Trolling? This thread was posted twice and in your "100W/* ohm doesn't work" thread you were shown why the amp won't work as drawn.
Those transistors are part of the protection circuit. Almost any small signal device will work there, but the BCs are well suited, why mess with it?
If you are truly trying to learn how amps work, read Nelson Pass' articles, the Leach Amp page and Self on the web.
Those transistors are part of the protection circuit. Almost any small signal device will work there, but the BCs are well suited, why mess with it?
If you are truly trying to learn how amps work, read Nelson Pass' articles, the Leach Amp page and Self on the web.
Re: didnt help
Funny, I read too and found a lot. Judging by your other posts you really don't know what to look for. I hope this short description about the complete circuit will help you.
The input stage is a differential amplifier circuit that amplifies the difference between its positive and negative inputs – thus it offers error correction. In an ideally balanced differential circuit currents through both transistors are equal. The current is “sampled” in a voltage amplifier stage, which is simply a transistor connected as common emitter. Its collector load is constructed of VBE-multiplier circuit (explained later) and a constant current source (explained later). Feedback to input stage is basically taken from it’s “output” (explained later). The input stage is connected to non-inverting configuration (similarly to operational amplifier circuits), which means the circuit’s input impedance is high. Also the gain is set up pretty much like in operational amplifier circuits.
Since the voltage amplifier stage (common emitter) has high output impedance it cannot feed enough current for small loads, therefore it has been buffered with a push-pull class-B amplifier stage (two emitter followers). In your circuit each emitter follower is a “super transistor” constructed of one pre-driver transistor controlling three parallel, higher power transistors. They are connected together in what is known as “Darlington” configuration. Since ideally the output signal from the buffer is pretty much identical to the one at the output of common emitter (only that it has higher current) this is an ideal point from where to take feedback to the input stage.
The buffer stage needs some quiescent current (bias) to keep transistors conducting during zero input signals – otherwise you will have a lot of crossover distortion due to voltage drops caused by base-emitter-junctions). The conduction is achieved with voltage difference between transistor bases. In your case, you use a circuit that is known with many names including bias-servo, VBE-multiplier or rubber zener. (Yours is incorrect in the schematic by the way). In short it (in a case of high beta transistors) generates a voltage drop equal to R1/R2, where R1 = resistance from collector to base and R2 = resistance from base to emitter. You should now understand why this part of your circuit is faulty. The VBE-multiplier also tracks temperature and varies “bias” according to it, which is a nice feature.
The protection stage (area circled) is known as VI-limiter. It samples the output current and if it raises too high the protection transistors open and shunts the driver transistors bases to feedback node. This is all I know about these circuits. Oh yes, and this shunting will cause your speaker load to feedback the current stored to its voice coil back to your amplifier. This will cause a high amplitude voltage peak that will “reverse bias” the BE-junctions in the output and destroy your transistors. You have to prevent this by connecting two diodes from the output nodes to both Vcc and Vee rail. Omit the circuit unless you know how to build it – it really can do damage instead of preventing it.
Why constant current source loads at voltage amplifier and input stages: To quote my own words from another forum: “1. Fluctuations in the supply voltage show up in the collector voltage and 2. Smaller voltage drop over a constant resistance means a smaller collector current Ic. The latter means that as the transistor’s collector voltage increases the collector current must decrease since the voltage difference between the collector and the supply becomes smaller. If the collector current decreases it means that the emitter current has to decrease as well and less emitter current means lower gain. The exact opposite happens when the output voltage decreases and as a result the waveform will become distorted having a flattened top half wave and a stretched bottom half wave.”
In short, current over constant current source is more constant than current over a resistor.
prorms said:
Funny, I read too and found a lot. Judging by your other posts you really don't know what to look for. I hope this short description about the complete circuit will help you.
The input stage is a differential amplifier circuit that amplifies the difference between its positive and negative inputs – thus it offers error correction. In an ideally balanced differential circuit currents through both transistors are equal. The current is “sampled” in a voltage amplifier stage, which is simply a transistor connected as common emitter. Its collector load is constructed of VBE-multiplier circuit (explained later) and a constant current source (explained later). Feedback to input stage is basically taken from it’s “output” (explained later). The input stage is connected to non-inverting configuration (similarly to operational amplifier circuits), which means the circuit’s input impedance is high. Also the gain is set up pretty much like in operational amplifier circuits.
