Impedance Matching

Discussion in 'Wireless & RF Design' started by PRS, Apr 12, 2010.

  1. PRS

    Thread Starter Well-Known Member

    Aug 24, 2008
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    I designed and built the attached circuit, and tested it. The experimental parameters are close to the calculated parameters and so I'm satisfied I know what I'm doing to that extent. However the input and output impedances surprised me.

    The input impedance, if this were low frequency, should be about 2.3kohms. But it is 560 ohms as measured with a series resistor at the input. Evidently the reactance of the capacitance seen at the input is producing this via a virtual RC filter.

    The output resistance is what really surprises me. Where is this 5.6 kohm resistance coming from?

    Last, I want to match the input resistance to 50 ohms via a transformer. Will a simple 1:3.3 step up transformer do this? Will the input be purely resistive?

    Same with the output. I estimate a 10:1 transformer there to do the trick. But will this 50 ohms output be purely resistive?

    Any help with this will be greatly appreciated. I'm still in the learning mode. Thanks!
     
  2. rjenkins

    AAC Fanatic!

    Nov 6, 2005
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    The input is basically feeding into a forward biassed diode to ground, due to the capacitor on the emitter.

    For a very crude guesstimation of the output impedance, I'd look at the DC bias current through the transistor and think what resistance would give the same current.

    It looks like roughly 1mA, so allowing for the emitter resistor another 10K.

    That would make the output either 10K or 2k5 depending if you are counting the whole primary or the lower half as the 2:1 ratio.

    It may also depend on the drive level, at higher levels I'd expect the apparent output impedance to go down (but again that is guesswork).
     
  3. PRS

    Thread Starter Well-Known Member

    Aug 24, 2008
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    First, I thank you for your reply, rjenkins. It seems 50 people have looked at this and decided not to reply. I appreciate the input.

    My thoughts about the input resistance are as follows. At low frequency the input resistance would be R1||R2|Rin, where Rin is the input resistance of the transistor which is just re times beta which is about 2500 ohms [(Vt/Ie)*100]. So it should be about 2.3 kohms. However, using the technique of puting a resistor in series with the input and experimentally finding the resistor that drops the voltage to 1/2 of the value without it shows a resistance of about 560 ohms. So why is this lower than the predicted Rin? At high frequencies the input capacitance plays a major role and the has to be included. This would create a circuit like figure 3, I think, which is a low pass filter. I could be wrong I'll admit it right now. But this would have the effect of reducing the input impedance. Guessing 2.3kohm is pure resistance, the rest of it must be due to the capacitance. There are impedance matching circuits but to use them you need to know the reactance and the real resistance. This is what I am looking for. I don't know how to experimentally determing the one from the other. Do you?

    It is 1mA.

    This output impedance is what really confuses me.

    As for the output impedance, I've had a crazy thought. What if the secondary winding is forming a tank with the stray capacitance at the output? This winding has an inductance (experimentally determined) of 3.7uH. And Cs is approximately 30 pF, which combine for a resonant frequency of 15 MHz. See figures 1 and 2.

    Now, since 5Mhz is on the left side of the center frequency (15Mhz), it is capacitive reactance (not inductive) and this reactance is large. But I still don't know how to seperate the capacitive reactance from the real resistance which is reflected to the secondary from the primary. And you need to know this (I think) in order to match the output to 50 ohms purely resistive. Any thoughts?
     
  4. rjenkins

    AAC Fanatic!

    Nov 6, 2005
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    For the input side, you have to allow for the forward biassed base-emitter diode and the emitter capacitor.

    Diodes are used for RF switching and a forward biassed diode is effectively a fairly low impedance connection.

    Look at the circuit part way down this page:
    http://www.radio-electronics.com/info/circuits/diode-rf-attenuator/pin-diode-switch.php

    The choke position in that is where the emitter cap and resistor are, so effectively you have a diode switch with the output shorted to ground.
    (PIN diodes have low 'off' capacitance but any silicon junction works basically the same).


    Can you clarify the turns on the output transformer?
     
  5. PRS

    Thread Starter Well-Known Member

    Aug 24, 2008
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    Thanks for the input, rjenkins, I think I know what you mean by the diode model from base to emitter. Looking into the emitter that resistance is Vt/Ie where Vt is a constant of about 25mV for silicon and in this case the Ie is 1 mA for 25ohms of resistance looking into the emitter. This reflects by beta to the base and beta is about 100 so we have 2500 ohms input resistance at the base. This is in parallel with R1 and R2 and the overall Rin ends up being about 2.3kohms. But the input capacitance parallels this resistance, reducing it to 560 ohms. I figured from this experimental figure that I could make the input match 50 ohms by using a transformer and taking the square root of 560/50 which is approximately 1:3.3 turns ratio. I built an input xformer on a T68-2 toroid core and, yes, 10 turns primary with 33 turns secondary created the desired 50 ohm match at the input.

    The experimental output resistance is 5600 ohms and I cannot explain why this is so from math. I am hoping someone reading this can. But knowing this resistance by experiment, I believe I can match the output to 50 ohms using a step down transformer and a 10:1 winding ratio. This comes from the square root of 5600/50 = 112 and its square root is about 10. So 10:1 turns which I hope to get from a T68-10 toroid core with 100 windings on the secondary and 10 windings on the primary.
     
  6. rjenkins

    AAC Fanatic!

    Nov 6, 2005
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    You can only multiply by the beta to give the base input resistance if the emitter has a resistive load, as it relies on the emitter voltage rising with the base voltage.

    You have an effective RF short on the emitter, the emitter voltage cannot follow the base so you get the diode effect without the gain multiplication (or rather you are multiplying something near zero).

    Adding a low value resistor (10R) between the emitter and existing RC load would drastically increase the base impedance.

    You've still not made clear what the turns on the existing output transformer are?

    If it's a hand wound part, can't you reduce the turns on the secondary to improve the matching?
     
  7. PRS

    Thread Starter Well-Known Member

    Aug 24, 2008
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    35
    When totally bypassed by the emitter capacitor, the input has a resistance of R(pi) = (beta + 1)re where re is the diode resistance between base and emitter. It is given by Vt/Ie where Vt is about 25 mV at Ie = 1mA and the ambient temperature is something like 25 degrees Celsius. This gives an input resistance of about 2.6 k. This is in parallel with R1||R2 and results in Rin = about 2.3 k. See figure 1.

    In the case of a partially bypassed emitter resistance (see figure 2) get Rin = R1||R2||(beta+1)(re + Re1)

    I experimented with a JFET source follower with a 1 Mohm input resistance at 1 kHz. As I increased the frequency Zin was reduced. See figure 3. This is, I think, a matter of the input capacitance being in parallel with the input resistance. The Xc of the resistor goes down as frequency goes up, hence the falling off of Zin with frequency.

    Yes, it is possible to reduce Zout by removing secondary turns. The transformer would be about 20:1 in that case. But this is where it gets complicated. I have in stock Amidon T68-2 toroid cores and T68-10 cores. I wound the original tank using a T68-2 with AL=57. This gave an inductance such that a resonablely sized capacitor in parallel with it would result in 5 MHz. The primary has 36 turns on it. For 20:1 there is a problem with the number of windings on the secondary. The coefficient of coupling is impossible to know and the winding would have to be done experimentally. There is no room for error. All the physical dimensions of the windings would have to be exact and unchangable by a careless movement of the turns with respect to one another. This problem might be corrected by a T68-10 core which would need more windings on the primary for an appropriate inductance and hence more windings on the secondary for the necessary 20:1 step down ratio.
     
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