fitting time domain pulse to impedance profile of a power delivery network

[yefj]

Hello , i have a general model of a power delivery network from my DC connector to the IC port shown below.
then i added a load and put a 0 to 6 Volt pulse.As you can see below i get 280mV at most which means that the inductors and capacitors didnt charge enough.
Could i see what will happen in the time domain pulse from the frequency domain plot?
Thanks.

 

[BradtheRad]

Series LC is able to shape a square wave into a sine (half sine). Starting from zero the inductor passes increasing current. Smaller Henry value lets current rise faster. The effect is similar to the rising half of a sine shape.

The capacitor goes into action by reducing current. The effect is to produce the falling half of the sine shape. Smaller Farad value makes current fall faster.

 

[D.A.(Tony)Stewart]

Here's my napkin analysis

The frequency domain requires the input spectrum and transfer function for dV/dt of 6V/20ns removed from a step function where f-3dB = 0.35/t(10~90%) = 0.35/(20ns*80%) .35 = 22 MHz half power BW

Vin= R*Ic=RC*dVdt= 3 * 0.67nF * 6 V/ 20ns ~ 22MHz LPF on a 6V step function.

Transfer function you can compute which appears to be a BPF @ 150 MHz with a half-power BW from ~ 74 to 287 MHz so the input step rate is too slow to to be captured by the low Q BPF at a much higher fo. So it appears well damped.

The source half-power BW from DC to 22 MHz while the BPF is centred at 150 MHz with a Q = 150/213 = 0.7 so you get no overshoot.
The R ratios give some slight additional loss.

But if you reduced the rise time to include fo= 150MHz to 287 MHz you might get a 3V triangular pulse.

With 1ns risetime on input, you ought to get the full 6V signal, minus the Rs 0.5 loss ratio to 3 ohm load or 3/3.5

If you are trying to use a 20 ns step only and maximize your output pulse then the BPF must be Q=0.7 and fo below half of 22 MHz

That means maintaining your L/C ratio for the same impedance and damping factor, but increasing the product of LC to lower fo to lower the resonant from 150 to about 10 MHz or a 15x lower frequency or about 225 bigger for each component because you know fo takes the inverse square root of the LC product.

Now show me your results.
.

 

[FvM]

I don't understand the model. Where's the IC pin? What's modeled by the capacitor?

 

[yefj]

Hello FVM, i dont have an accurate IC model.i only have data sheetof QPA2575 shown below.
at maximum i have 6V and 3A so can i model the load of the pdn to be 6V/3A=2 Ohm?

QPA2575 - Qorvo

www.qorvo.com

Thanks.

 

[D.A.(Tony)Stewart]

Why the series cap to supply DC to a 35 GHz RF Amp ?

You need to start over and define your requirements for a PDN spectral impedance.

 

[yefj]

Hello Stewart ,you are correct.the first thing i have problem with is the "LOAD".
my pulse described in the data sheet going to a load which is the IC.
at avarage my current is 3A and voltage is 6V.
Can i build my pdn based on the assumption that the final destination is 6V/3A=2 Ohm?

QPA2575 - Qorvo

www.qorvo.com

 

[D.A.(Tony)Stewart]

Use 2 Ohms for loss and 3 for ripple. It will be well-damped. More important is the geometry and mechanics of power delivery. Gold-plated for low-loss skin effects? or ? Risetime of pulses and the carrier f determine the critical spectral BW requirements for load regulation losses and AM distortion. Low inductance dielectric feed from many micro or large dia. via's and Teflon ground plane? These are the more difficult and expensive challenges.

 

[yefj]

Hello Stewart,could you please say the meaning of this sentence.what do you mean 3 Ohm for ripple and 2Ohm for loss?

"Use 2 Ohms for loss and 3 for ripple"

I have basicly a signal going threw a transmission line and the LOAD is very imprtant because it will create reflection and signals going back to the source.so if you could please elaborate on the advice of "use 2 Ohms for loss and 3 for ripple"

--- Updated Nov 25, 2023 ---

Hello Stewart,could you please say the meaning of this sentence.what do you mean 3 Ohm for ripple and 2Ohm for loss?
"Use 2 Ohms for loss and 3 for ripple"

I have basicly a signal going threw a transmission line and the LOAD is very imprtant because it will create reflection and signals going back to the source.so if you could please elaborate on the advice of "use 2 Ohms for loss and 3 for ripple"

--- Updated Nov 25, 2023 ---

Hello Stewart,could you please say the meaning of this sentence.what do you mean 3 Ohm for ripple and 2Ohm for loss?
"Use 2 Ohms for loss and 3 for ripple"

I have basicly a signal going threw a transmission line and the LOAD is very imprtant because it will create reflection and signals going back to the source.so if you could please elaborate on the advice of "use 2 Ohms for loss and 3 for ripple"

 

[BradtheRad]

My impression was you wish to increase amplitude of a square wave (square pulse) by sending it through a series LC network. With values from your schematic it results in 93 MHz resonant frequency. The waveforms stay in the range of 3V. (Leftmost simulation)

Increasing the L:C ratio (rightmost simulation) creates greater waveform amplitudes within the network. Further revising could improve performance.

 

[yefj]

Hello , My purpose is to see how the signal will reach the load.
In ADS simulation i got an interesting phenomena .
For TL with 30 Ohm caracteristic impedance i get under damped response where as for 300Ohm load i got over shoot and underdamped responce as shown in situation 1 below.

but when I put 0 ohm resistance with 50 ohm TL charactersitic impedance at the source and same load then i get under dompted response as shown in situation 2 in the end .
I know that we have a some of infinite number of reflection back and forth.
Is there some TL theory that could predict if i would have underdamped or overdamped response on the load?
Thanks.

Situation 1:

Situation 2: