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zorgrian
modified 1 year ago

Measure the output impedance of a transistor

11
20
507
07:51:41
This is a very crude method. But it does take into account the working voltage, the Z of the circuit (here about as simple a Class C amplifier that you can get). It also takes account of the working frequency. All of these factors affect the output Z Adjust the resistor until you see half the initial peak voltage reading. The output impedance of your transistor is equal to the ohmic resistance of the load resistor once adjusted to half the no load peak voltage. REMEMBER THIS IS SIMULATION The EC simulator is very good for many circuits. However, it doesn't adjust the reactance of simulation transistors, such that, supply voltage and frequency don't affect the results much. A real transistor will change its reactance with both operating frequency and importantly, with supply voltage. The latter is due to the variable capacitor effect of PN junctions. In simulation of many circuits I have deducted that there is an inbuilt fixed output capacitance of approximately 5pF when using the EC transistor model. IN OTHER WORDS 1) measure the peak output with the switch open. 2) close the switch. Then make the load resistor a value such that the peak voltage falls to 50% (half) of the peak voltage as measured when the switch was open. 3) The output impedance value of the transistor, is the value of the resistor, which when placed, caused the output RF voltage to fall by 50%. Note: Measuring is only safe with this method for circuits where the expected power is less than approximately 1W, for the no-load voltage will rise proportionally EXAMPLE A 1Watt example of a given transistor, at 50 Ohms might result in 10 Volts peak voltage. At 100 Ohms the peak voltage is at 40 Volts At 200 Ohms the peak voltage rises to 80 Volts Note that a BD139 does not support much more than that - see the data sheet. (Mentioned because this transistor is very popular with low power PA stages in QRP HAM radio circles - mainly because its very low cost) So this is important stuff! The longer way round this is to do loads of pointless calculation. This is very good for your brain and if you like that type of thing, great! There's a problem with this calculated method. It may not reflect the device under test at all for many reasons which I'm not going to delve into here. Suffice to say, its better to test the real device! The output impedance will determine the type of coupling to the filter or next stage. Be it a transformer or capacitor or more complex coupling topology. Please note that the choke (inductor) at the collector of the transistor will have to be adjusted for operation at different frequencies. EXAMPLE If you lower the frequency to 1 MHz, the choke must be something in the region of 8uH You can make a self supporting air core coil 20 mm diameter - plastic water pipe 7 mm length 6 turns 1 mm copper wire Or you could just buy one of those chokes that look like resistors. 1001001
published 7 years ago
BillyT
7 years ago
👍
2ctiby
7 years ago
@zorgrian ... in part 2) of your description you mentioned 'peak' voltage twice. Did you mean to say 'peak to peak' voltage instead in both of those?
hurz
7 years ago
Vp before switch closed and Vp with switch closed. Vp_open/Vp_closed=2
zorgrian
7 years ago
@hurz, please use more words. I am not sure i get what you mean to say...
zorgrian
7 years ago
Yes peak to peak voltage, but peak would also work here i think as would RMS (maybe more difficult to measure tho)
zorgrian
7 years ago
Basically, you measure the voltage with the switch open and then again with it closed ( obviously, having adjusted to half the first voltage reading) Actually, your reading is the value of the resistance that will make half the voltage obtained with the switch open. Problems of voltage arise due to the zero volt line being skewed on both loaded an not loaded positions
zorgrian
7 years ago
I use the highest positive peak value, which may not be correct, but seems to work. I ignore the negative swing completely
zorgrian
7 years ago
If someone has a better practical method, i am interested....????
zorgrian
7 years ago
@hurz... OK I get your meaning. However, same applies, your comment is correct but a little terse....
zorgrian
7 years ago
I have a rigol 100 MHz scope in the workshop. Even with X10 setting on the probe, on low power devices, the probe Z affects the readings. Recently, i built my own low Z RF probe. It has a 10 M ohm input impedance across DC to a gig (not that I can measure up there very accurately)
hurz
7 years ago
Whats irritating for me is, when you say - you measure voltage from peak to peak' which normaly means Vpp. EC default Vpp is 2V. But what you mean was clear to me, you measure two times a voltage at peak and compare them. Ok enough to this topic ;-)
zorgrian
7 years ago
OK
hurz
7 years ago
Just put one mega ohm resistor to the probe positive clip. It divides the by 11. In most cases its enough to use a 10kOhm and you even can ignore the error.
hurz
7 years ago
The idea is, I guess, @zorgrian presented here, the transistor pumps energy into to coil while its closed then it opens and the stored energy gets free and walks throug the cap to the resistor were the electrical energy is transformt into heat. If one go and measure the peak power at the resistor (simspeed 30ns/s) its around 300 Ohm a maximum which is around half the peak voltage while this circuit runs open. @simspeed 30ns/s Open voltage about 280V 200 ohm 61.6Wpeak 250 ohm 64Wpeak 300 ohm 65.4Wpeak 140V 0.467A 350 ohm 65.1Wpeak 400 ohm 64.4Wpeak http://everycircuit.com/circuit/4985817394315264 so the drive impedance gets its maximum power out at 300 Ohm which makes also the open output a close to 300Ohm one.
zorgrian
7 years ago
@hurz, thanks for the tip, (pun intended) on the measurement & oscilloscope probe. The probe i built, deals with the stray capacitance and other Z issues. Its an active design, which i suppose i could post. It works well and doesn't interfere perceptibly with the circuit i am measuring. It has an input Z of 10 M ohms. Its very useful for low signal levels such as seeing signals of 20 uV, from an oscillator or other reactive circuit. I suppose this post of mine should be rewritten... Its not well described as I'm not feeling so good with flu.
zorgrian
7 years ago
@2ctiby, yes you can go with your idea of Vpp. However your comment about sinusoidal wave forms is dimensionally unstable! (Makes no sense). The class C design presented doesn't output sinusoidal waves. It outputs a pulse of energy. Also, the reason i say ignore the negative component, is that its negligible. You are right to suppose that at other (non HF 1 - 30 MHz) frequencies, this method will have problems. However, the principal is sound. (Good and useful)
zorgrian
7 years ago
@hurz, Perhaps, for the probe, an oscillator isn't a good example. A better example might be the input to the front end of a radio receiver. However, oscillators of most analog types are very sensitive, such that the measuring probe often affects results. This is particularly true of low current oscillator design, which I prefer as these can be built to be highly stable with either a ceramic resonator or crystal. Cos low current means low heat through the transistor. Nominal 3 - 6 volts voltage. @ 5 - 20 uA
hurz
7 years ago
@zorgrian, get well soon!
zorgrian
7 years ago
Im OK, but now my 5 year old son has it!
zorgrian
7 years ago
He loves EC, its one of his favorite 'games'. That's how he sees it, which is great but he can read, so i don't let him on the EC world wide alternative mindset club just yet

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