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[Homework] Resonance and capacitor values.

D

Daniel Pitts

Jan 1, 1970
0
So, I have a circuit:


|-------------
___ |
D1 \,/ S
===== S
| S
VR ----+ S 2 mH inductor
| S
===== S
D1 /^\ S
--- S
| |
|----------+-+
|
-----
---
-

D1 and D2 are Varactors. I'm supposed to find the value of D1 and D2
for a resonance of 1MHz.

I went with 1MHz = 1/(2Ï€*2mH*C) which results in C=50.6pF

So, I'm not entirely sure if that is the entire capacitance of the
circuit, or the capacitance of of each of D1 and D2.

This homework exercise is about varactors, but I have skipped the
"Introduction to AC" class, so my basic understanding of the above
circuit as an LC circuit is lacking. If I understood that much of it, I
know I'd understand the varactor part.

Any hints, help, suggestions would be appreciated.

Thanks,
Daniel.
 
A

amdx

Jan 1, 1970
0
I get 12.7 pF.


The varicaps are in series, so each one needs to be about 25 pF *if* VR is
applied through a big resistor. If VR is literally a voltage source, the lower
diode is shorted for AC signals and only the upper one participates in the
resonance. So it's a trick question or a poorly stated problem.

And there are two D1s. Maybe it's a dual varactor.

In real life, 2 mH is a pretty big inductor to use at 1 MHz.
I was surprised when I found 2.2mH inductors with a SRF of 1.9Mhz and
1.7Mhz, that's only 3 or 4 pf. I'm not argueing against your "2 mH is a
pretty big inductor to use at 1 MHZ" I agree. I was just surprised by
the high SRF of 2.2mH inductors.

Mikek
 
M

Michael Black

Jan 1, 1970
0
So, I have a circuit:


|-------------
___ |
D1 \,/ S
===== S
| S
VR ----+ S 2 mH inductor
| S
===== S
D1 /^\ S
--- S
| |
|----------+-+
|
-----
---
-

D1 and D2 are Varactors. I'm supposed to find the value of D1 and D2 for a
resonance of 1MHz.

I went with 1MHz = 1/(2?*2mH*C) which results in C=50.6pF

So, I'm not entirely sure if that is the entire capacitance of the circuit,
or the capacitance of of each of D1 and D2.

This homework exercise is about varactors, but I have skipped the
"Introduction to AC" class, so my basic understanding of the above circuit as
an LC circuit is lacking. If I understood that much of it, I know I'd
understand the varactor part.

Any hints, help, suggestions would be appreciated.
Take out the varactors, and replace with regular capacitors. Then figure
what's needed for resonance. Keep in mind that the two capacitors will be
in series, so study the pertinent section of the book about capacitors in
series.

Then put back the right value varactors. Keep in mind that your homework
as you present it doesn't say anything about tuning range, so you can only
pick a varactor that tunes the coil to the 1MHz, rather than figure out
how much range it should have. It also leaves out what voltage range is
available for tuning the varactors, so you can't pick varactors either
since you don't know if you have a good range of voltage available, or
have to use some of those special varactors that have a high max to min
capacitance range with a small control voltage.

Michael
 
F

Fred Abse

Jan 1, 1970
0
The varicaps are in series, so each one needs to be about 25 pF *if* VR is
applied through a big resistor. If VR is literally a voltage source, the
lower diode is shorted for AC signals and only the upper one participates
in the resonance. So it's a trick question or a poorly stated problem.

And there are two D1s. Maybe it's a dual varactor.

In real life, 2 mH is a pretty big inductor to use at 1 MHz.

Poorly stated problems, and inappropriate values are symptomatic of
teachers who have no practical experience. Nothing new there.
 
F

Fred Abse

Jan 1, 1970
0
I went with 1MHz = 1/(2Ï€*2mH*C) which results in C=50.6pF

It's f=1/(2*pi*sqrt(L*C))

That's not the easiest way to remember it:

Just remember that 2*pi*f*L=1/(2*pi*f*C) at resonance....(1)

From which you can derive everything you need for whatever circumstance.

