Power Amplifier - Science topic

Hi.

The following books by ''Andrei Grebennikov'' can be useful.

1. Switchmode RF and Microwave Power Amplifiers

2. Broadband RF and Microwave Amplifiers

3. RF and Microwave Power Amplifier Design

The answer to your question is yes, 180 phase shifts are possible. This results in a push-pull amplifier. But I suggest you ask Prof PW van der Walt (he has responded to your original question) for more information on the passive circuit options as he is a grand-master in this regard. I know there are LC equivalents for when frequencies gets to low for transmission lines but have no hands on experience with these.

through power amplifer

You can do it using one of the methods

1) Use of orthogonal mode

2) Use of multiple patches

3) Use of reactive loading

Dear Anubhav Singh ,

In my book about the semicondcutor devices I defined very precisely the small signal and the large signal concepts. Any nonlinear device can be modeled by a small signal model and a large signal model. Small and large is related to the ac signal amplitudes either current or voltage compared to the DC operating voltages and currents. In small signal, the signal is so small that one can linearize the device I-V characteristics. Otherwise the one has to use the large signal model.

In small signal amplifiers analysis one can use the small signal device models otherwise one has to use the large signal models.

Small signal models are either physically based or black box two port networks such as the h. model, the y-model and the s-model.

For more information please refer to the book:

Best wishes

Dear Seohong Park ,

welcome, I will try to answer your questions one by one.

The requirements on a device is determined by its intended applications.

It is the power handling capability of the device which is also called the power ratings. It is limited by its maximum operating drain voltage VDSmax and it maximum operating drain current IDS. As they increase the power handing capability increases.

The factors that limit the maximum current is the the maximum on gate to source voltage, the channel width/ channel length ratio and the mobility of channel.

fT is the gain bandwidth product of the device and it is basically controlled by the transit time in the channel T= L/ vd where L is the channel length and vd is the drift velocity. The drift velocity itself is eqaul to mu times the electric field.

So T will decrease by making the channel shorter and the mobility higher. and operating the transistor at higher voltage.

The gain of the transistor operating as an amplifier is limited by its transconductance gm which depends basically on drain current and W/L ratio in addition to the mobility. So the factors increasing the the drain current increases also gm. The other factor controlling the highest gain of the transistor is its drain to source dynamic resistance.

Higher fT means either higher gain or higher bandwidth or both together.

fT is the gin bandwidth product.

If the transistor is used a as a switch its on resistance must be as small as possible ideally zero. This is required to make the on power loss as small as possible.

While working in various transceiver fitted in an airplane, often I found power amplifier goes unserviceable first which is undesirable. Please try to analyze the problem and make a solution to enhance the capability of a transmitter.

Dear P. Murugapandiyan,

You can use

Cadence AWR Design Environment

Further Readings

Power consumption is dependent on your specifications of the amplifier, GBW, slew rate ,etc.

Dear Fabrício Gerônimo Simões Silva ,

Thank you very much and I wish you also success!

My answer to you questions is yes for sure. In the implementing the rf front end one normally implement it off chip using resonance circuits. As the circuits introduced in these are based on conventional LC circuit configurations and switches, then all these circuits are practical and can be built in commertial radios.

We also implements the matching circuits as off chip LC circuits with suitable switches.

We are going to fabricate our rf circuits we will report their net performance and publish the results.

Best wishes

@Amit Sresht

I've decided to do the Rf power amplifier for boost the cellular signal within the range of 0.8GHz to 1GHz. And also decided to use the GaN FET model transistor. Can you help me with this type of transistor and this perticular design I choosed to do.

Dear Fabricio,

welcome,

The one dB compression point is chosen to set a limit for the linear operating range of the amplifier.

So in order to extend this point at higher output power, one has to operate the transistor at higher power supply voltage.

One also nay increases the limit of the power by operating the transistor at higher currents. In class A configuration it is the quiescent current.

In class B it is the peak drain or the collector current.

So, it is the transistor ratings at the intended operating frequency which are interesting in increasing the one dB compression point.

Best wishes

Electrical grid control (national electrical core network) may need a precision/latency below 1ms.

