EMRFD Message Archive 6734
Message Date From Subject 6734 2011-10-11 06:55:18 jnm659 Newbie Question - help with amplifier basics, please.
I blame Bill N2CQR, But i've been using "the book" as a learning text. I'm i'm stuck on Chapter 2 - just at the end of section 2.3 - Large Section Amplifiers.
I have the "revised first edition - (c) 2003 - 2009". It says "revised 1st edition" on the cover, so I'ts going to be the 3rd printing. I've checked the errata on Wes' website - nothing mentioned there, so clearly it's my fault (?!)
So, I've followed the maths as far as the 10Mhz Class A amp in Fig 2.34. My first problem is simply calculating the DC emitter voltage:
Here, the 2N3904 transistor base is biased through a potential divider comprising 10K and 3.3K resistors, the latter being to ground. VCC is +10V.
I understand that VE is VB - 0.6V.
So VB is (10V x 3.3K) / (10K + 3.3K) = 2.48 V
So, VE must be 1.88V (But the text says it's 1.64V) : I resorted to LT Spice - Clearly Wes is right. So what am I doing wrong?
OK, for the rest of my working, I use the books value for VB and the rest of the maths slots into place. IE, 're', VG, beta, and Rin are all good.
Now, I figure that it's the 50 ohm resistor in line with the ideal voltage source, followed by the the 100 ohm value for Rin that gives rise to the statement that only 2/3 of the open circuit input voltage appears at the base, because 100 * (100 + 50 ) = 2/3. First time i did this I went with the 100 ohms of Rin and the 200 ohms connected to the emitter! Wrong, i know, but the numbers work out so I didn't spot it!
I can then follow the maths behind what happens with a 10mV peak input signal. And with the 50mV signal. And I can see with LT spice what's going on when the input is cranked up to 0.5v peak.
But I loose the plot when the Power Output is mentioned: The text says "the output power with a 50 ohm load is about 2.5mW". How? I thought Pout was (Vout) ^2 / Rload ? With a load of 50 ohms, this would imply a Vout of the square root of (0.0025 * 50), or 0.35V Peak.
After about 3 microseconds, LT Spice shows that the circuit driven with .5V settles down and the output signal is between + 0.5V and -1.9V, (so 2.4 V peak-to-peak, half of which is 1.2V) None of those values is anywhere near 350mV - so where does the 2.5mW figure come from?
So then we try and apply equation 2.22 to fix the load. I plug in values of VCC=10V, VB=2.48V and IE = 8.2mA and get Rload=917ohms, which is just UNDER 1K, where the text predicts a load just OVER 1K.
The text then says the new load gives a 10MHz output of 11V peak TO PEAK. I had to go back to calculated the new Vgain of Rload/re, or 1000/3.2 = 312.5. This in turn gives a Vout of (2/3) X Vin X Vgain = 2/3 x 0.5V x 312.5 = 104V peak ( 208V peak TO PEAK) (pretty good going from a 10V supply!!!!) LT SPICE shows neither of us are correct, with a very distorted output of +20.5V to -8v, or 28.5V paek TO PEAK, half of which is 14.25V.
Then the text says that 11V peak to peak output, accros a 1000k load is a output power of 16mw. Well, I make it Pout = V^2 / R = (11/2)^2 / 1000 = 30mW.
Please can someone kindly give me the kick I need to realise what I'm doing wrong?
Jeff
G7TAT
[Non-text portions of this message have been removed]6735 2011-10-11 21:15:42 Wes Re: Newbie Question - help with amplifier basics, please.
The simplistic bias calculation outlined earlier in the chapter won't work here, for the base divider is not "stiff" enough. If we had used 1K and 330 instead of 10k and 3.3K, it might yield better accuracy. As it turns out, the numbers in the text were SPICE simulation values rather than numbers from a simple bias calculation. Even those are not exactly the same as what I got a few minutes ago with SPICE, for the model used when the simulation was done differ a bit from those used in today's simulation. But they are close.
When I used LT-SPICE with the 2N3904 model that is present in my version, I get dc bias numbers of vb=2.41, ve=1.70, ie=8.5. This is all the result of the computer model. This will vary with the details of the model that we use. Physical reality, which means what you would measure when you build it, will depend upon the parts you grab when you build. Beta will vary, etc. The 2N3904 model I'm using might not have the same parameters as present in the defaults offered by Linear Technology.
