2N3904 Voltage Amplifier Circuit

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elec_geek

Joined Jan 24, 2017
7
Schematic is for a CE Voltage Amplifier Circuit.

According to Heath-Zenith, this circuit was designed for a centered Q-point.

Component values are: R1 = 3.9K Ω, R2 = 1.5KΩ, Rc = 3.3KΩ, Re1 = 120Ω, RE2 = 1.5kΩ,
C1 = C2 = C3 = 47uF, Transistor = 2N3904.

Supply voltage = 20V
Gain = 10
Icq = 3 mA

Measured voltages are: Vc = 10.3V, Ve = 4.8V, Vb = 5.5V, Vceq = 5.4V.

What would the maximum, undistorted ouput voltage be? Heath claims that it should be 4.42V.

Thanks
 

Jony130

Joined Feb 17, 2009
5,145
According to Heath-Zenith
Who ??
his circuit was designed for a centered Q-point.
Component values are: R1 = 3.9K Ω, R2 = 1.5KΩ, Rc = 3.3KΩ, Re1 = 120Ω, RE2 = 1.5kΩ,
C1 = C2 = C3 = 47uF, Transistor = 2N3904.
Supply voltage = 20V
Gain = 10
Icq = 3 mA
Measured voltages are: Vc = 10.3V, Ve = 4.8V, Vb = 5.5V, Vceq = 5.4V.
R1 and R2 have a strange low values but ok.
What would the maximum, undistorted ouput voltage be? Heath claims that it should be 4.42V.
Without the load resistance (RL) the minimum voltage at collector will be equal around 5.2V

Vc_min ≈ Ve+Vce(sat) ≈ 4.8V+0.2V ≈ 5V ---> so the voltage at Vc can swing from 10.3V down to 5V----> 10.3V - 5V ≈ 5.3Vpeak = 3.74V RMS.

https://leachlegacy.ece.gatech.edu/ece3050/notes/bjt/bjtclipsu08.pdf
 

DickCappels

Joined Aug 21, 2008
6,198
Jony130 wrote:
Who?

Zenith, a consumer electronics company bought the struggling electronic kit maker Heath (as in Heath Kits) and went out of business soon after that.
 

Thread Starter

elec_geek

Joined Jan 24, 2017
7
Referring to this Heath-Zenith course material that I have (I question it's quality). It
says that this circuit will have an output that rides on a Vc of 10.3 V. So that the max
undistorted output would be 10.3 + 4.4 = 14.7V peak and a minimum undistorted output
of 10.3-4.4 = 5.9V peak.

This is so confusing.

Regards,
John
 

Dodgydave

Joined Jun 22, 2012
8,858
Referring to this Heath-Zenith course material that I have (I question it's quality). It
says that this circuit will have an output that rides on a Vc of 10.3 V. So that the max
undistorted output would be 10.3 + 4.4 = 14.7V peak and a minimum undistorted output
of 10.3-4.4 = 5.9V peak.

This is so confusing.

Regards,
John

Did you know that candles have been replaced by light bulbs, or are still using them!!
 

Thread Starter

elec_geek

Joined Jan 24, 2017
7
Jony130 wrote:
Who?

Zenith, a consumer electronics company bought the struggling electronic kit maker Heath (as in Heath Kits) and went out of business soon after that.
Why do you allow ignorant jerks like Dodgydave to remain members of this site. People are trying to learn and we appreciate
those that help us! We don't need the ignorance that some of these people are displaying!
 

MrChips

Joined Oct 2, 2009
20,299
Why do you allow ignorant jerks like Dodgydave and Jony130 remain members of this site. People are trying to learn and we appreciate
those that help us! We don't need the ignorance that some of these people are displaying!
If you do not watch your language and curb your temper you will be booted of this site pronto.
 

dl324

Joined Mar 30, 2015
10,072
It
says that this circuit will have an output that rides on a Vc of 10.3 V. So that the max
undistorted output would be 10.3 + 4.4 = 14.7V peak and a minimum undistorted output
of 10.3-4.4 = 5.9V peak.

This is so confusing.
This is what I calculated:
upload_2017-1-24_9-1-50.pngupload_2017-1-24_9-2-0.png
The output signal will be distorted when the positive portion approaches 20V or the negative portion causes the transistor to begin saturating.
 

AnalogKid

Joined Aug 1, 2013
8,310
We don't need
While I agree with your sentiments, your attitude is a bit thick for someone with only 4 messages. I recommend that you focus on the responses that help, and ignore the ignorables. Basic transistor circuit theory never goes out of style, but an over-exaggerated view of one's place in the universe never is helpful.

ak
 
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