Hmm
First we want to know the current in the first stage.
Suppose the PNP & NPN have identical characteristics.
The bases of the first stage will reach equilibrium @ 0V. Thus the voltage drop across the emitter resistors will be ~14.4V and the current 14.4/3000 = 4.8mA.
Now the emitter-emitter voltage is applied to the series-connected output stages bases. It attempts to drive a current in the output stage of 4.8mA. This creates a voltage drop across the emitter resistors of 4.8/1000 * 2 * 4.7 = 0.045V which reduces the 2 * 0.6 nominal voltage being applied to the bases of the second stage. This naturally reduces the current drive but not so much as to cut off the current.
But don't you only get about 2mA rms current in class A operation?
w
First we want to know the current in the first stage.
Suppose the PNP & NPN have identical characteristics.
The bases of the first stage will reach equilibrium @ 0V. Thus the voltage drop across the emitter resistors will be ~14.4V and the current 14.4/3000 = 4.8mA.
Now the emitter-emitter voltage is applied to the series-connected output stages bases. It attempts to drive a current in the output stage of 4.8mA. This creates a voltage drop across the emitter resistors of 4.8/1000 * 2 * 4.7 = 0.045V which reduces the 2 * 0.6 nominal voltage being applied to the bases of the second stage. This naturally reduces the current drive but not so much as to cut off the current.
But don't you only get about 2mA rms current in class A operation?
w
No need to be naggy or ironic in the answers, we are here to discuss and learn. I don't have Spice, just good old-school formulas, and I'm not an engineer, I'm a physician doctor specialized in biology. That's why I may need help, not sarcasms. Each one his skills. I just need someone to explain me clearly the stuff, may I be taken as a stupid one. Now, I explain my 2 points :
1) I say "tiny fraction of volt" because if we take the well known common collector formula, the emitter constant voltage is (Gm*Re)/((Gm*Re)+1) minus the Vbe of the transistor and as Gm*Re is huge compared to 1 we can assume the voltage drop is only 0,6 since this finite transconductance is not so finite after all. So, tiny it is I think.
2) In the scheme of post 13, the use of BD136-149 serie is not so evident for me : bigger dye means higher Vbe as long as I know and if you look at the datasheet of a BD136, you get a Vbe of 1V instead of 0,6-0,7 for small transistors! It's even harder to turn it on!
So, once again, probably I'm wrong and missing something. But I would like a better explanation than "build it yourself and see the light". I'm willing to learn and please respect my (usually steep) learning curve.
1) I say "tiny fraction of volt" because if we take the well known common collector formula, the emitter constant voltage is (Gm*Re)/((Gm*Re)+1) minus the Vbe of the transistor and as Gm*Re is huge compared to 1 we can assume the voltage drop is only 0,6 since this finite transconductance is not so finite after all. So, tiny it is I think.
2) In the scheme of post 13, the use of BD136-149 serie is not so evident for me : bigger dye means higher Vbe as long as I know and if you look at the datasheet of a BD136, you get a Vbe of 1V instead of 0,6-0,7 for small transistors! It's even harder to turn it on!
So, once again, probably I'm wrong and missing something. But I would like a better explanation than "build it yourself and see the light". I'm willing to learn and please respect my (usually steep) learning curve.
Not ment to be naggy but if you have a PC I do really recommend to try to use LTSpice because you don't have to calculate too much and you can also see what happens if you change things. Use 0.65 V as Vbe and hfe=300 when you do the raw calculations and then start to tweak in the simulator. It's very educational since you can see frequency response and distortion.
BTW: BD136 don't have 1 V as Vbe with reasonable currents.
BTW: BD136 don't have 1 V as Vbe with reasonable currents.
Sorry I got a little carried away in my answer but I recovered.
I guess there is no way I will ever understand this circuit without Spice. But people didn't have Spice before and were doing okay with formulas so I thought it would be enough.
Do you recommend any version of Spice which is light and graphically oriented more than type, type, type? I used to do some Spice simulations for transient behavior of analog filters but I got bored by the text interface.
That's still crazy I would need Spice to see how this 4 transistor circuit behaves...
But I'm still waiting for some old-school explanation of this Vbe paradox. With Spice I would observe the behavior, but understand is better, ain't it?
I guess there is no way I will ever understand this circuit without Spice. But people didn't have Spice before and were doing okay with formulas so I thought it would be enough.
Do you recommend any version of Spice which is light and graphically oriented more than type, type, type? I used to do some Spice simulations for transient behavior of analog filters but I got bored by the text interface.
That's still crazy I would need Spice to see how this 4 transistor circuit behaves...
But I'm still waiting for some old-school explanation of this Vbe paradox. With Spice I would observe the behavior, but understand is better, ain't it?
The transconductance for the bipolar transistor can be expressed as
gm = Ic/Vt
where IC = DC collector current at the Q-point, and VT = thermal voltage, typically about 26 mV at room temperature. For a typical current of 10 mA, gm ¡Ö 385 mS.
Spice is great to learn electronice.. hihgly recommended..
google for a circuitmaker2000 demo, or LTSpice..
gm = Ic/Vt
where IC = DC collector current at the Q-point, and VT = thermal voltage, typically about 26 mV at room temperature. For a typical current of 10 mA, gm ¡Ö 385 mS.
Spice is great to learn electronice.. hihgly recommended..
google for a circuitmaker2000 demo, or LTSpice..
There is no paradox and what you are asking for is basic knowledge which you learn in school. My school book was the brick "Microelectronics" by Jacob Millman and Arvin Grabel McGraw-Hilldarian said:
http://www.amazon.com/Microelectronics-Mcgraw-Electrical-Computer-Engineering/dp/007042330X
I didn't go to electronic classes as I said. I learned with the Horowitz book which is not to shabby as far as I know but he didn't mention that. It tends to be oversimplified and skips a lot of math in that book but I didn't need more at the time.
So it is elementary knowledge, I'm sorry for myself.
Can you still explain it please?
I'm downloading LTspice by the way...
So it is elementary knowledge, I'm sorry for myself.
Can you still explain it please?
I'm downloading LTspice by the way...
I know you don't need formulas most of the time. Good old U=RI and Vbe and Hfe, I know that. That s why I didn't get a version of Spice.
But again, what is your detailed explanation about how this circuit works? Don't be afraid to put me to shame! I am asking for it! How do you get this f@ng bias? I want to know!
And even more in the version of post 13 where the Vbe of the output device is higher! I really want to know! No kidding!
But again, what is your detailed explanation about how this circuit works? Don't be afraid to put me to shame! I am asking for it! How do you get this f@ng bias? I want to know!
And even more in the version of post 13 where the Vbe of the output device is higher! I really want to know! No kidding!
I read it several times and re-reading it makes me see this sentence :
"Their output current levels, along with the use of emitter-stabilization resistors R3-R4, work to indirectly set up the output-stage quiescent current."
And that is the absence of R3 and R4 in the first circuit I posted that I don't get. No resistors equals no bias or a very low one if you play around matching the Vbe of the transistors, which should be not required in a satisfyingly designed circuit.
Again, I may be wrong. Just show me how you see this circuit behavior using words. Please...
"Their output current levels, along with the use of emitter-stabilization resistors R3-R4, work to indirectly set up the output-stage quiescent current."
And that is the absence of R3 and R4 in the first circuit I posted that I don't get. No resistors equals no bias or a very low one if you play around matching the Vbe of the transistors, which should be not required in a satisfyingly designed circuit.
Again, I may be wrong. Just show me how you see this circuit behavior using words. Please...
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