Simple Amplifier

[StevenStanleyBayes]

Headphones do not require power. A simple common collector couple (standard boosting of an operational amplifier output) connected to the same power supply as the operational amplifier would do or just the operational amplifier driving 50mA max through the headphones as depends on the power supply and, bad news, the resistance of the headphone speakers. Thus 36V power supply to TL084 would allow TL084 to swing -18 to 16.5V, to keep the symetry: -16.5 to 16.5 which is 16.5*0.707~12V RMS. I do NOT know how big the resistance of the headphones is. In case of 240 Ohms, then you get the maximum of 0.6W. In case higher, you get less power. In case lower, you must lower the supply voltage (or play it risky and rely you don't push the gain too much for the applications you want) and still get less power.

Another thing I do not know is what "Diamond Buffer" is but this may be the same to what I referred to as "common collectors at the output of the IC". I do NOT want to look up right now because I do NOT like terminology.

[StevenStanleyBayes]

Sorry, in case of high resistance you best go to another supply but do you believe the resistance would be much higher than 240? Or much lower? In case of no, do you not believe 0.6W or even 0.1W is good enough for headphones?

I do not object your idea, I just do NOT know how big the resistance of headphones is (believe to be much higher than 4, 8, 12 even 16 Ohm speakers), so I cannot calculate the power and I do NOT know how much power is needed for headphones.

[StevenStanleyBayes]

I do NOT want to go search now, sorry.

[StevenStanleyBayes]

Information on headphones:

HeadWize - Headphone FAQs

[StevenStanleyBayes]

A simple zener couple to limit the output.

[StevenStanleyBayes]

A few sentences added on electrical protection designed by the use of electronics.

[sregor]

A couple of questions about the circuit. As I see it, the first pair of transistors will be "cut off" i.e. no current through them. Then all the current through the resistors rc1 and rc2 will flow through the bases of the second pair. The second pair must operate in "class A" mode, since there has to be enough base current drive to deliver current to the load. So will be hot. Also, as output devices heat up, Hfe goes up and current increases, ie thermal runaway. Still trying to "wrap my mind" about what goes on in the output - how much drive do you need to be able to get output to near the rails with a load. Interesting circuit ...

[StevenStanleyBayes]

A straight protection can be achieved with a couple of zeners just before the feedback. These will protect the speackers but will straight connect the transisrtors. Higher voltage zeners can be put just in case. The leackage is not important neither is their non linearity because they are before the feedback.

Originally Posted by StevenStanleyBayes

A simple zener couple to limit the output.

[StevenStanleyBayes]

Forgot the circuit. Need to recall.

Originally Posted by sregor

A couple of questions about the circuit. As I see it, the first pair of transistors will be "cut off" i.e. no current through them. Then all the current through the resistors rc1 and rc2 will flow through the bases of the second pair. The second pair must operate in "class A" mode, since there has to be enough base current drive to deliver current to the load. So will be hot. Also, as output devices heat up, Hfe goes up and current increases, ie thermal runaway. Still trying to "wrap my mind" about what goes on in the output - how much drive do you need to be able to get output to near the rails with a load. Interesting circuit ...

[StevenStanleyBayes]

What you say looks 100% true. However, the second pair of transistors are in common collector connection. Look at the upper half of the schematic only for simplicity. The left hand side transistor is in commen emitter. The right hand side is in common collector. The inverted voltage of the input of the common emitter would be limited ( " cut off " ). Thus the voltage supplied to the common collector gets limited ( " cut off " ) too. The 100% feedback of the common collector will not allow any high current to go through because the voltage of the output ( emitter) must be maintained to be the voltage of the input ( base ). The transistor would open and close accordingly to maintain this voltage. The end transistor, basically, is a common collector with all the load in the emitter as opposed to having the load in parallel to Re.

Originally Posted by sregor

A couple of questions about the circuit. As I see it, the first pair of transistors will be "cut off" i.e. no current through them. Then all the current through the resistors rc1 and rc2 will flow through the bases of the second pair. The second pair must operate in "class A" mode, since there has to be enough base current drive to deliver current to the load. So will be hot. Also, as output devices heat up, Hfe goes up and current increases, ie thermal runaway. Still trying to "wrap my mind" about what goes on in the output - how much drive do you need to be able to get output to near the rails with a load. Interesting circuit ...