In general the "dt" or time precision depends on the events tracked by the IoT system considered. Furthermore the "dt" can mean time resolution requirement, it can be the irreducible uncertainty on a time measurement = t, or the irreducible uncertainty on the difference between two times t1-t0.

Let me help you visualise when time is critical:

Example 1

-high frequency trading

Buy or sell dollars?

Do it 1000 times faster than your competitors, you could win big. Do it 10 times slower and you lose a lot.

Transmitter overshoot is due to transient response of biasing circuits as well as input and output matching circuits.

additionally there is thermal effect involved Typically Gm of transistors is higher when device temperature is lower...

Agree

Sarika Shrivastava can you please elaborate which filters are you talking about exactly ? I suppose filters on biasing voltage for the transistor ?

I am also having same issue but its not in complete frequency band.

The saturation power is the maximum output power Ps. The one dB compression point is the point at the Po versus Pi curve where the saturation output power is reduced by one dB. I would like that you refer to the figure of Po vs.Pi in the link: https://ebrary.net/78582/computer_science/compression_point

Best wishes

Amplifier is already in compression. You need to fine tune the amplifier for higher output power and P1dB.

The isolator is reducing the reflected power by 18 dB so for the worst case of an open circuit (no antenna), the reflected power is 18 dB less than ~42 dBm (15W) in absolute terms. This is a VSWR of ~1.3 as worst case that should be within the mismatch range of the PA. The PA specs may be checked though. And the Isolator must have capacity to absorb the heat loss, if this scenario is to be handled.

Just make sure that the S-parameter files are used in a way that makes sense. HB simulations are very useful to obtain a steady state response of non-linear circuit components (think a PA driven into compression). S-parameters are undefined in large signal operation, which means that you use S-parameters of a transistor, let's say, you might get nonsensical results. However, if you use the S-parameter files of passive components, everything should be fine. This comes from the definition of S-parameters, where no non-linear effects are taken into account.

There is a constraint in designing the wide band power amplifiers since the amplifying devices have a limited gain bandwidth product. In creasing the bandwidth will limit the gain if the amplifier.

It is so that one better design multi band amplifier rather than a wide band amplifier since one can optimize the amplifier operation in each band using resonant rf amplifiers.

At first you choose the transistor, then you set the operating point then you match the input source and the output load. You can use the Agilent tools ADS to help you design the amplifier. You have to take care of the stability of the amplifier.

For more information please refer to the thesis: https://orca.cf.ac.uk/32468/2/2012AlmuhaisenAPhD.pdf

there is a small note in the datasheet of CMPA0060002F which I think is the answer to this. It states that:

Note1: PSAT is defined as the RF output power where the device starts to draw a positive gate current in the range of 2-4 mA.

I believe that this behavior is the reason as to why the voltage drops.

Great! Thanks for the info.

There is a recommended gate R of 150 ohms. Are you incorporating this resistance in your gate BIAS TEE network? You must provide a low frequency gain reduction or stabilization network as low frequency gain is quite large. The drain bias tee network also requires low frequency gain suppression. Large decoupling C as well shunt inductance to supply drain voltage should be free of parasitic resonance.

Perfectly explained by Osman Ceylan , LDMOS devices have better linearity, but will give you less efficiency plus u have to be more careful abt harmonic control with GaN.

A good place to start would be understanding the behavioral model first by any scholar book such as :

https://www.cambridge.org/core/books/rf-power-amplifier-behavioral-modeling/53306384B5A017F4BE2B05E9D2C097EC

Dear Bhagyashree,

welcome,

The presentation in the attached link gives you the technique to solve for the rf power amplifier using ADA design tool: http://literature.cdn.keysight.com/litweb/pdf/5989-9594EN.pdf

After designing the amplifier you can measure its S-PARAMETERS which are the EM wave description of the amplifier.

Best wishes

Hello Zihong Tang,

As per my knowledge, for the DLP7970ABP TI does not have any recommended power amplifier designs or power amplifier board. So it's difficult to say why your power amplifier was not working. I will suggest Ask this in E2E Community forum or try another higher output power ISO/IEC 15693 Reader (Make sure its compatibility with RF430FRL152H). Think it is not only a problem of the transmitter power but also of the receiver sensitivity (RF430FRL152HEVM or RF430FRL152H Patch).