There's a further trick. 2*pi*f crops up so many times, that we allocate a
symbol to it, actually the lower case Greek omega, which I will type as a
lower case "w", here. That means you only have to do a single pi
approximation for the whole calculation, which improves precision.

Like wL=1/wC (same as (1) above)

Hence wC=1/wL

and C=1/w^2*L

giving 12.6651pF for the combination. The varactors are in series, so the
effective capacitance, assuming both are equal, will be half that of each.
I leave it to you to do the multiplication;-)


There's no point stuffing your head with lots of formulas, which you can
remember wrongly. Just remember one, and derive the others with a bit of
mental algebra.
 
J

Jamie

Jan 1, 1970
0
Fred said:
It's f=1/(2*pi*sqrt(L*C))

That's not the easiest way to remember it:

Just remember that 2*pi*f*L=1/(2*pi*f*C) at resonance....(1)

From which you can derive everything you need for whatever circumstance.

There's a further trick. 2*pi*f crops up so many times, that we allocate a
symbol to it, actually the lower case Greek omega, which I will type as a
lower case "w", here. That means you only have to do a single pi
approximation for the whole calculation, which improves precision.
Ok, but since "w" omega is kind of not a constant, because you still
need to define it at the start, may I add the part there to even shorten it

w= Ï„*f

Most likely the tau symbol didn't come out but some where in my math
history it once represented 2 * PI.

for some reason I can't find the lower case Omega (Ω ) in my chart.

Jamie
 
D

Daniel Pitts

Jan 1, 1970
0
Oops I wrote the formula wrong in here.
I get 12.7 pF.
Running through it the second time, I get the same answer you do. Not
sure where I messed up, but such is arithmatic.
The varicaps are in series, so each one needs to be about 25 pF *if* VR is
applied through a big resistor. If VR is literally a voltage source, the lower
diode is shorted for AC signals and only the upper one participates in the
resonance. So it's a trick question or a poorly stated problem.
VR provides the reverse bias on the diodes which "sets" the capacitance,
no? Or am I not understanding?
And there are two D1s. Maybe it's a dual varactor.
That was my error. Each one should have been labeled separately. D1 and D2
In real life, 2 mH is a pretty big inductor to use at 1 MHz.

Perhaps, though that is what the problem states.

Thanks,
Daniel.
 
D

Daniel Pitts

Jan 1, 1970
0
Poorly stated problems, and inappropriate values are symptomatic of
teachers who have no practical experience. Nothing new there.
This is straight from the text book:
Electronic Devices (Conventional Current Version) (9th Edition), Thomas
Floyd.

Frankly, my teacher has perhaps too little theoretical experience. You
may have seen my previous thread about a different question. I tried to
go over the calculus approach to solve the Average Power problem, but he
said it really wasn't that important, and he alluded to the fact that he
doesn't really remember calculus very well.
 
G

George Herold

Jan 1, 1970
0
EEs use tau to mean "time constant." We represent 2*pi as 2*pi.

I'm always just taking 2*pi = 10, (for approximations)
It's the metric system for frequencies... 1 cycle = 10 radians :^)

George H.
(I then remember there's a factor of ~1.6 floating around)
 
G

George Herold

Jan 1, 1970
0
There's a mistake in the wiki article. Does anybody see it?

Only one?
I didn't read the whole thing, and I don't know much about
varactors.
But I'd rather it read 'at DC only a very small reverse bias current
flows' rather than 'no current flows'.

What didn't you like?

George H
 
G

George Herold

Jan 1, 1970
0
George is suffering from Fortran dyslexia--I think he meant pi**2 not 2*pi.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510 USA
+1 845 480 2058

hobbs at electrooptical dot nethttp://electrooptical.net- Hide quoted text -

- Show quoted text -

You must have heard about the dyslexic agnostic with insomnia...

He stayed up all night wondering if there really is a dog.
George H.
 
