Characterisation of 3. To characterise the experimental samples of 3. This method allows the components to be evaluated at different test conditions in real operation without the need of several megawatt power supplies. Once the samples were characte rised, the applicability analysis of these components on specific applications related to the French ra ilway network SNCF is performed.
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A specific combination of 2 complementary transistors leads to an interesting self-latching behaviour, that can be compared with the behaviour of a thyristor aka SCR or, as the title insinuates, a programmable unijunction transistor, abbreviated to "PUT". Many different circuits can be build with this transistor combination, In this instructable i show you a far from complete collection of circuits that are built around the PUT :.
Of course this collection will never be complete, but the intention is to show you some circuit idea's that might be inspiring or just as learning examples to step into the analog world for a change. Did you use this instructable in your classroom? Add a Teacher Note to share how you incorporated it into your lesson. The configuration has a similar behaviour as a thyristor, but it is NOT a direct replacement for a thyristor!
Lets take a look at the behaviour of this configuration and compare it with the behaviour of a thyristor. What do we know about a thyristor? Well, it has a gate, an anode and a cathode. When a small current is injected into the gate milli-amperes , the thyristor will start to conduct, provided that the anode-to-cathode current is above a certain threshold, called latching current IL. Once the thyristor is conducting, the gate can be disconnected, and the thyristor will still keep on conducting.
The only way to make the thyristor stop conducting, is to make the anode-to-cathode current lower than the minimal holding current. So cutting off the power supply briefly will make the thyristor stop conducting again and return to the initial situation.
The circuit shown in the first figure of this step behaves like a thyristor. Q1 will not conduct, because it's base is pulled high by R1. Q2 will not conduct, because it's base is floating and does not receive any current. Because Q1 does not conduct, the OUTput will be high. The only way to reset the circuit back to the initial state, is to briefly disconnect it from the power supply.
This is the same self-latching behaviour as a thyristor. When Q2 conducts, the base of Q1 is pulled low and Q1 will conduct. The difference between an ordinary unijunction transistor and a programmable unijunction transistor is the fact that the trip voltage, at which the transistor starts conducting, can be programmed by a reference voltage, that is applied to the gate of the PUT.
In the second figure of this step, 2 gate pins are shown because the PUT can be used either with Gate 1 or with Gate 2 as a reference voltage input. When using Gate 1, the PUT will conduct when the anode voltage exceeds the reference voltage on Gate 1 with 0. When using Gate 2, the PUT will conduct when the cathode becomes 0.
In the figure practical circuits are shown illustrating the behaviour of the PUT transistor combination. The left and right circuits have exactly the same function. The right circuit is the "upside down" version of the left one.
The voltage at node 2 can be set with a potmeter. Another example of the behaviour of a PUT is shown in the figure. When the power is connected, the LED will illuminate, getting it's current via R1 from the power supply. So both transistors are not conducting.
From the moment that the switch is pressed, the base of Q1 receives current from the power supply via R3, which limits this current. Q1 starts conducting and pulls down the base of Q2, so current can flow through the base emitter junction of Q2 causing Q2 to conduct.
Q2 will now pull up the base of Q1, so the base of Q1 receives current via Q2. This means that the PUT combination is now latched into conduction, thereby "short-circuiting" the LED, which will extinguish. The following circuits show 2 implementations of a relaxation oscillator using the PUT transistor combination.
The oscillators can be used with a wide range of power supply voltages. The first circuit see figure generates a non-linear sawtooth waveform with a rising ramp.
At nodes 1 and 2, different waveforms are produced. At node 1, a ramping waveform is available and at node 2, a pulsating waveform is available. The slope, and thus frequency of the ramping waveform is controlled by R4 and C1 and depends on the power supply voltage.
Increasing R4 or C1 will decrease the frequency, same as when increasing the power supply The pulse width of the pulses that are generated when the ramp waveform resets is controlled by R3. How it works: Initially C1 is discharged, so node 1 is at 0V. Because Q2 is open, Q1 doesn't get any base current and will not conduct either. The voltage over C1 will increase while the capacitor is being charged by the current flowing from the power supply via R4.
At node 1, the voltage increases exponentially while C1 is charged. As a result Q1 will get base current via Q2 and will also start conducting, thereby pulling the base of Q2 low. Because R3 has a low value, the capacitor is discharged fast via the PUT combination. The voltage a node 2 will no drop to 0V.
When the voltage over C1 has dropped below 0. This process repeats itself over and over. When R1 is changed to f. R3 is added so the pulse width of the pulses at node 2 can be adjusted. When R3 is increased, the pulsewidth of the pulses will increase. The following circuit works in the same way as the first circuit, but generates a non-linear sawtooth waveform with a falling ramp.
