falstad constant current source with transistor - Page 1
Forum section: electronics / beginners

author: wiresaho
author_rank: Contributor
timestamp: June 02, 2024, 03:57:00 am
content: https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxEKSshAQFMBaMMAKAHcQ88aGAWPTtyhR2IYigF8BXHvxEdihHoV5il4QvM5hN2FNs0MVWvHo2czZyKJnnT+nSIAuYiSCmuBjmhCbRMYGiBBBg6CGRIMITMhBhKGMS4CNjYCORgcCAAJnQAZgCGAK4ANk4sAE6CNDS23uAVnIp2QnUZ1gBuYoF2TXU0vNXQAtVQ0AgsnYo+msS8qn0gA7TCg+PYGDWORqpT7sYQ1Q14jmb2IFYgaHAsQAWorks from 1 ohm -> 2k ohm, constant current delivery.Theres 2 powers there, the one in front of the transistor is the resistor sensor, and the one behind is the activating power, the resistor sensor opens the transistor itself by initially shorting it 100% on, so I'm pretty sure itll work in real life.

author: wiresaho
author_rank: Contributor
timestamp: June 02, 2024, 04:59:20 am
content: https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxEKSshAQFMBaMMAKAHcQ88aGAWPTtyhR2IYigF8BXHvxEdihHoV5il4QvM5hN2FNs0MVWvHo2czZyKJnnT+nSIAuYiSCmuBjmhD7QyaAgoCGDY3AQo2FDQhMyEGEoY4rGQxKnINAAmdABmAIYArgA2TiwAToI0NLbe4OWcinZCtWBwLABuYmAOmniNtTS8VdACVdEIHWo+msS8qgMgQ7TCwxPYGNWORqqKyvMrLGAYqvYUJ47GELhw0YggAEp0AM4Alk9OeQB2AMZ0No5CPCORrWDhAma9OQgmxQyECKww+H6U4IsGOOTgkCogyLaRmOSgiz6DFmYzWIAHere I added a potentiometre to it.

author: golden_labels
author_rank: Super Contributor
timestamp: June 02, 2024, 06:19:57 am
content: Hello. You forgot to ask the question.Note that the circuit is a standardZener diode current source, but replacing the Zener voltage source with a battery.While it may theoretically work IRL, in practice it will not beyond precisely fine-tuned lab experiment. To start with, it relies on a precisely 0.8 V battery with a truly flat discharge curve.

author: Ian.M
author_rank: Super Contributor
timestamp: June 02, 2024, 07:06:28 am
content: Its also not thermally stable, as both Vbe and hFEvary with temperature. To minimise these effects use a bias voltage >> Vbe (which unfortunately increases the min. voltage drop across the current sink) and a high gain transistor, preferably a Szlaki pair.

author: golden_labels
author_rank: Super Contributor
timestamp: June 02, 2024, 07:23:20 am
content: A picture to help with seeing the equivalence:Upper-left is the original, skipping the irrelevant 1 Ω resistor and rotated.Upper-right swaps VBand RL, to match the typical way current sources are being drawn.The lower part is not exact equivalent, but functionally does the same. VAprimarily powers the loop through the B-E junction and RR. In the original circuit it’s also in the loop with the load (RL), but the influence is relatively minor and unwanted. So, focusing only on the notable and wanted function, it can be seen as providing current across B-E junction only. To make it numerically follow the original, one may increase voltage on VAby what disappered with our VBmove.In the end: VAand RRform an emmiter-follower. With approximately constant voltage on RRyou get constant current. The current is independent of the load RL, because the transistor will regulate it to maintain constant voltage on RR. If the load resistance is lower, it would increase RRvoltage, but the transistor counters that and limits current to the desired value. If the load resistance is higher, it would limit RRvoltage, but the transitor reacts by opening more.

author: wiresaho
author_rank: Contributor
timestamp: June 02, 2024, 08:53:22 am
content: That's very nice of you to do that for me. I like the schematics.So one of the resistors isn't required on it!I think I could use this in real life fine, its two power sources to run it, but as long as u can run any amount of them off the two its not that bad, still practical.But I think its easier just to replace the whole thing with just a battery, then just sit under the ohms of its amp limit, and just short it all. Probably makes the battery go dead quickly tho, but I guess there could be a battery you could design which lasts longer with a lower current limit. (Maybe a dry cell.)And thats how a battery by itself, can act as if its a transistor in this case.

author: golden_labels
author_rank: Super Contributor
timestamp: June 02, 2024, 09:44:21 am
content: Quote from: wiresaho on June 02, 2024, 08:53:22 amThe simplified version is not absolutely identical to your circuit, as noted in last step’s description. But it’s functionally the same. If I included that 1 Ω resistor, it would be at the base of the transistor. Given microscopic current at the base, its effect is irrelevant.Quote from: wiresaho on June 02, 2024, 08:53:22 amIn the original schematic both power sources don’t share a common node. The 0.8 V would need to be floating, be very precise, not change with time, and all this while being in the current path of the load. A floating DC source is easiest to do with a battery, but it fails all the other requirements. And you need to replace it regularly.Please remember, that a circuit diagram represents asimplified, mathematical model. Not real components. It’s your task to make the model represent reality well enough. Mathematical equations may do many things, but it doesn’t mean it is reflected in physical circuits. The easiest test you can perform, when thinking of a battery in a circuit, is to replace the voltage source with a charged capacitor. Does the circuit still make sense?Quote from: wiresaho on June 02, 2024, 08:53:22 amA battery can’t regulate current.