Since the voltage amplifier stage (common emitter) has high output impedance it cannot feed enough current for small loads, therefore it has been buffered with a push-pull class-B amplifier stage (two emitter followers). In your circuit each emitter follower is a “super transistor” constructed of one pre-driver transistor controlling three parallel, higher power transistors. They are connected together in what is known as “Darlington” configuration. Since ideally the output signal from the buffer is pretty much identical to the one at the output of common emitter (only that it has higher current) this is an ideal point from where to take feedback to the input stage.
The buffer stage needs some quiescent current (bias) to keep transistors conducting during zero input signals – otherwise you will have a lot of crossover distortion due to voltage drops caused by base-emitter-junctions). The conduction is achieved with voltage difference between transistor bases. In your case, you use a circuit that is known with many names including bias-servo, VBE-multiplier or rubber zener. (Yours is incorrect in the schematic by the way). In short it (in a case of high beta transistors) generates a voltage drop equal to R1/R2, where R1 = resistance from collector to base and R2 = resistance from base to emitter. You should now understand why this part of your circuit is faulty. The VBE-multiplier also tracks temperature and varies “bias” according to it, which is a nice feature.
The protection stage (area circled) is known as VI-limiter. It samples the output current and if it raises too high the protection transistors open and shunts the driver transistors bases to feedback node. This is all I know about these circuits. Oh yes, and this shunting will cause your speaker load to feedback the current stored to its voice coil back to your amplifier. This will cause a high amplitude voltage peak that will “reverse bias” the BE-junctions in the output and destroy your transistors. You have to prevent this by connecting two diodes from the output nodes to both Vcc and Vee rail. Omit the circuit unless you know how to build it – it really can do damage instead of preventing it.
Why constant current source loads at voltage amplifier and input stages: To quote my own words from another forum: “1. Fluctuations in the supply voltage show up in the collector voltage and 2. Smaller voltage drop over a constant resistance means a smaller collector current Ic. The latter means that as the transistor’s collector voltage increases the collector current must decrease since the voltage difference between the collector and the supply becomes smaller. If the collector current decreases it means that the emitter current has to decrease as well and less emitter current means lower gain. The exact opposite happens when the output voltage decreases and as a result the waveform will become distorted having a flattened top half wave and a stretched bottom half wave.”
In short, current over constant current source is more constant than current over a resistor.
Hi,
to expand on that very comprehensive reply.
The schematic in post1 is a three stage LIN type with long tail pair (LTP) as the differential input stage.
The second stage is a voltage amplifier (VAS) with constant current sink (CCS) followed by an emitter follower (EF) but set up as a darlington pair with triplet output devices.
The amp (when it works) can be set up as a push pull (PP) ClassA or ClassAB or ClassB, depending on the quiescent current (Iq) passed by the output devices.
To provide the thermal compensation offered by the Vbe multiplier, the transistor must be thermally attached to the output device heatsink.
Have you found the Leach paper I referred you to yet?
to expand on that very comprehensive reply.
The schematic in post1 is a three stage LIN type with long tail pair (LTP) as the differential input stage.
The second stage is a voltage amplifier (VAS) with constant current sink (CCS) followed by an emitter follower (EF) but set up as a darlington pair with triplet output devices.
The amp (when it works) can be set up as a push pull (PP) ClassA or ClassAB or ClassB, depending on the quiescent current (Iq) passed by the output devices.
To provide the thermal compensation offered by the Vbe multiplier, the transistor must be thermally attached to the output device heatsink.
Have you found the Leach paper I referred you to yet?
Re: ...
See posts 2,4,5 and 11 for your answer - they all say the same thing: use the BC546/BC556 as shown. Any grade would be fine They are just a few cents at Mouser, why torment yourself over it?I also have a bunch of BC546C and BC556C, of which I would be happy to sell you up to 4,000 of each.
Since you'll probably find many of your other parts at Mouser (once you figure out appropriate values) that likely makes more sense.
prorms said:
See posts 2,4,5 and 11 for your answer - they all say the same thing: use the BC546/BC556 as shown. Any grade would be fine They are just a few cents at Mouser, why torment yourself over it?I also have a bunch of BC546C and BC556C, of which I would be happy to sell you up to 4,000 of each.
Since you'll probably find many of your other parts at Mouser (once you figure out appropriate values) that likely makes more sense.
No,
Bob thought that by buying up the world stock of C grade he could prevent all the rest of us from building any decent front ends for our amps.
His ploy failed when he offered them as part of a group buy for the benefit of our community.
It did stop all the others from improving their front ends!
Bob thought that by buying up the world stock of C grade he could prevent all the rest of us from building any decent front ends for our amps.
His ploy failed when he offered them as part of a group buy for the benefit of our community.
It did stop all the others from improving their front ends!
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