Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

[SOLVED] Some question about transistor analysis

Status
Not open for further replies.

ehsantech

Member level 3
Joined
Jul 29, 2014
Messages
67
Helped
0
Reputation
0
Reaction score
0
Trophy points
6
Activity points
492
Hello
Here is my circuit and dc bias point
T1.JPG
and here is ac output voltage
T2.JPG

question 1:the output amplitude is too much higher than Dc input voltage what is my mistake or spice mistake?

question 2:if we want to find upper and lower frequency ( cutoff) we should calculate (1/sqr(2))*Vmax and then find the frequencies?

question 3:in below output figure if we want to find upper and lower frequency ( cutoff) we should fine max DB of output voltage and then minus 3db?
t3.JPG

question 4:is there any simplest way to find upper and lower frequency in two different above questions(2,3)?
 

Borber

Advanced Member level 5
Joined
Dec 31, 1999
Messages
1,602
Helped
235
Reputation
472
Reaction score
111
Trophy points
1,343
Location
on third flor
Activity points
11,862
At simulated gain of 190+something (46dB) and input voltage of 1V your funny simulator shows 190+something volts of output voltage. If you divide that "output voltage" with 1V you will get amplification factor or gain.
Upper and lower frequencies are always 3dB (0.707) down the max gain and has to be read on graph.
 

Audioguru

Advanced Member level 5
Joined
Jan 19, 2008
Messages
9,284
Helped
2,141
Reputation
4,276
Reaction score
1,968
Trophy points
1,393
Location
Toronto area of Canada
Activity points
58,419
Your simulation software is defective. It should warn you that your input signal level is way too high at 1V. With a 1V input level then the output is a saturated squarewave, not a sinewave and the frequency response graph will be completely wrong.

The signal across the 1k load resistor should be a maximum of 13V p-p or 6.5V peak but with severe distortion because there is no negative feedback. Maybe the distortion will be acceptable if the output is reduced to 3V peak. Then the input level should be 3V/190= 15.8mV peak.
I do not know if your simulation software uses peak signal levels or RMS levels.
 

BradtheRad

Super Moderator
Staff member
Joined
Apr 1, 2011
Messages
14,186
Helped
2,808
Reputation
5,624
Reaction score
2,745
Trophy points
1,393
Location
Minneapolis, Minnesota, USA
Activity points
105,823
Your amplifier has one low cutoff frequency for the input signal, and another low cutoff frequency for the output.

Regarding the input signal, its AC portion acts through R6, C3, C2. You can calculate the time constant, and from that you derive its cutoff frequency.

Regarding the output signal, its AC portion acts through R3, R4, C1, and load R5.

Notice that if you adjust the bias carefully, you can center the output signal around 0V. Then C1 is not needed.

If you want larger output amplitude, reduce R3 & R4.
 

godfreyl

Advanced Member level 5
Joined
Apr 18, 2012
Messages
1,981
Helped
632
Reputation
1,266
Reaction score
624
Trophy points
1,393
Activity points
12,772
question 1:the output amplitude is too much higher than Dc input voltage what is my mistake or spice mistake?
No mistake - that's normal with AC analysis, because the similator does a small-signal analysis. That means it calculates the gain of the circuit with a very small input voltage, then multiplies that by the input voltage you gave.

So if you give 1V input and it shows 100V output, it doesn't mean your circuit can actually give 100V out. It just means the output is 100 times bigger than the input for very small inputs. It may be a bit confusing at first, but that's the way most simulators work.

If you want to see what the maximum output voltage is, and what the waveform looks like, you need to do a transient analysis with a sine wave souce, not an AC source.

question 2:if we want to find upper and lower frequency ( cutoff) we should calculate (1/sqr(2))*Vmax and then find the frequencies?
Yes, that's one way to do it.

question 3:in below output figure if we want to find upper and lower frequency ( cutoff) we should fine max DB of output voltage and then minus 3db?
Yes. I think this is easier - less calculation.


Your simulation software is defective.
No, it's normal. See above.
 

Audioguru

Advanced Member level 5
Joined
Jan 19, 2008
Messages
9,284
Helped
2,141
Reputation
4,276
Reaction score
1,968
Trophy points
1,393
Location
Toronto area of Canada
Activity points
58,419
I think simulation software should warn than the input signal level is much too high resulting in an error in the frequency response if the circuit is actually built and used with such a high input signal level.
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Top