use a few MOSFETs to drive them.

Pseudo_Prodigal_Son · 1 month

I'm confused. Are you driving both VG1 and VM1? These two nodes should be in phase (and offset by 2Vbe), not out of phase. Seems to me that this is your problem.

jlangli · 2 months

I think you're going to want to reconsider how you're charging the cap. I'm not sure what your supply voltage is but your current setup will essentially be operating the regulator entirely in current limit mode. At several amps and a few volts this will almost certainly be a thermal problem, not to mention the inefficiency. I would consider building a charging circuit around a buck converter for these power levels.

stanigator · 2 months

Here is how I go about analyzing a circuit like this.

First, for small signal analysis the ideal current source has infinite output impedance so can be ignored (for the time being). In addition we know that a capacitor is open at low frequency and acts like a short at high frequency. If we replace the capacitor with an open, we have something that looks a lot like a source follower with a resistor from drain to the supply. The drop across this resistor will be constant here since the current source is constant however, so the circuit acts like a source follow with a gain of 1.

Now at high frequency we can replace the cap with a short and see that our circuit now looks just like a common source stage. We all know that the gain of a common source stage is gmR so we know that at some frequency our gain increases to gmR.

The next question is at what frequency does our gain begin to change? Well, this will be at the frequency that the capacitor comes in which is generally 1/(2piRC). Here, the R is just 1/gm of the transistor, thus at a frequency of gm/(2piC) our gain will begin to move from 0dB to gmR at a rate of 20dB/dec.

The last question is what is the gm of the transistor? This is where the current source comes into play. Luckily we know K, Vt and I so we can easily do the arithmetic to calculate Vgs and gm.

funpowder_plot · 4 months

Uh yeah, thats why it's funny Einstein.

funpowder_plot · 4 months

How so?

funpowder_plot · 4 months

"phonies" - people who play on their phones when they're out with friends instead of engaging in conversation.

ab57 · 4 months

In that case you would get 3.0V. In fact you might get higher depending on the load. Dropout is typically spec'd at max rated load current so at lighter loads you would still get 3.3V. You would not get 0V.

jlangli · 5 months

All resistors contribute thermal noise. The amount of noise is dependent on the size of the resistor given the relationship sqrt(4kTR), i.e. proportional to the square root of the resistor value. If your application is sensitive to noise, you may wish to use smaller resistors to reduce this contribution.

jlangli · 5 months

Also be aware that larger resistors add more noise to your system which may be a second consideration.

somehacker · 7 months

Are you sure this is what your professor was asking for? This simulation doesn't make any sense. SPICE will linearize it's models around the DC point when doing an AC sim, so sweeping amplitude will give you no useful information. Worse actually, it will give you incorrect information (for large amplitudes). I don't even think it's possible to sweep amplitude with an AC simulation.

TuxingtonIII · 8 months

This is true. Though if you compare a PhD to a masters with 2-3 years of experience the gap is pretty small.

TuxingtonIII · 8 months

It really depends on what you want to do. As a manager, I oftentimes actually prefer a masters over a PhD student as PhD studies are quite specific and are rarely in the exact area that we'd like. The old adage is that PhD's learn more and more about less and less until they know everything about nothing. :) There is often very little advantage to hiring a PhD over a masters student, and the PhD is typically more expensive.

Not to knock getting a PhD however. If you want to do research / teach, or if you have a specific area that you want to work in then getting the PhD might be the only way to accomplish your goals. And of course, my experience is limited to the EE analog design world. Other industries may have a different perspective, but my opinion is that you need to ask yourself what it is you want to accomplish with the PhD to know if it is worth it to you.

zsgar · 8 months

99% of the time doodle77 would be correct in that you would want to use as small a sense resistor as possible to reduce power losses. However, in this particular circuit you've got 24V and 1A no matter what, so making the resistor smaller just shifts the power to the FET.

Now, if your 24V supply is directly from a battery, or otherwise poorly controlled, then using a smaller resistor might still be a good idea as it will give you greater headroom for the power FET to operate and thus your current source will continue to operate at lower supply voltages.

zsgar · 8 months

I'm not sure that I would advise using a smaller resistor. You'll need to dissipate 24W regardless of the size of the sense resistor. Using a smaller resistor just moves more of the burden onto the power FET. Spreading the heat around between the FET and the resistors might be a good idea to help keep things cooler in this application. Moreover, using 0.1Ω will move the amplifier input much closer to the rail which may be a problem for some amplifiers. Lastly, with a 0.1Ω resistor you'll add some inaccuracy to the current source due to trace and lead resistances.

zsgar · 8 months

The power will probably be a bigger concern here than the current per se. You've got 24W being consumed here, 4W across the top resistor, 6W across the power FET and 14W in your load. So long as the components you are using are rated for this much current and wattage, and you are able to dissipate the heat, I should think that you'll be OK with this circuit. What exactly is the load you're driving?

quadomatic · 9 months

Wow. I've stumbled onto the most humorless person on reddit. You must be a lot of fun at parties. Let's make a deal. You explain to me how taxes work, and I'll explain to you how humor does.

quadomatic · 9 months

Because they're rich people.

HumaneFlesh · 10 months

A quarter of the veiled women will be converts to Islam born in families with non-muslim cultures, traditions or religions.

poprocksncoke · 10 months

Hmm. I'd like to actually see some data on this topic. I read a lot of anecdotal accounts here, I suspect mostly from men, indicating that women have an easier time, but my personal experience is quite the opposite. I've always found that people are hired based on experience and qualifications and I've never encountered any kind of hiring quota for race or gender. One the other hand, as a woman I've often encounter this attitude (that I only got my job because of my gender) and it infuriates me to no end.

I'm certain policies vary from company to company and between different industries. I've worked for over 10 years in the semiconductor industry for both startups and large companies and am currently a design manager myself. But perhaps I've never seen this because I've only worked for competent groups.

Bring on the down votes.

insidiousParadox · 11 months

First, understand that voltage is the measure of the DIFFERENCE in potential between two different points. When someone says such-and-such a point measures 24V, what they really mean is "such-and-such a point measures 24V greater than GROUND". This is much the same as saying that Mt. Everest is 29,000 ft. tall. You need to know where they start measuring from in order to make sense of this measurement. Are they starting from sea level? From the base of the mountain? If so, how do they define the "base"?

In much the same way, GROUND is simply used as a reference to compare other voltage in the circuit too. You can define the ground to be WHEREVER you like. For convenience, it is usually, though certainly not always, defined as the lowest potential in a circuit. Having a common reference point is important as it allows us to compare different potentials easily. Similar to how sea level is used as a common reference point to allow us to compare the height of two mountains easily.

mepper · 11 months

Yes, but how does he know there is no oxygen up there if he can't go to the sun to measure it?

jlangli

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