You can use below reference document for the same

http://literature.cdn.keysight.com/litweb/pdf/5991-0306EN.pdf

Applications of MESFETs were initially in microwave circuits for high frequency performance, generally in the range of 1 to 10 gigahertz. As hole mobility in GaAs is 5 to 10 times higher than it is in Silicon and allows MESFET operation at frequencies higher than can be achieved with Silicon devices.

You can also have a look at the following conference paper: P. Saad, R. Hou, R. Hellberg and B. Berglund, "Ultra-Wideband Doherty-Like Power Amplifier," 2018 IEEE/MTT-S International Microwave Symposium - IMS, Philadelphia, PA, 2018, pp. 1215-1218.

What is the waveform of your PEM Ref In signal? Your LIA can use 3 types - sine, TTL +ve edge and TTL -ve edge.

Some PEMs reference output can be set to twice the modulation frequency. Have you checked with an oscilloscope or frequency meter that the signal is 50kHz?

Kindly mail me with your proposal to shub@nitm.ac.in or

Call me +91-9485186003.

Dear Phan Huu Lam

Actually, there are many ways to do it. However, you need to consider your budget and what is your workspace. You can find many publications, from that you can follow them. If you have details questions, you can email to me.

You can find my email in my publications.

Thanks

This question iv very interesting. To answer this question clearly one has to define exactly what is the memory effect in the power amplifier. There is a principle concerning the measurement of the quantities. The quantities are measured as they are defined.

Memory means to preserve information when the time elapses. So, it is a direct time domain effect. It is so that one can preserve a analog sample for specific time in a capacitor. So, if the devices are containing energy storage effect they will show some memory effect. The memory effect in an amplifier can be tested by applying a pulse on the amplifier and then observe how long this pulse can last after the input pulse is removed.

If there is no memory effect the two pulses will follow each other in the time variations. This is straight forward from the transfer characteristics of the strictly linear amplifier Vo= AVi with A is the gain, Vo is the output voltage and Vi is the input voltage. So Vo is a replica of the output and no memory effect at all.

So, we have to define the memory effect on time domain and then determine it in the frequency domain.

Best wishes

The transistors are narrow band and the stubs create a short circuit at the operating frequency on the bias line; this is transformed by the quarter wave transmission line to present an open circuit on the main output line. By using two at slightly different lengths & offsets you can broaden the bandwidth to be close to the operating band.

The first step is to determine the specification of the rf power amplifier according to the transceiver specifications.

The most important parameters of the transmitter power amplifier is:

The maximum rf output power

The carrier frequency

The bandwidth of the amplifier for tuned amplifiers

The one dB compression point

The third order intercept point

The power conversion efficeincy

The DC input power

The power supply voltage

The input impedance which is normally 50 ohm

and the output impedance which

is normally 50 ohm

After that you can select an amplifier which matches your requirements and specs from the that available in the semiconductor market

According to the power rating you can find:

integrated circuit amplifier,

amplifier modules

Or you can yourself design one using the proper transistor circuit with proper biasing and tuned circuits.

Best wishes

Use Harmonic balance simulation (HB) in ADS, Sweep your input power , Calculate your output power and DC power

then you can plot your PAE= Pout-Pin/Pdc

Dear Tim,

as you correctly stated, the methods to extract  the parameters of the small-signal equivalent circuit of active devices rely on  S-parameter measurements at different bias points. In my experience, for what concern bipolar devices and in particular HBTs for RF applications, most of the time measurements on a single device in different operating conditions are enough without resorting to measurements of current-mirror topologies.

To answer your question, the capacitance levels for sub-micron InP/InGaAs DHBTs are in the fF order of magnitude and are commonly measured as a function of collector current. The frequency range of measurements with a VNA could be up to 140 GHz or higher. 

I hope this answers your question and in case I'm open for further discussion.

Hi Rifki,

If you want to get Power Added Efficiency in ADS using HB simulation, you should do these steps:

1. put P_Probes (power probes) from probe pallette at input and output of your PA schematic with names for example P_Out for output and P_In for input.

2. use a frequency source i.e. P1_tone. at input and define a variable for input power so you can sweep that.

3. simulate the design.

