Timeline for How can I solve this simple BJT circuit?
Current License: CC BY-SA 4.0
22 events
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| Aug 3, 2021 at 8:05 | comment | added | jonk | @Dat Because \$I_{_{\text{E}_1}}=I_{_\text{CBO}}\$ in case 2 is a lot less than \$I_{_{\text{E}_1}}=I_{_{\text{C}_2}}\$ in case 4. Shockley equation, again. A lower diode current means a lower diode voltage drop. On the order of about \$60\:\text{mV}\$ change for each factor of 10 in the current. A \$600\:\text{mV}\$ difference implies about one million times difference in current. Say, the difference between milliamps and nanoamps. Which is about right for the difference between case 2 and 4. There's nothing surprising to see here. Just learn the Shockley equation. | |
| Aug 3, 2021 at 7:40 | comment | added | Dat | In case 2 and 4, Q1 is active in both cases but the Vx is different: in case 2, Vx = 5.8V ; in case 4, Vx = 5.2V. Why is that? | |
| Aug 2, 2021 at 9:24 | comment | added | Dat | Let us continue this discussion in chat. | |
| Aug 2, 2021 at 9:23 | comment | added | jonk | @Dat You break things up into pieces, work out the pieces, then assemble them back together again. Divide and conquer, etc. | |
| Aug 2, 2021 at 9:20 | comment | added | Dat | it has only two BJTs and has very complicated interaction between two BJTs. Now let's try adding 3, 4 BJTs in series like this AND circuit and solve @@ | |
| Aug 2, 2021 at 8:52 | history | edited | jonk | CC BY-SA 4.0 |
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| Aug 2, 2021 at 8:47 | history | edited | jonk | CC BY-SA 4.0 |
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| Aug 2, 2021 at 8:45 | comment | added | jonk | @Dat Cartoons added. Perhaps they help. Perhaps I made some mistake. But there they are, anyway. | |
| Aug 2, 2021 at 8:41 | history | edited | jonk | CC BY-SA 4.0 |
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| Aug 2, 2021 at 8:02 | comment | added | jonk | @Dat In cases 2 and 4, you can consider Q1 as "just a diode" from Vcc. In case 2, there's no emitter current to speak of, so the Q1 emitter voltage will be only a few hundred millivolts below Vcc, at most. In case 4, though, Q1's emitter will be a full diode drop below Vcc because Q1 is now pulling down on Q1's emitter. Other than that minor difference, Q1 is still "just a diode" in both cases. | |
| Aug 2, 2021 at 7:54 | comment | added | Dat | I am doing these circuit simulations in CircuitLab.com, changing VA, VB to see what happens. I am struggling with BJTs now, there are FETS too. Thank you for the cartoon thought :) it is great way to learn. | |
| Aug 2, 2021 at 7:49 | comment | added | jonk | @Dat Have you put this circuit into a Spice program to see about verifying my above predictions? (By the way, I completely empathize. Don't get me wrong. I struggled with BJTs for years. You will probably master them in far less time than I spent at it.) As a cartoon for Q1 in case 2, since I grew up on a dairy and filbert nut farm, I can almost "see" the emitter as a cow's nipple that you can "pull down on" to get milk. Some of that milk comes via the base. But most of it from the collector. All you have to do is "pull down on it." Oh, well. It's a thought. | |
| Aug 2, 2021 at 7:36 | comment | added | jonk | @Dat I won't disagree with experimentation and experience. There's no substitution for that. That said, there still remains "theory," too. And theory works pretty darned well. You can deduce theory into specific circumstances when you see them and make predictions. These predictions will comport with theory, applied and observed in th e correct light. The key is knowing how to deduce from theory to circumstance. | |
| Aug 2, 2021 at 7:34 | comment | added | Dat | I don't know exactly, there are so many mechanisms about BJTs and this makes me confused. I feel like we can't predict what will happen, we must do the experiments to have experiences then we can predict other circuit after | |
| Aug 2, 2021 at 7:34 | comment | added | jonk | @Dat In case 2 you can almost imagine that Q1 is "diode-connected" if you just stop worrying about R1. If R1=0 then it is diode-connected and it's just a flurgen diode from Vcc. So of course the emitter is "hot." The fact that there's a resistor in the base path doesn't much alter this fact. The Q1 emitter is hot in case 2. Its emitter can, if asked, source current. Can't you see that fact? | |
| Aug 2, 2021 at 7:23 | comment | added | Dat | At second case, how do you know Q1 is active when Vx is not known? I am so confused with a lot of things can happend in this circuit. | |
| Aug 2, 2021 at 7:23 | comment | added | jonk | @Dat There's nothing to cause an emitter current in Q1. There is Icbo from collector to base. But that's not going to trip over to the emitter, not for any reason. So tell me: from what exact mechanism do you see emitter current in Q1? Who is sinking that current, and how? How does it source in the first place? There's no Q1 emitter current in case 1. | |
| Aug 2, 2021 at 7:02 | comment | added | Dat | A lot more questions: at the first case A = LO, B = LO, how do you know the emitter current of Q1 is essentially zero? I thought ICBO = 50 uA when the emitter is open (to ensure IE =0) , but in this circuit, emitter of Q1 is not open. | |
| Aug 2, 2021 at 1:30 | history | edited | jonk | CC BY-SA 4.0 |
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| Aug 1, 2021 at 22:55 | history | edited | jonk | CC BY-SA 4.0 |
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| Aug 1, 2021 at 22:47 | history | edited | jonk | CC BY-SA 4.0 |
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| Aug 1, 2021 at 22:41 | history | answered | jonk | CC BY-SA 4.0 |