D

Daniel Pitts

Jan 1, 1970
0
I get 12.7 pF.
Yup, after I reworked it, that is exactly what I got.
The varicaps are in series, so each one needs to be about 25 pF *if* VR is
applied through a big resistor. If VR is literally a voltage source, the lower
diode is shorted for AC signals and only the upper one participates in the
resonance. So it's a trick question or a poorly stated problem.
"About 25 pF" was the correct answer for this problem. I'm not sure how
the lower diode is shorted, since it is reverse biased.
And there are two D1s. Maybe it's a dual varactor.
My bad, D1 is the upper, I meant to label the bottom diode D2.
 
F

Fred Abse

Jan 1, 1970
0
EEs use tau to mean "time constant." We represent 2*pi as 2*pi.

Indeed.

Usenet doesn't necessarily support 8-bit extended character sets (of
which there are a few). Better to stick to standard ASCII.

That goddamn Greek mu screws up LTspice listings over Usenet, if you
don't watch out for it, for example.
 
F

Fred Abse

Jan 1, 1970
0
Yup. 1m and 1M are both 0.001. 1meg is 1e6.

"1meg" works fine, though. It's been like that since Berkeley Spice
started, and it was all on punch cards and Teletype.

Heck, we still talk about model "cards".

Spice was always *supposed* to be non-case-sensitive.
 
B

Bill Bowden

Jan 1, 1970
0
Yup. 1m and 1M are both 0.001. 1meg is 1e6.

Yes, I discovered that. Also the LTspice default value for resistance
of an inductor is 1 milliohm and so the Q is too high. I'm playing
with a couple LC tank circuits for IR 36KHz use (6.8mH and 2700pF) and
need the rise and fall times to be only a couple cycles out of 32.
What sort of inductor Q would you recommend?

-Bill
 
G

George Herold

Jan 1, 1970
0
Yes, I discovered that. Also the LTspice default value for resistance
of an inductor is 1 milliohm and so the Q is too high. I'm playing
with a couple LC tank circuits for IR 36KHz use (6.8mH and 2700pF) and
need the rise and fall times to be only a couple cycles out of 32.
What sort of inductor Q would you recommend?

-Bill

Bill I'm not sure I understand your question. But it will take about
Q cycles for an oscillator/ filter to either build up or decay back
down.

George H.
 
B

Bill Bowden

Jan 1, 1970
0
Bill I'm not sure I understand your question.  But it will take about
Q cycles for an oscillator/ filter to either build up or decay back
down.

George H.

The problem is to receive 32 or 64 cycle bursts of 36KHz from a hand
held IR remote control and output a DC level representing the time
duration of the bursts, either 890uS or twice that of 1.778 mS. It
looks like a tradeoff between gain and response time. If the Q of the
LC circuit is low, the response time will be fast with low gain, and
visa versa. How do I approximate the Q of the inductor for a
reasonable compromise of gain and response time?

-Bill
 
B

Bill Bowden

Jan 1, 1970
0
Why not just count the number of cycles in a burst and assign the
length of the burst accordingly?

Yes, could do that. But I want to roll my own input circuit and
receive weak IR signals from 10 feet or so using a regular IR LED. So,
the idea is to amplify weak signals and provide varying DC time
lengths to a processor for decoding. I realize there are (3 terminal)
IR receiver modules to do the job, but I wanted to roll my own using
tuned circuits.

-Bill
 
J

Jamie

Jan 1, 1970
0
Bill said:
Yes, could do that. But I want to roll my own input circuit and
receive weak IR signals from 10 feet or so using a regular IR LED. So,
the idea is to amplify weak signals and provide varying DC time
lengths to a processor for decoding. I realize there are (3 terminal)
IR receiver modules to do the job, but I wanted to roll my own using
tuned circuits.

-Bill

You have back ground IR to worry about, sun light contributes to ~50%
of IR out side. Being inside you have other sources that can saturate
the detector.

If you put a detector in operation in a linear state, you can see a
lot of this noise on the scope. If you have a scope with FFT on it you
can also locate or should be able to, signals from a remote.

I suppose one could make a regenerative oscillator with the IR
detector element as part of the circuit in the gain loop.

Jamie
 
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