The orihginal circuit has the DC voltages penciled in.
An immediate indication of what's wrong can be had by measuring those and let us no.
That should be done before starting to connect a signal generator.
If the DC's are OK, I guarantee it will work as it should.
If not, it should be clear where the issue is.
Jan
An immediate indication of what's wrong can be had by measuring those and let us no.
That should be done before starting to connect a signal generator.
If the DC's are OK, I guarantee it will work as it should.
If not, it should be clear where the issue is.
Jan
ok its finished, and does work but with distortion. 100mV get 4.5Vp2p out.
i made jasons first schematic, only because i have npn's to hand right now.
i made jasons first schematic, only because i have npn's to hand right now.
hi hugo. i think i've damaged my function generator without realizing it at the time in the past. because when i drop it down to low voltage its giving out a 'funky' waveform.
i started trying to trace back the circuit to find where it was coming from and its coming from the generator 🙁 .
i started trying to trace back the circuit to find where it was coming from and its coming from the generator 🙁 .
i guess the circuit probably works very well.
the generator produces a 6v sine wave nicely so maybe i'll just add a resistor , for the time being.
the generator produces a 6v sine wave nicely so maybe i'll just add a resistor , for the time being.
Oh..
You could try with a generated sine wave from a DAW.
I'm curious how much unclipped audio you get out of the circuit.
Hugo
You could try with a generated sine wave from a DAW.
I'm curious how much unclipped audio you get out of the circuit.
Hugo
Digital Audio Workstation.
Audacity is a free one.
You then generate a sine (or other wave) on the computer and play it back via the soundcard, feeding the input of your circuit.
Hugo
Audacity is a free one.
You then generate a sine (or other wave) on the computer and play it back via the soundcard, feeding the input of your circuit.
Hugo
i did google it and saw stuff like that, my computer dosent have a sound card(only onboard sound).
just opened the generator to see if theres anything obviously wrong. its one of those chinese copies of the a feeltech.
thank you hugo, and everyone else for trying to help me out. i'm gonna get my kit sorted and then i'll be back.
warm regards
gazza
just opened the generator to see if theres anything obviously wrong. its one of those chinese copies of the a feeltech.
thank you hugo, and everyone else for trying to help me out. i'm gonna get my kit sorted and then i'll be back.
warm regards
gazza
The circuit as given has a gain of 80dB, or 10,000 times. So an input of 0.1mV will give a 1V output.
That is confirmed by modelling using a 100uF input coupling capacitor. A 100uV RMS input at 1kHz gives an output voltage of 1.05V RMS output, but with a harmonic distortion of 5.8%
That is pretty much the maximum input voltage before (even more) serious distortion and clipping occurs.
That is confirmed by modelling using a 100uF input coupling capacitor. A 100uV RMS input at 1kHz gives an output voltage of 1.05V RMS output, but with a harmonic distortion of 5.8%
That is pretty much the maximum input voltage before (even more) serious distortion and clipping occurs.
If it has a headphone output, it has an (internal) soundcard.
You can use that as source with Audacity or a multitude of free software.
Did you measure the DC points?
Jan
Sounds right. A crude estimate of the ac gain is (18k/1k) * (1800/680)=47 (assuming I did that right).
Also, if the base-emitter junctions have 0.6v drop, then a back-of-the-envelope estimate tells me the dc operating point of the collector of Q2 should be at about 9.9V. This means that at most you can get about 2V peak AC signal out before it hits the rail. So you are probably clipping. Lower the input by a factor of 2 and i would expect the output to be cleaner.
As Jan said, you should measure the dc voltages at each terminal of the two transistors. It will help you understand the circuit and verify it is working right. What do you measure for these voltages? Is the collector of Q2 near 10 volts?
(Edit: if the base-emitter voltage drop is 0.7 volts then it may be close to 8.5 volts. It is so different because the design has a small voltage across the 1k resistor, which isn’t the best design).
Also, it is worthwhile to learn how to calculate what you expect to measure.
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No - the emitter resistors are bypassed, so the individual gains are RC/Re, where Re is the emitter resistance of each transistor. Hence the high gain of 80dB
That is true of the circuit in post 7, but he indicated that he built the first circuit in post 13.