4. at data display sheet, create an equation and write PAE equation. don't forget to calculate DC power from your bias variables (using p_probe or i_probe)

good luck

Dear Ekta and Sajeed,

a totally resistive input impedance is a good theoretical assumption but in practice it is very ideal.  In addition, in order to have the maximum power transfer from amplifier to antenna, the requirement is to have Z(input antenna) = Z*(output amplifier).

In my opinion (also considering the type and low complexity of the yagi design) the way to follows is to simulate the antenna and its changes by means of NEC Numerical Electromagnetic Code

 https://en.wikipedia.org/wiki/Numerical_Electromagnetics_Code

Regards,

Massimo

The oscillators can be built by a positive feedback amplifier which satisfies the Parkhausen criteria. The oscillating open loop gain AB must be equal or greater than one, which means magnitude AB=1 and the phase= 0, 2 pi,..etc. Whenever this condition is satisfied the circuit will oscillate. It remains that one has to make the circuit oscillate at certain frequency. This is normally accomplished by frequency selective circuit where the above condition is satisfied at certain specific frequency. These circuits are called timing circuits of the oscillators. It is so that the amplifier odf the oscillator is made flat which means that its gain is independent of frequency. Then the feed back network which is called the B network must be frequency selective. This means that it acts as a narrow band pass filter with the center frequency is the most preferred frequency for oscillations. This filter may be passive like an LC tuned circuit, or a phase shift network or a lag lead circuit. An alternative, one can use an active B-network, say acting also as a band pass filter.

The active filters themselves are used preferably at low frequencies where using the LC components will  make the filter bulky and lossy.

So, the active filter tuned oscillators are advantageous at low frequencies and extremely low frequencies.

Best wishes

.

Hello Nuri gulmez.

in my opinion The best way to do that is to measure the S11 of the diode alone and from the results obtained you design the matching circuit to ensure a maximum transfer of power to the input of diode.

Best regards

Thank you so much:)

DEAR SIR AND MADAM PROFESSOR

MAYBE BELOW IS THE ANSWERS

HOPEFULLY WE CAN DERIVE

THE EQUATIONS WITH THE

HELICAL ANTENNA PICTURE

DESIGN

BEST WISHES

DR MOHAMAD ARINAL

Dear Chaouki,

Thank you so very much. I would go over these articles.

Best Regards!

make all the components of the same kind equals (i.e. R1=R2 and C1=C2) it can helps.

Thank you so much for your complete answer:)

Dear Zahra,

Ninad has touched the answer. The data rate depends on the double sided bandwidth and the modulation type. In order to make the answer clear assume that you have a carrier fc and you modulated this carrier by a data stream fb and the type of digital modulation is 4 QAM. Then according to the theory of the digital modulation techniques the double sided bandwidth that is the two side bands around the carrier 

will be equal to fb/2. For 16 QAM it will be fb/4 and so on. Assuming now that the resonance frequency coincides with the carrier of the input modulated wave that will be amplified by the amplifier then the bandwidth BW of the amplifier will set the maximum limit on the data rate such that BW of the amplifier= fb max/2 in case of 4-QAM. This results in the relation fbmax=2 BW,

In principle the bandwidth of the amplifier must accommodate twice the baseband bandwidth. This is the rule. The baseband bandwidth is related to the bit rate and type of modulation.

So, it is the pass band width of the amplifier that limits the bitrate of the data. The reactive elements in the circuit together with the parasympathetic capacitances that limit the BW of the amplifier.

Best wishes 

Dear Zahra,

As Far I understand you need low power high speed (Not low power dissipation) Amplifier for modulation for this purpose I think current mode amplifiers are best.

Thanks

Hello : You need to define function first (definition >>>> functions >>>> then you can select your function shape). knowing that you need to used function name for next steps.

Regards,

ABBAS JASIM

Dear Nicolas,

Kindly discuss your view point on  the parasitic capacitance effect  on  current mode  circuits

Thanks

Dear Gandhiraj,

Please till us what is the signal at channel 1 and channel 2. Can you display the input of the amplifier?

The sources of noise may be the power supply, the ground, the amplifier itself or added to the input before entering the amplifier. It could be contaminated by the data acquisition process of the digital storage oscilloscope. The contamination by the sampling pulses is clear in your signal.

In order to avoid contamination of the signal by sampling you may better use analog oscilloscope.

