Follow along with the video below to see how to install our site as a web app on your home screen.
Note: This feature may not be available in some browsers.
I should've noticed that problem before.bananasiong said:I've changed Q3, it is still getting hot (without any resistor at the emitter)
You might be lucky, this time your transistor might have a slightly lower gain, there again if you turn the heating on or use it on a hot summer's day it will probably blow again.bananasiong said:I haven't tried to add the resistor yet.
Did you calculate the heating when the oscillator is stopped?Hero999 said:I don't think so audioguru.
I disagree. The pulses from the oscillator are from a low impedance emitter follower so have plenty of current to charge the coupling capacitor so that Q3 is cutoff over some of its cycle.Q3 is being operated in class A, it's always conducting some current because pulses from the oscillator stage simply aren't strong enough to turn it fully off at any part of the cycle.
Lots of circuits blow up if their oscillator fails, but I show above that this one just gets hot (not too hot). What about an inverter that makes mains AC from a car battery? If its oscillator stops then hundreds or thousands of DC amps flow.Even if it was operating in class C while the ocillator is running, this is still very dodgy because if the oscillator decides to stop the amplifier won't fail safe, it will just blow up.
Yes 200 is typical but what if it's the maximum which is 300?audioguru said:Did you calculate the heating when the oscillator is stopped?
The 47k base resistor will have about 8.2V across it.
At room temperature the 2N3904 will have a typical gain of 200 so its collector current is 34.9ma and its power dissipation is only 307mW.
So the at the higher temperature gain has increased by 40%, the gain will go up to 420. Now it's passing 73.3mA, giving a power dissipation of 659.5mW causing the transistor to rise to 131.9°C which is above it's maximum rating of 125°C and given you're supposed to derate by 5mW/°C the transistor is toast!audioguru said:If it is at its max temperature then its gain is 280 and its dissipation is only 430mW. Its max rated dissipation is a lot more at 625mW so it is fine.
I doubt it because when I built a simple oscillator like this the output was less than 2Vpp or so and the emitter voltage was even lower, far too low to completely turn the transistor off.I disagree. The pulses from the oscillator are from a low impedance emitter follower so have plenty of current to charge the coupling capacitor so that Q3 is cutoff over some of its cycle.
Firstly the MOSFETs will turn off, unless the oscillator jams on in which case that's what fuses are for!Lots of circuits blow up if their oscillator fails, but I show above that this one just gets hot (not too hot). What about an inverter that makes mains AC from a car battery? If its oscillator stops then hundreds or thousands of DC amps flow.
That is at a collector current of only 10mA. The gain drops at higher current.Hero999 said:Yes 200 is typical but what if it's the maximum which is 300?
It is rated to have a dissipation of 625mW in an ambient temp of 25 degrees C. So its current can be 625mW/8.8V= 71mA then its current gain would need to be 413. I don't think any will have a current gain so high at the high current, even when hot.So the at the higher temperature gain has increased by 40%, the gain will go up to 420. Now it's passing 73.3mA, giving a power dissipation of 659.5mW causing the transistor to rise to 131.9°C which is above it's maximum rating of 125°C and given you're supposed to derate by 5mW/°C the transistor is toast!
My 'scope can't show 100MHz but the base of the output transistor needs a voltage swing of only -0.8V to turn it off. The current through the coupling capacitor needs to be only 191uA (in the 47k base resistor)to turn it off.I doubt it because when I built a simple oscillator like this the output was less than 2Vpp or so and the emitter voltage was even lower, far too low to completely turn the transistor off.
No, most transistors won't but some might, it's the same old story, typical values on a datasheet aren't reliable especially when a transistor's gain is concerned.audioguru said:I don't think any will have a current gain so high at the high current, even when hot.
The voltage swing on the emitter is lower than the collector, so that 2V will easilly become 1Vpp which is only -0.5V - far too low to turn off the transistor.My 'scope can't show 100MHz but the base of the output transistor needs a voltage swing of only -0.8V to turn it off. The current through the coupling capacitor needs to be only 191uA (in the 47k base resistor)to turn it off.
The oscillator transistor operates at an average current of 9mA. So the emitter resistor has an average of 1.98V across it and the emitter's voltage swing must be pretty high.
Yes, i've replace a new transistor. After adding the resistor and capacitor, Q3 is not as hot as previously, still warm. But I still couldn't get any signal at the antenna side.Hero999 said:Did you replace Q3?
Are you getting a signal from Q2?
The 56hm: resistor won't reduce the range much if it's bypassed by 1nF but it will keep Q3 cool and stable under a wider temperature range.
audioguru's concern about your antenna bieng to short is valid because the voltage across the LC circuit will be well above the supply voltage due to resonant voltage magnification if the Q is too high. The emitter resistor should reduce this problem a little but never the less try increasing your antenna's length to about 750mm.
750mm is a quater right?audioguru said:A half-wavelength at 100MHz is 750mm. The antenna on my cell phone is only 15mm long but most of its total length of about 195mm is inside, and it operates at 824MHz to 1990MHz.
The range will be much less with a shorter antenna and the output's tuned circuit will develop a high voltage that might damage the circuit.