# How to calculate the value of Zin and Zout???

Discussion in 'General Electronics Discussion' started by samy555, Aug 12, 2012.

1. ### samy555

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May 11, 2010
Can you please, help me in the calculation of input and output impedance of the following circuit: This stage operates as the oscillator in a FM transmitter, and I want to put a matching Pi network between it and the 50 Ω antenna, so I need to calculate its input and output impedances.
Thanks

2. ### duke37

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Jan 9, 2011
Zin consists of 47k in parallel with 1n. Below about 4kHz the resistor will predominate and above this frequency, the capacitor will predominate. In addition there will be loading due to the transistor, since this is swtching, I would not know how to calculate its input impedance.

In a similar way the output impedance will be difficult to calculate. I would point out that if you match the output impedance, then you will stop oscillation and get no output. Transmitters of any reasonable power are not matched (to give maximum output), instead they are loaded to give an output which the final device can manage.

3. ### john monks

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Mar 9, 2012
Duke37 is exactly on the mark.
The output impedance does change with the characteristics of the antenna.
However if you wanted to calculate this type of a circuit that was merly an amplifier a good shortcut is to figure the frequency to be about 100MHz, calculate the reactance of all the components, place place all these values into a TI-83 calculator in their imaginary forms, and calculate the impedances you want.

Inductors and capacitors calculate exactly the same as ordinary resistors except you must include the imaginary component with them. To do this long hand might take you a half an hour or more without any mistakes. With a mistake it might take you all night.

So put the values in imaginary form into your TI-83, then calculate.

But Duke37 is correct, you cannot solve this one this way because you do not have the characteristics of the antenna.

Last edited: Aug 13, 2012

63
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May 11, 2010
5. ### duke37

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Jan 9, 2011
The 27MHz transmitter will suffer the same problem as the earlier circuit. If you load the circuit to get higher power out, then you may stop oscillation.

The pi match should be tuned so you will need adjustable capacitors.

6. ### samy555

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May 11, 2010
Ok I want to walk with you, Look this circuit from the link:
http://www.talkingelectronics.com/projects/Wasp/Wasp-P1.html Here, how the designer maching the second stage with the final stage? Sure he calculates the output and the input impedances between them? Is not it?

thanks alot

7. ### duke37

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Jan 9, 2011
Perhaps he did, but I doubt it.
He has taken an output from Q2 emitter which will be low inpedance and coupled it to Q3 base with a small capacitor. This was probably sized to give sufficient drive to Q3.

8. ### samy555

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May 11, 2010
Ok, you are right duke37, but I need a way to calculate the output impedance of a transistor has a tuned LC circuit.
Or at least give me a title or a useful link to read from.

thanks alot

9. ### samy555

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May 11, 2010
Is the following true?
1- The 2nd transistor acts as an oscillator, so the 1n cap made it a common base?
2- Since the 1st stage is a voice preamplifier so it supply a low frequency signal to the 2nd, that is the 1st see the 2nd as a common emitter?

thanks

10. ### john monks

693
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Mar 9, 2012
Are you doing an academic exersize or is this something you are trying to build?
If you want a practical circuit you need to replace the 330 ohm resistor with a coil and you will need some kind of an emitter resistor, maybe 50 ohms in parallel with a capacitor.

11. ### samy555

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May 11, 2010
Thank you john monks for response
I do not do anything these days
My goal is to understand these circuits because I hope to become a designer of electronic circuits
I guess that's understanding the circuit more important than implementation
I do not want to be merely an imitator, I am building the circuit and I do not understand any of it

Do you want to help me and answer my questions?

12. ### john monks

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Mar 9, 2012
What you started out with was very complicated because you had an andenna of unknown characteristics connected to the output of an oscillator. Loading affects the characteristics of an oscillator. Now you have a much easier circuit to analyze I will be very happy to analysize it. But I need to understand the characteristics of the antenna. Maybe if you simply replaced it with a resistor we could come up with a first approximation of the circuit. Maybe 300 ohms or so.
But analyzing these things is exactly as I said in my first post. You can treat every reactance as a resistor but in complex form and then you can simply inter that into a TI-83 calculator and punch out your number . I can show you how to do that.

13. ### samy555

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May 11, 2010
consider a 50 ohm antenna

14. ### john monks

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Mar 9, 2012
Code:

What do we know about the circuit.
We the collector we'll call Rc
We have a base resistor we'll call Rb
Now there are two ways we can look at the circuit. One way is to say that the oscillator only drives the output transistor in it's linear state or class A operation. The other way is to say that the output transistor is being driven to it's extremes, on and off or class C operation.
For simplicity let's consider class A operation first.
So now we have a 3 volt supply and about 0.7 volts at the base or about 2.3 volts across Rb.
Therefore we know that the current going through Rb is 2.3 volts divided by 150Kohms giving about 15.3uA of current.
One helpful piece of information that has been discovered experimentally is that the impedance of a silicon junction is 25millivolts divided by the forward current. Therefore we calculate the impedance to be 0.025/.0000153amps or about 1.63Kohms.
The output impedance of an NPN transistor is usually very high so we will only consider the collector resistor Rc or 330 ohms. Normally we would consider the collector capacitance. From the data sheet we can conclude that the capacitance between the collector and the base is about 10pF. This capacitance is amplified by the transistor and goes up by a value of beta or in this case to about 1nF. At 90MHz this computes to about 1768ohms capacitive. This is in parallel with the 330 ohms.
This causes the output impedance to be closer to about 325 ohm slightly capacitive.
This also causes a slight decrease in input impedance also to about maybe 1.5Kohms slightly capacitive.