It seems me that the hundred Herz pulses are superimposed on your signal.

Please show us the input signal.

wish you success

Worked at Wilcoxen many years.  Solder two low temp coefficient disk cap (X7R +/- 15%, -55-+125 C) face to face to reduce non-linearity and reduce X-Y response. Bias the stack with a few volts and glue assembly to a bolt head for mounting and solder a low profile ref mass to the top. Pick signal off junction at the middle of the caps. Good approach up to about 5 gs for 10 K hours lifetime. The connecting wires must be thin and glued in place to prevent them from resonating. Ref mass and cap ER determine to first order output. The response can be calculated but cut and try will get you close faster.

For audio frequency range, I think class D amplifier is the best solution for high-efficiency application. And there are many integrated chips that can be used to implement easily.

thanks Prof. Yeom.(also for Send attached file)

my challenge is in wideband matching!

in wideband matching, harmonic impedances in any frequency point inside the bandwidth determined in intrinsic plan (arranged in increasing frequency)

but when this datas transferred to package plan (by using 2-port parameters of package), the arranged form of these, will be degenerate!

that means datas in different frequency point, are Back and forth and to cluttered! in this case design the wide band matching network that can  follow this clutter frequency is very difficult & in some cases impossible!

This method is not used in electrochemistry. As in electrical engineering group.

It is true that the peak-to-average ratio might be the same in a massive MIMO system as in a conventional single-antenna system, although there are methods to reduce it (such as low-PAR precoding).

However, the average transmit power can be greatly reduced. The total transmit power per base station can be reduced, thanks to the array gain, but in particular the total power is divided over 100 antennas or so. With 1 W total power, you will get only 10 mW per antenna.

I am not an expert in circuit design, but in some situations you might not even need a power amplifier, but can feed the antenna directly from CMOS.

Have you tried matched filter?

Dear Norhamiza,

You can increase the phase margin by making a dominant pole nearer to the zero frequency origin. This is accomplished by compensating the op amp through adding a shunting capacitor in the highest impedance node of the amplifier. This is a very well known technique which is used commonly to increase the phase margin. However, this will be on the cost of the amplifier bandwidth. The amplifier bandwidth will be decreased. The other method is by using pole zero compensation, where one adds a zero which cancels one of the poles of the op amp. Again this technique is known and you can consult the relevant literature.

Wish you success

The lock in amplifier is intended to measure the amplitude of a sinusoidal signal contaminated by white noise. In your case of biosignals, the wave-forms are complex which means that it is require to analyse it using spectrum analyzer. This can be done by the lock in amplifier as explained in the following paragraph.

In your case you want to recover the biosignal from noise, so , it is so that we have to select the frequency to be recovered and obtain it  from a reference oscillator say this frequency is wr.  So, one has to input a sinusoidal wave from with wr from a standard laboratory oscillator with amplitude say one volt. Then you input your signal to be recovered and set the required gain at the input say G. Then according to the operation of the instrument you will have two outputs: Y output  and an X output. After the low pass filter with very narrow bandwidth the DC values will represent  the sine and cosine components of the signal. amplitude.

This is if you want to recover and analyse your signal into frequency components.

However you can use the amplifier section of the instrument to amplify your signal.

wish you success

Welcome friends Zhou and Dreher,

Radio frequency choke is an inductor thought to prevent the radio frequency current from passing in the power supply. It acts as a low pass filter allowing the dc current to pass to the amplifier circuit while preventing the radio frequency to pass to the power supply in German means ( Drossel). If the rf current is allowed to pass to the power supply which is not an ideal voltage source in the sense that it has a source resistance, it will leak to the other RF building blocks powered by the same power supply. It acts as a decoupling element between the RF component and the power supply. It has a lumped element with metallic coil and sometimes with ferrite core to reduce its size.

So, the inductor is named according its function.

for more information please see the link:

Best wishes

Hello,

I'll attach the 1902 MK III manual. It is stating " The 1902 requires approximately 220mA on the +5V, 75mA on the +15V and 20mA on the –15V power rails" - which are the same voltages as the MK IV. And there are some hints about power supply compatibility.

Regarding safety you should use a power supply that is fully isolated according to the needs for medical instruments with direct body connection, but this should be doable. I did not yet find a connector drawing, but I only cursorily read the manual.

