Why is the Schottky diode being used in control circuit?

I would need to know more about the function of the circuit to make an informed comment.

The two transistors are high voltage parts (250V and 500V).

What are the inductor connections and what is the working voltage on emitter of T104 ?

D108 is only rated for 40V reverse voltage.

I suspect that the diode is to protect T103 from negative voltages at TP131 - but without more information its only a guess. 

MK

The two transistors are high voltage parts (250V and 500V)- fuse & inductor path is max60V from Lead-acid battery. This a random selection. 100V part cab used.

What are the inductor connections and what is the working voltage on emitter of T104 ? -  ferrite bead used for High Impedance at high frequency (EMI/EMC) 

D108 is only rated for 40V reverse voltage. - D109, after 100k and 300k, 60V will drop...so they have used 40V schottky

I suspect that the diode is to protect T103 from negative voltages at TP131 - do we get -ve voltage in between?not sure, I thought it's only at product out (Connector/GND...).

I think I can see what the ciruit is attempting:

T103 switches T104 on and T104 connects R108 to the supply rail.

R108 and R119 attenuate the signal by a factor of  (180 + 10.2)/10.2.

C111 makes a low pass filter with the source resistance of R108 and R109 in parallel (1/(1/R108 + 1/R109)) about 9.6k.

The -3dB point of the filter will be at 36Hz.

The two diodes are there to limit the maximum voltage range on the output to -0.6V -> +2.4V (roughly).

The whole circuit is a bit odd - it could be much simpler:

One low voltage transistor - the logic sense of the input is different. You could use a second low voltage transistor to fix this it you wanted.

There is no need for any of the diodes unless it is expected that the power supply may go very wildly out of range.

If you apply 60V on the power supply the output will be about 3.2V (the diodes in the original would have limited this to 2.4V)

MK

Yeah, sounds like you got it.  If the frequency is only 60Hz  I think a larger cap than 470nF should be used to lower the corner frequency below 36Hz.

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 

geralds said:

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).

geralds said:

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.

geralds said:

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.

Thank for the efforts and discussions️.