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# Input matching network design

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#### adnan012

##### Advanced Member level 1
Hi,
ADS simulation for PD20015 gives me input impedance of (1.33+j*0.193)ohm at 1850MHz.
ADS smith chart utility generates matching network with a series 680pH inductor value under auto matching. how can i realize this small inductor value? is it possible to design the inductor using transmission lines on standard FR4 sheet?

the matching network is

50ohm(source)----shunt capacitor----series inductor----ldmos input

Are you sure that the total impedance is less then 1.5 Ohm?

I'm no expert in matching by any stretch of the imagination, but if you find an unrealizable or difficult to realize component value, I was under the impression that you could attempt to use a different matching network type. Does ADS allow you to choose other match network types? Shunt capacitor->series inductor I think is an L-match.

Also, what do you have available to do the match? .68 nH=680 pH inductors are readily available to order and if you don't have a Q specification then they'll probably do the trick.

---------- Post added at 11:06 ---------- Previous post was at 11:03 ----------

I would think at 1850 Mhz a small (possibly normalized?) input impedance would not be surprising.

It doesn't seem likely that a slightly inductive antenna can gain anything with another serial inductor at 0.68 nH ~7 Ohm, normalized values or not.

Isn't the inductor in question part of the matching network not a characteristic of the antenna?

(1.33+j*0.193)ohm => inductive
680pH => inductive

Right. But it sounds like adnan012 is trying to do the following :

50 ohms -> matching network -> Z=(1.33+j*0.193)

Where the matching network in question is a shunt capacitor and an inductor in series which would be an L match. Assuming the L match values were computed correctly (to match 50 ohms to Zout), an inductor value of 680 pH doesn't seem unreasonable.

According to the L match equations (if we consider exact values and not feasible values), an inductor of 675 pH and 10.4 pF results in a match between the input and output impedances. The 680 pH was probably an output since it is a standard inductor value.

adnan012

### adnan012

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Yes it is a possible that Z really is 1.33+j0.193 Ohm, and if so is it then possible to match against 50 Ohm with your values.
I would however not even think about to implement a 680 pH inductor as a component. Maybe thinkable as 2 mm length of pcb trace but it will still be sensitive to capacitive unwanted coupling.

adnan012

### adnan012

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What about a thin film inductor from Digikey? The listed 680 pH inductor has a Q of 50 at 1.9 GHz (which is basically the frequency of interest).

adnan012

### adnan012

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Assume that we implement this component on a FR4 PCB.
For this do we need to add traces and soldering pads on both sides of the inductor with a minimum length of 2-3 mm.
It is likely that these traces adds more reactance then what the component actually do.

adnan012

### adnan012

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Ah! Yes, I can see where the trouble lies. I guess I assumed one didn't have to use a PCB for the circuit, just if one could. My mistake.

---------- Post added at 16:00 ---------- Previous post was at 15:53 ----------

So, I don't know much about inductance caused by PCB lines, but this ADI document: Analog Devices : Analog Dialogue : PCB Layout : seems to suggest that PCB lines on the order of 2-3 mm would still be well under the hundreds of pH range. Unless I'm assuming too much...

adnan012

### adnan012

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A PCB trace can be low loss if it is impedance matched in both ends. That is not the case here but much of the problems are reduced if the trace is as short as possible. Stray capacitance due to component body is another problem.
Test this one.
With trace values 2*2.5 mm length, 1 mm wide, 0.5 mm height did I got 2.5 nH.

adnan012

### adnan012

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680pH is possible for around 1.8GHz.

You need to care about trace inductance. parasitic inductor may be around 0.1nH to 1nH. Try to make the trace at short as possible to reduce the trace inductance.

parasitic cap is not a big deal. Once you test the matching network, you can modify the cap value.

adnan012

### adnan012

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Well can't argue with Kafeman this time

But I'm glad dragonolji is optimistic!

You'd better not use FR4, you can use other rogers substrate, because FR4 Er is between 4-4.9, that can change your design every time you make PCB.

There is a mistake on input impedance of LDMOS...
LDMOS transistors have almost capacitive input impedance as expected by MOS word.So, this impedance can not be accurately simulated/measured.
Eventough we take the bonding wires inductances into account, dominant imaginary part will still be capacitive..
Double check it...

### E Kafeman

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### tony_lth

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### adnan012

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The 680 pH inductor in series with input impedance of 1.33+j0.193 ohms at 1850 MHz yields a parallel equivalent of 50.6|| +j8.32 ohms.

So you need a 10.3 pF shunt across 50 ohm input.

0.68 nH and 10.3 pF at 1850 MHz are somewhat unrealizable as discrete components. A chip component will have about 0.5 nH of series stray inductance and trying to connect and ground a 10.3 pF at 1850 MHz will have significant series stray inductance (about 1 nH). On capacitor you can alway fudge it down in value to cancel out the stray series inductance. Something like 4.3 pF but it will depend on PCB layout.

adnan012

### adnan012

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To succeed with impedance matching, you must be in a optimistic mood due to all error sources. Main thing is to check that start values are likely.
I didnt reflect over that it was a cmos component, shame on me and thanks BigBoss.

adnan012

### adnan012

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hi,
Sorry for late reply, i was busy in making pcb.

On more question , what is simultaneous conjugate matching? I tried equations given in Microwave Engineering by POZAR , in ADS. The result was strange. How can i implement "simultaneous conjugate matching" in ADS

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