Regards

Hasula,

I worked in communication system all my career and never heard of a charge amplifier so it must be a very specialized terminology. Often, the transmitting transducer need an amplifier that has the capability of a high peak to average power. This means the power dissipation in the amplifier does not need to be too high but the voltage levels on the output do need to be high. On the receiving transducer, a good low-noise voltage amplifier with a high gain is needed. I am not familiar with the amplifier numbers you gave.

good luck

Steve

thanks for the helpful documents. I was wondering if this could be found out from transient simulation output of PA. Thanks.

Hello,

you may provide a row of examples, but I don't think that the above is generally true.

I'd assume that you are comparing different "generations" of log amps. During evolution power consumptions tends to go down while "performance" (here expressed in 'dynamic range') tends to go up. If you are comparing "equals" (ICs of the same generation - eventually from different suppliers) the picture tends to change. But log amps are to some extend "specific" as they are a "rare species". So you may not be able to find more than a few to validate my assessment.

Regards

Hello,

I am sorry, but it is hard to answer your question without a schematic diagram showing all the values of the passive components surrounding the opamp.  I t would also help if you attached an actual photograph of the circuit.

When you say that you see a 4MHz sine wave with 15 V, do you mean that the 4 MHz sine wave is riding on top of a +15 VDC bias, i.e., that the average value of the sine wave is +15 VDC instead of zero?  When you say that adding a 100 ohm resistor between the opamp output pin and the 'scope probe gets rid of the +15 VDC bias, does that mean you don't see the 4MHz.  Or do I have that backwards, and the 4MHz signal is the amplified signal and the +15 VDC bias disappears with the 100 ohm resistor?

I think you can see what I'm getting at.  You need to pose the problem so we don't have to read your mind. 

Regards,

Tom Cuff

Hi Nishil,

For a short introduction you can visit

The most useful comment above is the shortest, your choice shall be based on your mission requirements. Both SSPA and TWTA have dramatically improved in the last decade. Better efficiency for SSPA, allowing to reach higher power levels 5GaN technology) and flexible TWTA (check Thales Electronic Devices website, output power can be adjusted in a 0 to 4 dB range while keeping almost a constant efficiency for such devices, allowing flexible output power per transponder) or mini-TWTA for example. In most traditional C-band COMSAT applications TWTA would be the only choice due to output power requirement (derived from needed EIRP). But as said it depends on the mission requirements, type of modulation used (=> linearity requirement). Do not forget that eventually solutions shall be compared taking into account system parameters, e.g. TWTA efficiency shall be considered at operating point, it will not be the same if your mission requirements impose an operation at 3 to 10 dB back-off or allow to operate the TWTA at saturation (which is usually the operating point point used for stating the efficiency by the TWT manufacturer). Then for TWTA you will have the option of conduction or radiation cooled devices... 

To make your choice, you will need to perform at least a short system analysis.

P. Pierrot

I do hope it will help 

Hoping you used the word "screw" for "drill":

With a mini drill (Dremel or other brand - near-to-perfect ), a drill press and cemented carbide drills 0.5 mm should be achievable.

It is not exactly "with hand" (due to the requirement for a small drill press) but it is still low-cost.

In addition I'd recommend to use collets to achieve the smallest drill eccentricity possible. Otherwise you risk to break the drill which is sensible vs. radial forces.

you must have the model library of your technology in ADS

As Prof. AR Reddy mentioned feedback generally used to improve the bandwidth and stabilize the circuit but it affects the gain since gain bandwidth product is constant value. you can  review Analog integrated circuits for Razavi chapter 8  

Please find the attached files:

Titile: A low-power CMOS power amplifier for Ultra wideband (UWB) Applications

Dear Gagendra,

Strictly peaking, the inductor L1 is connected to the point Vgg and R1 in parallel with R2 are connected in series with it at the input side of the equivalent circuit. But , practically L1 must be chosen to be high enough to make this path AC open circuit at the operating frequencies. So, the divider resistor together with the L1 can be omitted from the equivalent circuit. Likewise, one can eliminate the path through L2 as i have aid before. So, you need to jut eliminate R1 and R2 from your equivalent circuit and your circuit is okay.

wih you ssuccess