Now for the second case. Suppose we have a class C amplifier.
For impedance purposes we can look at the output voltage without the antenna and then with an antenna that has 0 ohms of impedance and do a thevinin equivalent to find out the output impedance. The antenna must be looked at at a 50 ohm resistor in series with a very large capacitor of 0 ohms.
So once you put the antenna on the output you will notice a big decrease in output voltage.
The input impedance depends on how much the oscillator drives this point. With a low drive the impedance will be about 1.5Kohm to close to 75Kohms. This is because the input of a PN junction looks close to a short when it is forward biased and like an open when it is reversed biased.

Now how much signal is coming from the oscillator?
I will dig into this more later after I sleep but in general I can tell you the limiting factor is usually when some goes out of a linear region. Sometimes that is the base emitter junction. Sometimes this is when the transistor saturated or completely turn off during part of it's cycle.

This oscillator is complicated because it is taking advantage of phase-shift in the capacitors. The numbers are complex and can be very confusing.

Last edited: Aug 21, 2012
15. ### samy555

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May 11, 2010
There is no RC, there inductor and capacitor
You are talking about the circuit other than what I mentioned
I meant the following circuit: This is not true, you calculated base current = .0000153amps, and the rule states to divide 0.025 or 0.026 by the collector current, not base current!

The first time I hear such talk, especially: "This capacitance is amplified by the transistor and goes up by a value of beta or in this case to about 1nF" are you sure??????

To your knowledge, I do not like a loose speak
We are discussing a scientific issue The figures must be accurate and the mathematical expressions is master of the situation.

No it is not a class C amplifier, a class C amplifier hasn't any base bias.

Generally, I thank you for trying.

16. ### john monks

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Mar 9, 2012
Sorry, I was using your third schematic that has a 330 ohm resistor as a pull-up resistor.
I can go through this again but I would dare say that this oscillator will not oscillate with a 50ohm antenna. Maybe you could put in a power amplifier and I will be happy to recalculate.

17. ### duke37

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Jan 9, 2011
A class C amplifier can have bias, either positive or negative. The drive has to be high enough to cut off the device over part of the cycle. Normally no bias is used as this is the simplest.

An oscillator is an amplifier with the output fed back to the input. The loop gain must be more than one for it to oscillate and the amplitude will rise from rest. At some stage, the amplitude will be limited either by a control circuit or by the supply voltage. If a load is placed on the oscillator, the regulating circiut will activate the limiting less and the output voltage will drop little so the output impedance looks low. If the load is increased so that more power is taken out than the amplifier puts in, then the oscillator will abruptly stop.

You could look up Miller effect, neutralisation and unilaterisation for the effect of transistor capacitance.

18. ### samy555

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May 11, 2010
OK
All books of electronics when they talk about oscillators, they just give you how to calculate the frequency, Do not talking about the analysis of the transistor Model,etc.
I do not know if my thinking is of the wrong foundation or not??!!
You speak as if it is not allowed to connect anything to oscillator Lest of loading

Ok,,, I'm going to change my question
The following circuit, from:
http://www.electronics-diy.com/tx500.php Q1 is a VCO, followed by Q2 and Q3 which are RF amplifier,between Q1 & Q2 there is a matching network (C12, C13, L2 and L3)
Do you think that the owner of the circuit designed this network/filter without known output impedance of Q1 stage and input impedance of Q2 stage???
If there is a way to design this filter without knowing Zin & Zout, please tell me?

Sorry this is so lengthy and thank you

19. ### duke37

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772
Jan 9, 2011
Q1 is a voltage controlled oscillator, the frequency is varied by the varying voltage altering the transistor capacitance.
The oscillator is very lightly loaded by the 2.2pF capacitance, so looking into the oscillator, it will have a high impedance.
This is stepped down by C12, C13, L2 and L3 to provide a low impedance to drive Q2. Note that C12 and C13 are variable to tune and load the circuit for maximum. If the designer knew all the parameters accurately one or both of these adjustments would not be necessary.
There is no matching between Q2 and Q3, they are just capacity coupled.
Q3 output has a similar circuit to Q2 input but the other way round. Once again, the values are set by experiment, not calculation.

A good book is "Solid state design for the radio amateur" by Hayward and DeMaw, ARRL 77-730-93. This is dated 1986, there are probably newer books which deal with newer devices.  