I'm trying to get to grips with simple electronics for workshop projects. I came across this design for for a bistable multivibrator:
The accompanying text says:
"Switching between the two states is achieved by applying a single trigger pulse which in turn will cause the “ON” transistor to turn “OFF” and the “OFF” transistor to turn “ON” on the negative half of the trigger pulse. The circuit will switch sequentially by applying a pulse to each base in turn and this is achieved from a single input trigger pulse using a [sic] biased diodes as a steering circuit."
I don't understand what the author means by 'a steering circuit' and no explanation is given.
If anyone can explain more clearly how this circuit works I'd be grateful. The diagram comes from an electronics tutorial which I have found generally useful.
Robin.
PS – I should have said that I know this cat can be skinned other ways, but my enquiry is specifically about understanding this circuit.
PPS – also, I don't understand what the writer means by 'the negative half of the trigger pulse'. It looks pretty negative throughout to me. I'm aware that the falling side of a pulse can be steeper than the rising side, so might get through the 1nF capacitors more easily, but I'm just guessing.
It is not really a conventional bistable as usually drawn that has each state individually triggered by two individual lines. It is configured as a divide by two.
I suggest you redraw it with one transistor 'on' and the trigger input high. Then mark all the lines in red and blue depending on whether they are high or low. You will see one capacitor is +ve on both sides, and the other is +/-. Then take the trigger low and you should see how currents will flow in different directions in each capacitor and hence affect the base current in their associated transistor.
Why do we need an idea of planned use? Robin merely asked how THAT circuit works…
my enquiry is specifically about understanding this circuit.
One of the two transistors will conduct more on power on. Make that TR2. When it conducts fully ( base biased via R1 &2) its collector is low, pulling TR1 base low, so TR1 is off, collector hi.
So, TR2 base – 0.7v, as is the top of D2. TR1 base = 0v.
A negative pulse ( meaning, the pulse edge goes from HI to LOW) passes through the two capacitors connected to D1 and D2 – the pulse does naught via D1 since the top of D1 is LO, since TR1 Base is LO and OFF already
That negative pulse on the bottom of D2 causes D2 to conduct, since its top is HI ( TR2 base = HI, since TR2 is ON).
That neg pulse pulls TR2 base low, now turning it off. Its collector goes 'off' or open circuit and voltage appears on TR1 base via R3 and R4, turning TR1 ON. TR1 ON set its collector LO ( at gnd) pulling TR2 base LO via R2. Now TR2 is OFF and TR1 is ON.
A new pulse swaps the state again – the diodes 'steer' the pulse to turn off the base of the transistor whose base is HI
Many years ago, certain Tech lecturers attempted to teach me the basics of electronics.
Then I transferred to a job where basic electronic design knowledge was a good idea – except that my tech training proved pretty useless!
Finally, one of my colleagues found a book in the local library, copied a part of it, showed it to me (that wasn't the reason for copying by the way), and suddenly I found I could understand what was going on.
The book was Electronic Designer's Handbook, and it had a follow up, Circuit Consultant's Casebook, and the writer was T.K.Hemingway. Both books are old, say 50 or so years, but may still be available via your local library. Highly recommended, indeed I used the knowledge gained from those books to design fairly basic test equipment, and to find design problems in other peoples work.
Joseph – thank you very much for your clear and detailed explanation. I now see what's going on and understand what the author of the article means by the diodes 'steering' the trigger pulse.
Bazyle – thanks for your reply, but Joseph explained it, so no need to dig out felt-tips!
Peter – thanks for your recommendation of Hemingway's books. Electronic Designer's Handbook is available used on Amazon for a mere £125.50 ! Plus postage. I shall try the library.
So far as intended use goes, I confess that I'm not entirely sure – it's more about the journey than the destination at this stage.
Robin
Edited By Robin Graham on 20/02/2023 23:17:37
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