hi all,
i would like to construct an experiment on coupling two stages of Common emitter amplifier by using transformer coupled method. But i dont know which ratio to choose or how many resistance for the primary and secondary winding.
Ok.Now i have the first stage CE amp with Zin= 1.6kohm and Zout= 3.9kohm
second stage CE amp with Zin =1.7kohm and Zout=3.9kohm
so i will use an impedance matching transformer to coupled the first stage's Zout to the second stage Zin. then i stuck here. what ratio and resistance of matching transformer should i buy?
Can someone help me on this?thank you
i would like to construct an experiment on coupling two stages of Common emitter amplifier by using transformer coupled method. But i dont know which ratio to choose or how many resistance for the primary and secondary winding.
Ok.Now i have the first stage CE amp with Zin= 1.6kohm and Zout= 3.9kohm
second stage CE amp with Zin =1.7kohm and Zout=3.9kohm
so i will use an impedance matching transformer to coupled the first stage's Zout to the second stage Zin. then i stuck here. what ratio and resistance of matching transformer should i buy?
Can someone help me on this?thank you
the first stage CE i have a 3.9kohm resistor connected from Vcc to Collector pin. so when i want to use a transformer to coupled the signal should i replace the 3.9K resistor with the primary side of the transformer?
thank you.
thank you.
The CE stage is usually designed to drive a high impedance load.
If you have Zout=3k9 then normally the Zin of the next stage is much higher than 3k9.
What is the minimum impedance that your CE stage can drive and still meet it's performance specification?
Once you know the minimum required load impedance you can then determine the turns ratio of the coupling transformer.
Let's suppose your CE stage can drive 10k, in addition to the 3k9 load on the collector.
Your second stage is actually 1k7. Impedance ratio is ~7:1.
The turns ratio required is sqrt(impedance ratio). sqrt(7) ~2.6.
The input has say 260T the output has (260/2.6) ~100Turns.
The voltage delivered to the second stage will be reduced to 1/2.6
Now I'm at my limit. If you replace the 3k9 collector load with the input winding of the transformer, then you are driving a quite different load by that first stage. Let's see what other replies you get.
If you have Zout=3k9 then normally the Zin of the next stage is much higher than 3k9.
What is the minimum impedance that your CE stage can drive and still meet it's performance specification?
Once you know the minimum required load impedance you can then determine the turns ratio of the coupling transformer.
Let's suppose your CE stage can drive 10k, in addition to the 3k9 load on the collector.
Your second stage is actually 1k7. Impedance ratio is ~7:1.
The turns ratio required is sqrt(impedance ratio). sqrt(7) ~2.6.
The input has say 260T the output has (260/2.6) ~100Turns.
The voltage delivered to the second stage will be reduced to 1/2.6
Now I'm at my limit. If you replace the 3k9 collector load with the input winding of the transformer, then you are driving a quite different load by that first stage. Let's see what other replies you get.
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The transformer can replace the collector resistor, or it can be used via capacitor coupling in parallel with the collector resistor. The two situations are quite different, so you would need to decide which method you want to use. The intended application is relevant. What type of signal? How much gain? How are you biasing the stages? Is this a paper exercise or do you intend to build it and test it?
anyone have any idea on how to calculate what is the spec i need to buy for this circuit? How is the calculation. i can see that the Zin for the 2nd stage is 1.06kohm (correct me if i'm wrong). then how about the Zout for the 1st stage?
thank you.
thank you.
How did you calculate 1.06k? There does not seem to be enough information for that. Similarly Zout for the first stage. Both depend on the transistors to be used. It is possible you are confusing Zout with Zload. I assume this circuit is from a textbook.
Sorry for the wrong answer. i had made the correction as below. (i'm using 2N3904)
Zin(2nd stage) = rb || (Beta)(re)
= 3.33k || (160)(0.0167)
= 2.67ohm
i get this circuit from google search. Another question is can i use a ferrite core to substitute the transformer if i cant get an impedance matching transformer?
Zin(2nd stage) = rb || (Beta)(re)
= 3.33k || (160)(0.0167)
= 2.67ohm
i get this circuit from google search. Another question is can i use a ferrite core to substitute the transformer if i cant get an impedance matching transformer?
Where did you get 0.0167 for re? What do you calculate for the current in stage 2?
Whether a ferrite cire is suitable depends partly on frequency. What is the intended application of this circuit? At present it looks like a standard textbook circuit - nothing like what is commonly used in the real world.
Whether a ferrite cire is suitable depends partly on frequency. What is the intended application of this circuit? At present it looks like a standard textbook circuit - nothing like what is commonly used in the real world.
I see you guys are struggling with this.
Now, should I chime in here and explain it all, or do we suspect that this is a student cheating on their homework?
Now, should I chime in here and explain it all, or do we suspect that this is a student cheating on their homework?
OK I get re=26/0.9=28.88r*hfe=160
Q2 input impedance=4,622.22r ohms.
Voltage at R6=(R5/(R4+R5)*9)-0.6V=0.9V Q2 collector current=0.9mA.
Q2 input impedance=4,622.22r ohms.
Voltage at R6=(R5/(R4+R5)*9)-0.6V=0.9V Q2 collector current=0.9mA.
ehm.. why would u say it is a student cheating on homework? since we can have open discussion here and i do not ask for the straight forward answer.i do provide all the calculations and the ideas.
and to show it the following is my calculations (if refer to the figure i shown):
Zin(2nd stage) = rb || (beta)(r'e)
rb= 20k||4k
= 3.33k
Vb= (4k/ 24K)* 9V = 1.5V
Ve = 1.5 - 0.7V
= 0.8V
Ie = 0.8V / 1k
= 8 x 10-4
r'e = 25mV / 8 E10-4
= 31.25ohm
Thus, Zin(2ns stage) = 3.33k || (160)(31.25)
= 2k ohm
and to show it the following is my calculations (if refer to the figure i shown):
Zin(2nd stage) = rb || (beta)(r'e)
rb= 20k||4k
= 3.33k
Vb= (4k/ 24K)* 9V = 1.5V
Ve = 1.5 - 0.7V
= 0.8V
Ie = 0.8V / 1k
= 8 x 10-4
r'e = 25mV / 8 E10-4
= 31.25ohm
Thus, Zin(2ns stage) = 3.33k || (160)(31.25)
= 2k ohm
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FYI, before that i calculate the Zin of 2nd stage is 1.06kohm is because i take the beta value inside the figure which is 50. For the latest calculation i take the beta value as 160 (which is my 2N3904 beta value)
I have been careful not to provide motbuddy with the answer, but to guide him towards finding the answer himself. I believe this is what he wants. You will note that I have asked a lot of questions. If he was simply looking for an answer he would probably have given up by now.
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