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As I understand it the output impedance approximates to Vcc*Vcc/Po, where Po
is output power and Vcc is the supply voltage It looks like your measurements point to 50 ohms as about right. How were you measuring the voltage? - this might affect the measurement Richard Gary Morton wrote in message ... Over the Xmas break I constructed a simple 2 transistor QRP transmitter using the Tuna Tin II design. Using a power meter and a 50 ohm dummy load it appears to show around 200mW, a little less than the article suggested. Then again I had substituted the final stage 2N2222 transistor for a beefier (size wise) BFY51. I'm not too sure what exact difference this will make. In the article it says that the 200 ohm output impedance of the final stage is transformed down to around 50 ohm (using a 2:1 turns ratio untuned transformer wound on a toroid). I have added a 7th order low pass filter to reduce the harmonics. I am interested in verifying the output impedance. If I understand the theory, when the output is terminated with a resistor which matches the output impedance then the power transferred to the load will be maximised. I set about loading the output will a series of resistors and measuring the peak to peak voltage across them in order to calculate the r.m.s. voltage and hence the power dissapation. Resistors of value less than 10 ohm gave strange results. load r.m.s. power (V*V/R) voltage 10 1.272 0.162 15 1.767 0.208 27 3.005 0.334 33 3.253 0.321 39 3.676 0.346 47 4.066 0.352 56 4.384 0.343 68 4.666 0.320 100 5.303 0.281 The numbers work reasonably well and point towards 50 ohms-ish. The power value looks too high - so I might have made a mis-calculation somewhere! Can anyone pass comments on the above? I'm interested in where the 200 ohm figure comes from. Any suggestions as to other methods of measuring the output impedance? regards... --Gary (M1GRY) |
#2
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On Sat, 3 Jan 2004 21:18:33 +0800, "Richard Hosking"
wrote: As I understand it the output impedance approximates to Vcc*Vcc/Po, where Po is output power and Vcc is the supply voltage I'd be interested to see the derivation of this. It looks like your measurements point to 50 ohms as about right. How were you measuring the voltage? - this might affect the measurement I did the same thing (using different resistor values and then calculating by V^2/R like the OP has done. It worked for me. The Zout of my particular design was 140 ohms, though - just right for a folded dipole. I used 2n3904 transistors; they appear to give a better gain than 2n2222s. -- "I expect history will be kind to me, since I intend to write it." - Winston Churchill |
#3
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Richard Hosking wrote:
As I understand it the output impedance approximates to Vcc*Vcc/Po, where Po is output power and Vcc is the supply voltage It looks like your measurements point to 50 ohms as about right. How were you measuring the voltage? - this might affect the measurement Vcc * Vcc / Po = 13.8 * 13.8 / .352 = 540 ohms (for a 47 ohm load). This value is certainly in the right ballpark. I guess that I should also measure the current taken by the final stage and see if this ties in too. I connected various 1/4w resistors across the output using a couple of adaptors, but being aware to keep wire lengths as short as possible. I then simply used a scope probe and scope to make the measurement. Scope probe was a standard x10 10Mohm impedance and presumably small C. Interestingly I was surprised at the magnitude of the voltages for such a tiny power. regards.. --Gary |
#4
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On Sat, 3 Jan 2004 21:18:33 +0800, "Richard Hosking"
wrote: As I understand it the output impedance approximates to Vcc*Vcc/Po, where Po is output power and Vcc is the supply voltage I'd be interested to see the derivation of this. It looks like your measurements point to 50 ohms as about right. How were you measuring the voltage? - this might affect the measurement I did the same thing (using different resistor values and then calculating by V^2/R like the OP has done. It worked for me. The Zout of my particular design was 140 ohms, though - just right for a folded dipole. I used 2n3904 transistors; they appear to give a better gain than 2n2222s. -- "I expect history will be kind to me, since I intend to write it." - Winston Churchill |
#5
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Richard Hosking wrote:
As I understand it the output impedance approximates to Vcc*Vcc/Po, where Po is output power and Vcc is the supply voltage It looks like your measurements point to 50 ohms as about right. How were you measuring the voltage? - this might affect the measurement Vcc * Vcc / Po = 13.8 * 13.8 / .352 = 540 ohms (for a 47 ohm load). This value is certainly in the right ballpark. I guess that I should also measure the current taken by the final stage and see if this ties in too. I connected various 1/4w resistors across the output using a couple of adaptors, but being aware to keep wire lengths as short as possible. I then simply used a scope probe and scope to make the measurement. Scope probe was a standard x10 10Mohm impedance and presumably small C. Interestingly I was surprised at the magnitude of the voltages for such a tiny power. regards.. --Gary |
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