The input resistance for the small signal AC model will be beta x re. (Actually beta+1, but no need to worry about too much detail.) Beta is approximately 300/10=30, for 300 MHz is the F-t and 10 MHz is the operating frequency. At 8.5 mA, re is 26/8.5, or 3. Hence, Rin is 90. When I did a small signal SPICE simulation with an input generator voltage of 50 mV peak, I simulated a base voltage of 41 mV peak. This implies Rin a bit higher than the 100 Ohm Rin that was mentioned in the text and the 90 Ohms above. The F-t in the computer model must be a bit higher than 300.
I repeated the text TRAN analysis in SPICE. An open circuit drive of 50 mV was used, and not the 500 mV that you mentioned. The 50 mV peak drive level comes from Fig 2.36. The result I got from the repeat analysis is virtually identical to Fig 2.36. The signal on the collector was about 950 mV pk-pk. It is not a sine wave, but is quite distorted. Still, just for grins, let's assume that it was a sine wave. This signal voltage appears across the 50 Ohm load. The peak to peak value of .95 volt is a peak value of 0.475. Dividing by root 2 yields .336 volt. The power can now be calculated using the formula you quoted and it comes out to be 2.25 mW.
The numbers that are provided from an AC analysis in SPICE are peak values. But we look at wave forms and usually extract peak to peak values. RMS values are used with your simple formula. You can't just hop around from one form to the next and use but one formula.
When I did today's SPICE TRAN analysis, I got 11.4 v pk-pk at the collector. The signal at the load was 2.548 v pk-pk. This calculates to be 16.23 mW, which is a major improvement over the 2.25 mW without the pi network. Must be something to this impedance matching thing.
The original reas6736 2011-10-11 21:15:46 Chris Trask Re: Newbie Question - help with amplifier basics, please.
> So, I've followed the maths as far as the 10Mhz Class A amp
> in Fig 2.34. My first problem is simply calculating the DC
> emitter voltage:
>
> Here, the 2N3904 transistor base is biased through a potential
> divider comprising 10K and 3.3K resistors, the latter being to
> ground. VCC is +10V.
>
> I understand that VE is VB - 0.6V.
>
> So VB is (10V x 3.3K) / (10K + 3.3K) = 2.48 V
>
> So, VE must be 1.88V (But the text says it's 1.64V) : I
> resorted to LT Spice - Clearly Wes is right. So what am I
> doing wrong?
>
Chris Trask
N7ZWY / WDX3HLB
Senior Member IEEE
http://www.home.earthlink.net/~christrask/6737 2011-10-11 21:15:47 cbayona Re: Newbie Question - help with amplifier basics, please.
>Hi,The base is also a resistor that is in parallel with the 3.3K
>
>I blame Bill N2CQR, But i've been using "the book" as a learning
>text. I'm i'm stuck on Chapter 2 - just at the end of section 2.3 -
>Large Section Amplifiers.
>
>I have the "revised first edition - (c) 2003 - 2009". It says
>"revised 1st edition" on the cover, so I'ts going to be the 3rd
>printing. I've checked the errata on Wes' website - nothing
>mentioned there, so clearly it's my fault (?!)
>
>So, I've followed the maths as far as the 10Mhz Class A amp in Fig
>2.34. My first problem is simply calculating the DC emitter voltage:
>
>Here, the 2N3904 transistor base is biased through a potential
>divider comprising 10K and 3.3K resistors, the latter being to
>ground. VCC is +10V.
>
>I understand that VE is VB - 0.6V.
>
>So VB is (10V x 3.3K) / (10K + 3.3K) = 2.48 V
>
>So, VE must be 1.88V (But the text says it's 1.64V) : I resorted
>to LT Spice - Clearly Wes is right. So what am I doing wrong?
resistor lowering the voltage at the base of the transistor.
Cecil
k5nwa
< www.softrockradio.org/ > < http://thepartsplace.k5nwa.com/ >
< http://parts.softrockradio.org/ >
Never take life seriously. Nobody gets out alive anyway.
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