Why is the Schottky diode being used in control circuit? and why is the 470nF cap is added at that voltage divider output? how is the Cap value 470nF calculated?
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Why is the Schottky diode being used in control circuit? and why is the 470nF cap is added at that voltage divider output? how is the Cap value 470nF calculated?
Yes... but maybe use the 9.6k parallel equivalent like michaelkellett
Hi,
I would say that the T103 is a simple switching transistor in a higher voltage-rated circuit. A BC846 would be too small at 60V.
The microcontroller's output switches it on to measure the voltage at the output of the "voltage sense" module.
The T103 then switches on the T104, which applies a voltage source to the voltage divider. The two diodes D105 are freewheeling, protection diodes, diodes that eliminate both positive and negative over voltages and dissipate voltage spikes against the 1.8V supply voltage (the voltage of the controller board).
The Schottky diode acts as a voltage barrier; it helps transistor T104 switch off very quickly (T103), much faster, because T103 is somewhat slower.
The capacitor at the output of the voltage source is simply a storage capacitor that temporarily stores the switched-on voltage, approximately 2V according to the RC propagation delay.
Furthermore, it helps to eliminate and smooth out some spikes present on this measuring line.
The choke above is a low-pass filter that suppresses RF interference when power is switched. It also limits inrush current.
The voltage divider R118/R119 has a ratio of approximately 17.6470; therefore, with a 36V power supply, the output voltage is approximately 2V (1.93V depends on the actual conditions).
There is no AC voltage at a battery (normally; however, superimposed voltage can unfortunately occur).
Best Regards
Gerald
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The
The choke above is a low-pass filter that suppresses RF interference when power is switched. It also limits inrush current.
It's an 0603 ferrite bead with an impedance of 120R at 300MHz and a resistance of approx 0.03ohms, it won't do much to limit inrush current. And it wont need to since the only path for inrush current is through R113 or R118 (300k and 180k respectively).
The Schottky diode acts as a voltage barrier; it helps transistor T104 switch off very quickly (T103), much faster, because T103 is somewhat slower.
Could you explain this - and maybe illustrate with simulation because I can't see how a diode of any kind in series with 300k will help T104 switch off quickly.
The two diodes D105 are freewheeling, protection diodes, diodes that eliminate both positive and negative over voltages and dissipate voltage spikes against the 1.8V supply voltage (the voltage of the controller board).
The diodes in the ciruit are not freewheeling diodes according to the generally accepted use of the term.
https://en.wikipedia.org/wiki/Flyback_diode
And the action of the diodes is to clamp the signal at between -0.6V and +2.4V - one clamps to the ground rail and the other to the 1.8V source.
MK
Hi
simulation with Schottky diode:


simulation without Schottky diode:


The Schottky diode is important during switching off the transistors.
Without Schottky it can produce very high voltage spikes during switching off.
Yes, the two diodes on the measure line are clamping diodes. sorry, it's my English is not very well. I translate it all from German to English. I forgot the word clamping.... "..to you know Mr. Alzheimer? - no, I forgot him." In the 70's I learned about them, telecommunication - telephones.
Thanks for the efforts️.
Without Schottky it can produce very high voltage spikes during switching off. - you are refereing T103(npn)?.
as we can see in the 2nd waveform (VC-Q1 - red) there is high voltage spike...in-turn leading Vc-q2 (orange) -ve spike....
But any how this -ve spike will be suppressed by clamping diode right? so can we remove schottky?...
And can you please share the Simulation settings. If you can share the .asc file that would be great.
Hi,


With Schottky - the current through the ferrite bead is minimal few micro ampere.
Without Schottky - the current is much more - it produce a current spike, that produce a spike voltage on the Q1 transistor during switching off.
This diode blocks the back current, which will stored in the induction during switching off, it separates Q1 from Q2.
refer what michaelkellet told - flyback diode
-->>> see also switched power supplies, such circuits looking similar.
copy this .docx file to your folder and rename the extension docx to .asc.
Best Regards
Gerald
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Ok now its clear. Once again thanks for the support
Thank for the efforts and discussions️.
Hello Gerald,
I have replicated your model quite closely in LTSpice and it shows no sign of the huge spikes.

I used a Wurth ferrite bead:

MK
original ferrite bead in the original circuit, Murata BLM18KG101TH1D:

I've selected the Würth ferrite bead 74276051:

Well, may be your simulation have other setting in the software.
I just installed the software and worked with default settings, in the simulation settings.
Just drag and drop, I selected components that have "real" parameters, I did not use ideal components.
In the fused wire draws < 3A at 36V, not mA. This is the power rail.
The measurement circuit is just a small part of a bigger circuit.
You can ask the developers who inserted the Schottky.