PLL_Seminar_2.php 12955 Bytes 15-10-2024 18:18:51
Mini PLL Seminar #2 - Practise
PLL Demoboard Single Frequency
The PLL Demoboard consists of a Crystal Oscillator with a Divider. You can select Reference Frequencies from 250 Hz to 4 kHz.
As PLL and VCO, a '4046 is used. Together with three programmable Decade Counters a handy Platform is available to study the
nature of a PLL.
Let's design a PLL !
Requirements :
Output Frequency : 77.5 kHz
Step 1 - The Divider
As the Demoboard offers Reference Frequencies of 250Hz, 500 Hz, 1kHz, 2 kHz and 4 kHz we obtain the following values for N:
Reference Frequency | Divider Value (N) | Remarks |
250 Hz | 310 | possible, integer value |
500 Hz | 155 | possible, integer value |
1 kHz | 77.5 | not possible |
2 kHz | 38.75 | not possible |
4 kHz | 19.375 | not possible |
As usual there are more possibilities. We will see later, which value to choose and why.
The 'not possible' values may need a different approach: dual modulus prescaler ...
Step 2 - The VCO (4046)
As the Tolerances are high - and depend on the Supply Voltage, the Temperature and other
Parameters, we dimension the VCO to work at 77.5 kHz +/- 20%.
Using the formulas/graphs in the datasheet or
this website, we find:
R1 = 100 kΩ
R2 = 120 kΩ
C1 = 1.5 nF
This results in : F
min = 60.8 kHz and F
max = 97.7 kHz whereas 77.5 kHz is somewhere in the middle.
The actual Frequency Range is somehow larger, if the Tuning Voltage goes below 1.1 V or above 3.9 V.
Other Solutions are possible as well !!!
Step 3 - The Loop Filter
The Loop Filter is somehow tricky. If we demand a very fast lock-time, the suppression of the Reference-Clock may be insufficient - and
if the Filter is too slow, Oscillation may occur as the Phase shift is to large.
The natural frequency (ωn) is a measure of the response time of the loop whilst the damping factor (ξ) is a measure of the overshoot and ringing.
Ideally, the natural frequency should be high ( 10% * REFCLK ) and the damping factor should be near 0.707 (critical damping).
Attempt 1 - Simple R-C-Lowpass
The Schematics |
(1) The natural Frequency : ωn |
(2) The Damping Factor : ξ |
|
|
|
Kφ : Phase Detector Gain K
VCO : VCO Gain K
N : Divider Gain, K
N = 1 / N
To proceed further, we have to decide which Phase Detector we use, as their Gain is different.
We choose PD2 and therefore get (from the Datasheet) : Kφ = V
cc / 4 * π , Volt/rad
We decide to use REFCLK = 500 Hz and N = 155.
As ω
n = 10% of REFCLK yields : ω
n = 50 Hz * 2 * π ≈ 314.159 rad/sec.
K
N = 1 / 155 ≈ 0.0064561
K
VCO = 13.185 kHz / V ≈ 82.84379 * 10
3 rad/sec/V
Rewriting Formula (1) : (and calculating it)
yields in R
1 * C
1 = 2.1547 * 10
-3
We choose C
1 = 1 µF and R
1 = 2.2 kΩ (other Solutions are also possible !)
Now, let's see what damping factor we have.
Using Formula (2) we calculate ξ = 9.5308 * 10
-3.
Ooooops ! This value is far away of the desired 0.707 !
Using this Filter may result in a very long time to get the VCO locked !
Attempt 2 - Simple R-R-C-Lowpass
The Schematics |
(3) The natural Frequency : ωn |
(4) The Damping Factor : ξ |
|
|
|
Kφ : Phase Detector Gain K
VCO : VCO Gain K
N : Divider Gain, K
N = 1 / N
Again, we choose PD2 and therefore get (from the Datasheet) : Kφ = V
cc / 4 * π , Volt/rad
We decide to use REFCLK = 500 Hz and N = 155.
As ω
n = 10% of REFCLK yields : ω
n = 50 Hz * 2 * π ≈ 314.159 rad/sec.
K
N = 1 / 155 ≈ 0.0064561 and ξ = 0.707
K
VCO = 13.185 kHz / V ≈ 82.84379 * 10
3 rad/sec/V
Rewriting Formula (3) :
We choose C
1 = 10 nF and calulate this expression.
We get : R
1 + R
2 = 215.4709 * 10
3 Ω.
Rewriting Formula (4) :
Calculating this expression yields : R
2 = -200.7 Ω.
Ooooops ! This is not possible. We decide to increase ξ to ξ = 2.43 as this is still acceptable.
Now, recalculating, we get R
2 = 15 kΩ and therefore R
1 = 200 kΩ.
LOOPFILTER: C1 = 1 µF, R1 = 200 kΩ, R2 = 15 kΩ, ξ = 2.43, ωn = 314.159 rad/sec
VCO: R1 = 100 kΩ, R2 = 120 kΩ, C1 = 1.5 nF
This is used in the Project :
Homebrew DCF-77 Signal Generator
Let's design another PLL !
Requirements :
Output Frequency : 1280 MHz, REFCLK : 10 MHz, N=128
As we use the VCO
ROS-1310C+ from
MiniCircuits™ we obtain K
VCO = 5 MHz / V
(from Datasheet). This equals K
VCO = 5 MHz / V * 2 * π = 31.41592 * 10
6 rad/sec/Volt
and ω
n = 10 MHz * 0.1 * 2 * π = 6.2831 * 10
6 rad/sec.
With N = 128 we obtain K
N = 1/128 ≈ 7.8125 * 10
-3.
Now, as the VCO delivers 1280 MHz at a Tuning Voltage of around 6 Volt, we have no chance
to reach that in using 5 V Logic. We must insert an Amplifier with V
u = 2 (approx.).
Therefore we get another K
Amp = 2.
to be continued ...
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