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I have seen many designs where the antenna/PA output is fed DC voltage through an inductor. Why not leave it out and feed the DC bias directly? It is L1a in the diagram below:

enter image description here

There are some values in the table below for some working frequencies. Is L1a also part of a filter/matching network? What is the general method of arriving at a certain value for it?

enter image description here

Source: MAX4146x EV KIT

EDIT

In the light of the answers below I shall attempt to answer the last question myself.

An LC pass-band filter centered around the operating frequency, and the other parameters pretty much made up:

IN:
Number of LC pairs:              1
Cutoff frequency (Fc):           314.8 MHz
Passband:                        315 MHz
Impedance (Zo):                  50 Z
Ripple:                          3 Z

OUT:
L:             ~ 50 nH
C:             ~ 5 pF



IN:
Number of LC pairs:              1
Cutoff frequency (Fc):           433.8 MHz
Passband:                        434 MHz
Impedance (Zo):                  50 Z
Ripple:                          1.5 Z

OUT:
L:             ~ 24 nH
C:             ~ 6 pF

Values calculated at https://www.allaboutcircuits.com/tools/bandpass-filter-calculator/

Maybe the values from datasheet were calculated similarly? Or maybe I did not understand all that much.

more edit

Probably miscalculated there... anyway, a bit more on the issue:

Generally,is stated that the output impedance of the bias circuitry should be kept small, in order to increase the linearity of theoutput bias stage. However, the output impedance is typically designed to have alargeresistance in order to reduce the noise contribution from the bias circuitry, and to avoidsignificant loading on the RF input port. For example, to use an inductor in the bias circuit to form lowimpedancenear DC,and highimpedance near the RF signal band

and

https://www.youtube.com/watch?v=lxgpm-UXTNY

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    \$\begingroup\$ It's crucial that the inductor has a high impedance at the transmitter's carrier frequency. That prevents the output disappearing into the supply instead of the load (antenna). A direct connection to the DC supply would be low impedance. \$\endgroup\$
    – user16324
    Sep 20, 2021 at 16:55

2 Answers 2

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Applying DC supply straight to the output pin would cause a dead short when the amplifier attempts to sink current to drive the output in an open-drain fashion, as documented in the datasheet:

enter image description here

This means that you need to present some series impedance between the output and your positive supply rail. In the case of RF devices, the impedance is often a resonant load, comprised here of your provided inductor and the internal capacitances mentioned in the datasheet above. This gives a nice large impedance at the resonant frequency, thus giving a high gain, while still allowing a DC bias current to flow as needed (here that second point is less relevant because the amplifier operates in a switching mode). As Bimpelrekkie mentions in the comments, the resonance improves output power and power efficiency, and is almost mandatory at frequencies in the GHz range and up.

Compare this with the drain resistance used in a simple common source/common-emitter amplifier used at baseband frequencies.

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  • \$\begingroup\$ What gets affected and how if a resistor was placed instead of that inductor, same impedance as the inductor @ operating frequency ? \$\endgroup\$
    – kellogs
    Sep 20, 2021 at 15:39
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    \$\begingroup\$ What gets affected and how if a resistor was placed instead of that inductor, same impedance as the inductor Consider the DC biasing current that is needed for the amplifier, through a resistor you will get a voltage drop, limiting the RF output voltage swing. With an inductor, the DC voltage drop will be much lower (almost zero). \$\endgroup\$
    – Bimpelrekkie
    Sep 20, 2021 at 15:41
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    \$\begingroup\$ @kellogs An additional factor I suspect (but perhaps someone can correct me as I haven't worked with switching PAs) is the filtering provided by the resonant circuit. A switching PA introduces a lot of high-frequency artifacts, which would be better attenuated by rolloff off resonance, than just the resistive load you propose along with the internal and trace capacitances. \$\endgroup\$
    – nanofarad
    Sep 20, 2021 at 15:59
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    \$\begingroup\$ @nanofarad You're right, often those inductors are part of a resonating (LC) tank. This increases output power and power efficiency. Sometimes the inductors have a value that is too high to be able to resonate, then RF designers call them "chokes" and their sole purpose is to deliver DC to the PA. There could then still be other inductors present to form a resonant tank. At frequencies above 1 GHz I would say that having such a resonant network is almost a must. At 10 GHz and higher it is impossible to do anything without a resonant circuit, even inside an IC. \$\endgroup\$
    – Bimpelrekkie
    Sep 20, 2021 at 16:23
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Pullup inductors offer high impedance to RF and double the AC voltage swing for RF/AC, centered around the DC voltage.

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  • \$\begingroup\$ should I take it that the DC bias is there in order to prevent negative voltage swings at PA output ? \$\endgroup\$
    – kellogs
    Sep 20, 2021 at 15:32
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    \$\begingroup\$ No . The voltage on an RF amp does not swing below 0V until a series cap blocks the DC. The DC + INductor provides the power and the transistor modulates that power out with RF voltage according to the ratio of RF load to transistor current (BJT) or resistance (FET) swing.. \$\endgroup\$
    – Tony Stewart EE75
    Sep 20, 2021 at 15:35

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