ECE3400 Lecture 28: One-BJT Current Sources & Jensen 990 Op Amp Analysis (Analog Electronics)
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- Опубліковано 8 лип 2024
- Reference: leachlegacy.ece.gatech.edu/ec...
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Jensen 990 Op Amp Datasheet (John Hardy): www.johnhardyco.com/pdf/990.pdf
Jensen 990 JAES paper: www.technicalaudio.com/pdf/Je...
Jensen 990 patent: patents.google.com/patent/US4...
0:00 -- Introduction
0:43 -- Resistor divider biasing
3:06 -- Output impedance
5:08 -- Zener biasing
5:55 -- Regular diode biasing
6:43 -- Current mirrors
7:29 -- Jensen 990 op amp
Just because it's kind of interesting to comment on. CR11 and CR12 are there specifically for over-current protection together with R13 and R14. This will all be familiar to everyone here but just to state it again the current into the base of the output transistors, Q8 and Q9, is controlled by the bias current into the collector Q7 and the input current. If the collector current of Q7 is lower than the input current the voltage at that node rises until the additional current is able to flow into the base of Q8, which has the effect of pulling the output up. If the bias current is greater than the input current then the voltage at that node will drop instead until current is able to flow out of the base of Q9 which pulls the output down.
The diodes offer a third path when the output current gets high enough. The voltage across R13 rises in proportion to the emitter current from Q8. Once the voltage across R13 exceeds two diode drops then instead of flowing exclusively toward the output through the base of Q8 a new path allows current to flow to the output across CR9, CR10, and CR12 instead bypassing Q8 entirely. Something similar happens on the low side. If the collector current of Q9 were high enough for the voltage across R14 to exceed two diode drops then current would flow through CR11, CR9, and CR10 into Q7 bypassing Q9. This has the effect of starving the output transistors of base current when the output current gets too high.
With 3.9 ohm resistors for R13 and R14 and assuming a forward voltage of approximately 1.3V across two diodes the current across Q8 and Q9 would be limited to around 333 mA.
Thanks for looking at the JE-990 circuit! I've been trying to figure out what those latchup diodes were for. Have you considered looking at the Neve BA283 microphone amplifier circuit? He implemented some clever biasing and current sourcing schemes in his design.
I've spent a LOT of time looking at the 1073 and related such things, and never really made much headway. It's terrifyingly complex.
@@Lantertronics Same. I have a vague understanding of some of the circuit. Lol.
it's mentioned in various datasheet of opamps that the contain some sort of circuitry to prevent latch up.
Excited for this one! ✌️🙏
(And I would have been even if that 990 was just for the thumbnail 🤭)
I try to be be somewhat honest with my thumbnails. ;)
CR11 and CR12 is protection diodes
Do you know this part: LM334MX? ... and you can place it on top of a differential amplifier instead of a pnp/npn current mirror, add 2 resistors below to the npn differential pair and it works just fine, probably even better.
Sure, wish i could afford going to your class to learn more about the "variables/names" in da math.
Not familiar with that part -- it looks interesting.
@@Lantertronics i got the MIT amplifier "running", i changed the input, put a new low pas filter and C9 is absolutely necessary (although the documents says you can skip it) to prevent oscillation.
a square looked like it was low pass filtered, chaotic like if resonance was in place and started playing with the values of the caps, but nothing helped. also due to the choosing the right inductor, removing it make the amplitude drop.
at first playing with frequencies in the original state,i noticed a amplitude drop at frequencies above 10K. i played around with the input and presto. now around 20 kHz you have a slight drop.
Falstad simulator, right click A/C source and add slider.
Now it's the trouble of finding a 2 x56 Volt transformer locally.
Um, C2 being only 62pf in parallel with 182 ohms has a corner frequency of 14MHz! You really can’t say the emitter is “fully bypassed” can you? 🤔
Please apply real numbers to calculate, but not algebra formulas. I am completely lost and don't know what you are talking about!!!
You can plug whatever numbers you want into the formulas.
@@Lantertronics Plug and Play is not so easy for a new fresh man. Only engineer can do that. 🤣🤣🤣