[SOLVED] Power Amplifier Classes

Status
Not open for further replies.
N

natnoraa

Full Member level 1
Joined
Jun 19, 2011
Messages
98
Helped
12
Reputation
24
Reaction score
11
Trophy points
1,288
Visit site
Activity points
2,108
Hi everyone,

Am questioning the reason behind these discrepancies and have yet to convince myself so here it is:

Linear amplifiers are classified as such according to theory:


I ran a PDK transistor and this is what I'm getting:


Question: Looking at what I'm getting, shouldn't a Class A biasing be at about 2.5V gate voltage if it should be about 50% of the peak current? However, at about 1.1V I'm picking up a Class A transient current waveform at the drain terminal. Could anyone please explain? Any help is appreciated. Thanks



F: 1GHz
Vdd = 1.2V

Natnoraa
 

Honestly, for me it doesn't look like class A, especially the waveform at the bottom, it's more class AB, maybe simulate with VCC/2 and post your result with the same simulation setup.
 

Hi johnjoe,

Thanks for your reply. By meaning of simulating with VCC/2, i suppose it's simulating Vdd to be 0.6V and everything else the same. Here is the waveform:



P.S: What's the purpose of reducing the VCC by 2 here?

Natnoraa
 
Last edited:

Sorry, I expressed myself not correct. You have to increase your biasing to set the operating point correctly. I don't know your schematic, but let's assume your using a emitter or source circuit with resistor in the collector/drain branch. Your dc-operating point should be at Vds=Vce=Vcc/2.

Consider the picture on page 2 of the attached pdf.
 

Attachments

  • ch06.pdf
    863.6 KB · Views: 67

Hi johnjoe,

Now i understand where your vcc/2 comes from. However it's inductor bias as shown:



Natnoraa
 

Ok, than Vds=Vcc (neglecting any ohmic resistance of the coil), nevertheless you have to increase your bias. You can also draw a loadline in your output characteristic and set the dc-operating point based on your loadline.
 
Last edited:

Hi johnjoe,

I understand your point. Yes it's an ideal inductor still so there's 0 ohmic loss. However, my question is as per my first post. According to theory, half point towards max Ids will be my class A biasing point ie. 2.5V Vg. However, at 1.1V (or safely assume 1.2V) I'm already getting class A waveform. Why is that so?

Natnoraa
 

First of all, your first pic of your waveform is not class A, class A means a sinusoidal input signal -> sinusoidal output signal, it's not in your case, it has some distortion. Second (now, I got it with Vg), if you use Vg=1.2V and your input signal is small enough, you still have a linear characteristic. Take your output plot and draw the output signal with large input signal and your output signal with small input signal. Because the transition between AB and A is smooth.
 

Hi johnjoe,

Alright I'll look into this first. I've seen too many documents on load-line and output signal = input signal. The example if i were to increase Vg to 1.4V I will be getting it but still why it's << 2.5V that's the question. Thanks for now.

Natnoraa
 

Maybe because your input signal is very small and you're using some harmonic filter, aren't you?
 

Hi johnjoe,

You're right. My input signal should range between -20dBm to 10dBm typically, base on the IEEE standard and research standard for 60GHz. And yes, there's a harmonic filter. So small input signal could result in the deviation?

Natnoraa

- - - Updated - - -

Hi johnjoe,

It's okay, the problem is solved now. The reason is that for the 2.5V i've been looking at, it should be the AC signal + the DC signal. so input signal does matter, which you gave me the hint. not just looking at the DC signal. Thanks again.

Best Regards,
Natnoraa
 

Status
Not open for further replies.

Similar threads

Menu
Log in
Register
Cookies are required to use this site. You must accept them to continue using the site. Learn more…