I'm trying to build a BJT amp. I have this so far, but I can't figure out why there is a voltage drop in the Darlington stage. The speaker should be seeing the signal with an amplitude of about 2.4V but instead it sees a signal with around 200mV amplitude with a 500mV DC offset. Any ideas?
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add a R from each output/driver Q11,12 emitters to the negative rail as pull-down/current sinks
Q12 pull-down resistor has to provide all of the current to the load at negative peaks, size the Q11 pull down ~ 20x higher in value
output coupling cap between Q12 emitter and 8 Ohm load R is advisable if it is a speaker - they really don't like any DC
Q12 pull-down resistor has to provide all of the current to the load at negative peaks, size the Q11 pull down ~ 20x higher in value
output coupling cap between Q12 emitter and 8 Ohm load R is advisable if it is a speaker - they really don't like any DC
some thoughts ...
your dc offset is coming from the fact that you have about ~6.6V at the emitter of Q12. So 6.6V across 108 ohms is about 61mA idle current. put 61mA through 8 ohms and you get 488mV ... so there is the 500mV you specified.
check your ac swing at the collector of Q10 both with and without the darlington connected. you should see approx a 4V swing with the 50mV input shown ... Vout ~ ((RC/RE) * Vin). If you get 4V without the darlington and significantly less with the darlington, then you know the 1.2k source impedance of the first stage is too high to drive the output.
If you can't get a 4V swing at the collector of Q10 either way, check and make sure you've got a full 50mV at the base of Q10 ... the ~1500 ohm input impedance may be excessively loading the source which is driving the circuit.
if all else fails try subbing in different transistors ... you may just have a flakey and/or low hfe one in there.
edit: also are you measuring output from ground or across the 8 ohm resistor only? if only the 8 ohm then you've got the rest of your 2.4 volts (~2.375V) output across the 100 ohm resistor.
your dc offset is coming from the fact that you have about ~6.6V at the emitter of Q12. So 6.6V across 108 ohms is about 61mA idle current. put 61mA through 8 ohms and you get 488mV ... so there is the 500mV you specified.
check your ac swing at the collector of Q10 both with and without the darlington connected. you should see approx a 4V swing with the 50mV input shown ... Vout ~ ((RC/RE) * Vin). If you get 4V without the darlington and significantly less with the darlington, then you know the 1.2k source impedance of the first stage is too high to drive the output.
If you can't get a 4V swing at the collector of Q10 either way, check and make sure you've got a full 50mV at the base of Q10 ... the ~1500 ohm input impedance may be excessively loading the source which is driving the circuit.
if all else fails try subbing in different transistors ... you may just have a flakey and/or low hfe one in there.
edit: also are you measuring output from ground or across the 8 ohm resistor only? if only the 8 ohm then you've got the rest of your 2.4 volts (~2.375V) output across the 100 ohm resistor.
to get rid of (most of) the dc offset you could increase the idle current of Q10 from ~3.3mA to ~8.8mA, by adjusting R25 to 750k. that would put Q10's collector at ~1.4V instead of ~8V, giving you more room for signal swing while at the same time putting ~0.0V at the emitter of Q12. To pull this off you'd have to tie the bottom of R34 to negative rail, and use the value of R35 + R34 to set the idle current in the darlington. then, capacitively couple Q12's emitter to the speaker with ~470uF, as jcx suggested. this should give you ~0V across the speaker terminals. you could even put a small trimmer connected as a rheostat in series with R24, and use that to 'dial in' the dc offset.
since the AC gain of the first stage remains unchanged at ~80, a 50mV AC input should still produce a ~4V AC swing at the collector of Q10, and thus approx +/- 2V AC swing at Q12's emitter relative to ground. Now at AC, the 470uF coupling cap will of course disappear (look like a short) so the darlington will essentially see the 8 ohm speaker in parallel with R34 + R35, which will just look like ~8 ohms. so, the input impedance into the darlington will be approx 20k at AC (Zin ~ hfe1 * hfe2 * RE ~ 50 * 50 * 8), varying slightly with frequency and output current, but should remain plenty high for the 1.2k output impedance drive of the input stage.
as a final thought you may want to consider upgrading the outputs! no offense intended my friend, but TIP31, imo ... ewwwwww.
since the AC gain of the first stage remains unchanged at ~80, a 50mV AC input should still produce a ~4V AC swing at the collector of Q10, and thus approx +/- 2V AC swing at Q12's emitter relative to ground. Now at AC, the 470uF coupling cap will of course disappear (look like a short) so the darlington will essentially see the 8 ohm speaker in parallel with R34 + R35, which will just look like ~8 ohms. so, the input impedance into the darlington will be approx 20k at AC (Zin ~ hfe1 * hfe2 * RE ~ 50 * 50 * 8), varying slightly with frequency and output current, but should remain plenty high for the 1.2k output impedance drive of the input stage.
as a final thought you may want to consider upgrading the outputs! no offense intended my friend, but TIP31, imo ... ewwwwww.
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