The waveforms of this circuit are now related to Vdd instead of ground, because the circuit is "upside-down" compared with the previous circuit. When R2 is changed to f. The following circuit uses a bootstrapping constant current source to charge C1 with a constant current, so we achieve a linear sawtooth waveform at node 1. The bootstrap is formed by Q3, that will conduct more current away from R5 and into the ground when the base emitter voltage exceeds 0.
So when the current through R5 increases such that the voltage over R5 exceeds 0. So it Q3 acts as a kind of overflow valve that draws abundant water into a well at the moment that the water flow is too high. Suppose the current through R5 would increase.
Then the base-emitter voltage of Q3 increases, because the voltage over R5 increases. The result is that Q3 will conduct more, pulling it's emitter "lower" closer to ground by adding extra current through R4. When the voltage over R4 increases, this means the voltage over R5 must decrease, so the current through R5 decreases. This way Q3 keeps the current through R5 constant and C1 will be charged with a constant current, resulting in a linear rising voltage.
As soon as the voltage reaches the threshold voltage at node 2, that is set by R1 and R2, the PUT will latch into conduction and fast discharges C1 via R3. The voltage at node 2 goes down to 0. When C1 is fully discharged, the PUT does not receive enough current to stay latched, and unlatches. When the PUT unlatches, the voltage at node 2 jumps up to the threshold voltage again which is close to the supply voltage and C1 is charged again by the constant current source.
The cycle repeats. A simple voltage controlled current source can be made using a current mirror. In the circuit, the current mirror is formed by Q3 and Q4.
The control voltage is converted to a reference current, that is flowing from the power supply via the base-emitter junction of Q3 and through R4. Q4 mirrors a copy of this current, that is then used to charge C1. The lower the control voltage, the higher the current that is used to charge C1. So the higher the frequency of the sawtooth waveform. The frequency varies linearly with the control voltage.
The control voltage can not be as high as the power supply because the transistors of the current mirror need at least 0. It is possible to generate symmetrical square waves using the PUT configuration. To obtain a symmetrical square wave output at node 2, diode D1 and R5 have been added. By adding D1, the junction between C1 and D1 is not fixed to the ground potential anymore, but can be pushed below zero, going negative.
When the voltage at the junction of C1 and R4 reaches the trigger voltage of the PUT, Q2 will start conducting and will pull the junction between R4 and C1 down to the ground potential.
When this happens, the voltage at the junction of C1 and D1 will go negative, because C1 is still charged and D1 became reverse biased, so it will not conduct anymore. C1 will then be discharged via R5, which is about twice the value of R4, so the discharge rate of C1 via R5 is equal to the charge rate via R4, resulting in a symmetrical waveform at node 2.
Discharge of C1 happens via R5 and discharges C1 from a negative voltage via R5 to the positive supply. To have an equal discharge and charge current, R5 should be twice the value of R4, so the discharge and charge currents are equal. With higher power supplies, the LED will blink faster because the charge current for C1 will be higher when the power supply is higher.
With lower power supplies, the LED will blink slower, because the charge current of C1 will be lower when the power supply is lower. The second picture is the same circuit, but turned "upside down". The circuit behaviour is exactly the same. The next circuit is based on the PUT symmetrical square wave generator to make a LED flasher, that generates a very bright flash approximately every second. At first sight, the LED seems to be the connected upside down.
But it is connected correctly. This is because the way we use C2 and D2 to generate a voltage at node 3 that rises above the power supply voltage. We want to generate a flash at the moment that node 2 pulled up to the supply rail by Q2 at the moment that the PUT latches into conduction. When Q2 is not conducting, C2 will be charged via D2 and R2. When Q2 starts conducting, node 2 will be pulled up to the supply rail, while C2 is charged. So at first instant, the voltage on node 3 will be the power supply voltage plus the voltage over C2.
A specific combination of 2 complementary transistors leads to an interesting self-latching behaviour, that can be compared with the behaviour of a thyristor aka SCR or, as the title insinuates, a programmable unijunction transistor, abbreviated to "PUT". Many different circuits can be build with this transistor combination, In this instructable i show you a far from complete collection of circuits that are built around the PUT :. Of course this collection will never be complete, but the intention is to show you some circuit idea's that might be inspiring or just as learning examples to step into the analog world for a change. Did you use this instructable in your classroom? Add a Teacher Note to share how you incorporated it into your lesson. The configuration has a similar behaviour as a thyristor, but it is NOT a direct replacement for a thyristor!
Variateur à triac : réalisation
On peut choisir 3. Eventuellement 1MOhm,. Ce sont des tensions dangereuses. Donc parfait pour Bonjour, Est-ce que le montage est valable pour une lampe halogene de watt?? Si la plaque fait plus de Watts, il faut choisir un triac plus gros 16A comme le BTB par exemple. Bonjour, oui, cela fonctionne.