DIY - Constant current sink - Active Load

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FesZ Electronics

FesZ Electronics

Күн бұрын

Пікірлер: 20
@ibanix2
@ibanix2 2 жыл бұрын
These early videos have a nice rawness not seen in later ones
@FesZElectronics
@FesZElectronics 2 жыл бұрын
I guess in the newer videos I put more effort into scripting them so they seem more well fleshed out. With the earlier ones I just had a general idea of what to do and then turned on the camera.
@q12x
@q12x 4 жыл бұрын
Esti tare ! Please, when you have time, or as a future video project, explain more about transistor capacitance (problems and benefits) as you encountered in your project here with your darlington tr. I'm interested from you to present more as a general information, to learn about it, but is ok to include your case here as an example as well. Thank you, you are the best!
@dominikworkshop6007
@dominikworkshop6007 5 жыл бұрын
Cool !!!
@EletroMakersBrasil
@EletroMakersBrasil 4 жыл бұрын
Very Good !!!
@FesZElectronics
@FesZElectronics 4 жыл бұрын
I'm happy you enjoyed it! Thank you for watching!
@andreimalai2735
@andreimalai2735 6 жыл бұрын
Good job! Congrats! :)
@FesZElectronics
@FesZElectronics 6 жыл бұрын
Thank you! I hope you enjoy my other videos also
@unregistereduser2
@unregistereduser2 2 жыл бұрын
Nice design. Handy, minimal, functional. I built my own version closely following your design. I think, though, that I may replace the LDO with a CR2032 coin cell to power the opamp and reference. Should get a few years runtime, I think, and also might get a little more range for low voltage inputs. I used an old CPU cooler, so I have the option of using the fan too, if I find I want more capacity/less heat.
@t1d100
@t1d100 5 жыл бұрын
I would have liked a better look at the x2/Final PCB board, too, please... Slowly, with various camera angles.
@FesZElectronics
@FesZElectronics 5 жыл бұрын
I'm sorry about that, next time I build something, I will make sure to show the boards in a better angle
@mrsc1914
@mrsc1914 4 жыл бұрын
3d printer changed my life. Rostock max delta
@michaelschwager786
@michaelschwager786 2 жыл бұрын
Where did you get your heat sink calculator from? Thanks.
@FesZElectronics
@FesZElectronics 2 жыл бұрын
I did not calculate the heat sink, by pure calculation, but rather measured it. I dissipated a known amount of power, and measured the temperature achieved as well as the ambient. From this I worked out the thermal resistance. Rth=(t_sink-T_amb)/Pdissipated
@MegaCadr
@MegaCadr 5 жыл бұрын
At 5:07, when you say “added wire inductance” - where do you add those, and how did you figure out the values? Thanks!
@FesZElectronics
@FesZElectronics 5 жыл бұрын
Hello @MegaCadr ! The wires I am referring to are the wires going from the circuit to the power supply. In the case of this simulation I represented that by L1. The value is based on the length and thickness of the wire, you can google a "Wire inductance calculator" to get an idea of what self inductance a certain length of wire has.
@MegaCadr
@MegaCadr 5 жыл бұрын
FesZ Electronics Thanks! I never thought wire had that much of an effect until very high frequencies. Learning so much!
@FesZElectronics
@FesZElectronics 5 жыл бұрын
If you go to high enough frequency all parasitics start to matter. Usually parasitics are ignored just to make examples or theory easier to understand, but in real life you can't really ignore its effects.
@blackarrow8683
@blackarrow8683 4 жыл бұрын
Dear Sir, I think you made a mistake. You calculated everything right for the Op Amp integrator as a low pass. But you connected it wrong in the final schematic. The input for a buffer is connected to the output. So you also need to connect the input of the Op Amp integrator to the output to make a buffer. Therefore the Resistor 2.2k and the capacitor 1nF should be both in parallel connected to the shunt resistor. So you 1nF capacity is wrong connected. It should be connected to the shunt resistor and not the output of the OP Amp. But then again both resistor and capacitor make no sense for this configuration as the input is high impedance. I don't understand why you connected it like this. Maybe just make a cap in parallel to the shunt and finish? As far as I understand you want to reduce the noise over the shunt resistor coupled into the feedback. But I really don't understand why you couple the noise (!) input into the output of the OP amp as it will add up. Therefore a positive feedback loop which is unstable. Maybe I am completely wrong. But please explain, maybe in a video?
@FesZElectronics
@FesZElectronics 4 жыл бұрын
Hello, well let me try to take it step by step - the integrator is built with the op amp, the 1nF and the 2.2k (see fig 11 of - www.ti.com/lit/an/snla140c/snla140c.pdf?ts=1591207495320 as example); the input signal that is to be integrated is the voltage on the shunt, and the output drives the transistor. There is no dedicated buffer, the point of the integrator was to slow down the response of the op-amp to any voltage change on the shunt when fast transients are present, and with DC signals, the capacitor can be ignored and the circuit does work like a buffer. The main difference between this schematic and the one in the AppNote is that here the positive input is not at ground but rather a different potential. You can also check out fig 36-38 from the AppNote on this type of circuit. The logic behind the schematic is that even though the opamp input is high Z, the op amp output is not high Z. So the signal coming into the opamp is filtered in reference to the op amp output not ground or something else. Let me know if this makes sence.
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