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How this circuit works ?

KrisBlueNZ

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Nov 28, 2011
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Two-transistor relaxation oscillator - description

This circuit is called a "relaxation oscillator".
To understand how the circuit works, you need to consider the voltages on Q1 base and Q2 collector, and the OFF/ON states of Q1 and Q2.
Start with both transistors OFF and C1 discharged. In this state, voltages in the circuit will be:
Q1 base: 0V initially, rising towards ~0.6V as C1 charges through R1
Q1 OFF; Q2 OFF; no current flow in the transistors; LED unlit
Q2 collector: 0V (pulled to 0V by R3).
C1 will start to charge (with the opposite polarity from that marked on the schematic) through R1. As C1 charges, Q1's base voltage rises. After a time that depends mainly on R1 and C1, Q1's base voltage will approach 0.6V and Q1 will start to turn on. As this happens, current flow in Q1's collector circuit into the base of Q2 will start to turn on Q2. As Q2 turns on, Q2's collector voltage rises, so the voltage at the right side of C1 rises too, and C1 couples this rising voltage back onto Q1's base, reinforcing the change from OFF to ON for Q1 and Q2.
Because of this positive feedback, once Q1 starts to turn ON, Q1 and Q2 will very quickly turn on and saturate. Q2 will conduct fully, so the LED will illuminate. In this state, voltages in the circuit are:
Q1 base: 0.7V (receiving current from Q2 collector through C1 and R2)
Q1 saturated; Q2 saturated; LED illuminated
Q2 emitter ~10V (12V supply minus the forward voltage of the LED)
Q2 base ~9.3V (emitter voltage minus ~0.7V)
Q1 and Q2 saturate very quickly, and Q2 pulls its collector voltage up to near its emitter voltage, which is ~10V. R3 limits the current through the Q2 collector-emitter circuit to around 20 mA (10V / 470 ohms). This current flows through the LED. The LED will also receive current directly from Q1, through the base-emitter junction of Q2.
R2 limits the current from Q2 collector through C1 into Q1's base due to the regenerative feedback.
C1 now starts to charge up through R2, in the positive direction (polarity as marked). As C1 charges up, current through C1 and R2 tapers off. After a time that is mostly related to the values of R2 and C1, the current into Q1 base drops to the point where Q1 starts to turn off (un-saturate). This causes Q2 to start to turn off as well, and Q2's collector voltage starts to fall. This change is again propagated through C1 and R2 into Q1's base, causing Q1 to turn off more quickly. This positive feedback works the same way as it did before, but instead of causing Q1 and Q2 to rapidly turn ON, it causes them to rapidly turn OFF.
Q2's collector voltage falls quickly from ~10V to 0V, and C1 has a voltage of ~9.3V across it; this drives Q1's base negative (below 0V) and causes D1 to conduct, clamping the Q1 base voltage at ~-0.7V. If D1 was not present, the base-emitter junction of Q1 would see ~-9.3V and would start to conduct in the reverse direction, due to the inherent zener diode characteristic (typical zener voltage: 7V) of the base-emitter junction of bipolar junction transistors.
If the reverse base current was high enough, Q1 could be damaged (the current is limited somewhat by R2). Even if the reverse base current was not high enough to permanently damage Q1, it might cause misbehaviour. I don't know what happens to a BJT's collector circuit in this state; it's just something you're not supposed to do, unless you deliberately want to use a BJT base-emitter junction as a zener diode - for example as a noise generator.
D1's action causes C1 to discharge quickly (through R3, R2 and D1) to ~0.7V (with polarity as marked). From this point (Q1 base at ~-0.7V), C1 begins to charge with opposite polarity from marked, as at the start of this description, and the cycle repeats.
So the circuit flips quickly between both transistors OFF and both transistors saturated, due to positive feedback through C1 and R2. The time spent in the OFF state (the "relaxation" state) is determined mostly by R1 and C1 (C1 charging in the reverse direction up to ~0.7V to start to turn Q1 on) and the time in the ON state is determined mostly by R2 and C1 and is much shorter than the OFF time.
In case you're interested, I've attached the schematic for a white LED torch that uses a similar circuit with an inductor replacing R3 to drive a white LED from a single 1.5V cell. In this circuit, the PNP and NPN are exchanged, but the principle of operation is very similar. The diode isn't needed because the voltages are low enough that they don't cause conduction of the base-emitter zener.
 

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dssteven

May 9, 2012
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How can I become a very good circuit analyzer ?

Practice :) you can use tools like falstad.com/circuit to see current flow in circuits too. Then you can try to understand each component and what it's purpose is in each circuit.
 

Rleo6965

Jan 22, 2012
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Your lucky that simulation software and datasheet readily available in internet today. Before we have to memorize every component and circuit and simulate the circuit in our memory. Sometimes our brain also hang up or temporarily blackout if circuit was too complicated.:D
 

vick5821

Jan 22, 2012
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Hahaha..yea..have to practice and appreciate what I have..Have to understand in detail what each components do :)
 

KrisBlueNZ

Sadly passed away in 2015
Nov 28, 2011
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You're welcome. I hope you followed it and found it interesting.
Different types of circuits can be explained in different ways. In this case, the circuit is fairly straightforward and switches between two clear states, so I described the states and what causes the state changes, and went into some detail on the behaviours of bipolar junction transistors that allow the circuit to work, since that was the most useful way to explain it, as a complete unit with a step-by-step breakdown.
Not all circuits are best explained that way. Many circuits are made up of standard configurations which can be understood separately first, then their interconnection and interaction can be explained as an overview of the design.
 
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