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PUT Oscillator?

Discussion in 'Electronic Basics' started by zero, Apr 5, 2005.

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  1. zero

    zero Guest

    The schematic I am studying comes from page 116 of Mims' Timer, Op Amp &
    Optoelectronic Circuits & Projects book. The circuit is a relaxation
    oscillator which pulses an LED 240 times per second.

    A +6V source is attached to the positive end of the LED and the emitter of a
    2N2907 PNP. The collector of the 2N2907 leads to the base of a 2N2222 NPN.
    The 2N2222's emitter is grounded and its collector goes to the negative side
    of the LED. The junction between the LED and the 2N2222 leads through a
    0.02uF capacitor and a 22k resistor to the base of the 2N2907, which then is
    grounded through a 2.2M resistor.

    This is the netlist I created in PSPICE:

    D_D1 N00292 N00346 D1N4148
    R_R2 N00573 N00543 22k
    R_R3 0 N00573 2.2m
    C_C1 N00543 N00346 0.02u
    Q_Q1 N00261 N00573 N00292 Q2N2907A
    Q_Q2 N00346 N00261 0 Q2N2222
    V_V1 N00292 0
    +PULSE 0 6 0.01 0 0

    I used a VPULSE because I don't know how to set the initial charge on the C
    to 0V through Capture. I also used a D1N4148 because there were no models
    for an LED.


    A. Is the configuration of the PNP and NPN transistors a programmable
    unijunction transistor?

    B. When I ran this circuit in PSPICE there was no oscillation. How is this
    circuit supposed to work, and what changes should I make to the netlist to
    cause the correct results?
  2. I haven't examined the netlist topology yet, but is it possible that
    this "2.2m" is interpreted by your spice as "2.2 milliOhms" instead of
    "2.2 MegOhms?"

  3. FWIW, the circuit oscillates OK in CircuitMaker (using
    Spice3f5/XSpice), with no added PULSE elements. Here are the waveforms
    using both an LED and a diode (I chose 1N914, as I've experienced odd
    results with CM's 1N4148 model).

    I also tried it after adding an .IC (initial condition) with V=0 on
    the LH side of the cap, and that also oscillated OK, starting after
    about 30 ms.

    Apart from not using a PULSE, note that my simulations used a very
    short Step Time of 1u.

    In case it helps, here also is the Spice Netlist for the diode case:

    *Spice netlist for Circuit:
    D1 V1_1 C1_1 D1N914
    V1 V1_1 0 DC 6V
    Q1 Q2_2 Q1_2 V1_1 Q2N2907
    Q2 C1_1 Q2_2 0 Q2N2222
    C1 C1_1 C1_2 20nF
    R2 0 Q1_2 2.2meg
    R1 C1_2 Q1_2 22k
    ..SAVE C1_1 C1_2 V1_1 Q2_2 Q1_2 @d1[p] @d1[id] @v1[p] v1#branch @q1[p]
    ..SAVE @q1[ib] @q1[ie] @q2[p] @q2[ic] @q2[ib] @q2[ie] @c1[p] @c1
    @r2[p] @r2
    ..SAVE @r1[p] @r1

    * Selected Circuit Analyses :
    ..TRAN 1u 50m 0 1u

    * Models/Subcircuits Used:

    *1N914 100V 80mA Si Switching Diode pkg:DIODE0.4 A,K
    ..MODEL D1N914 D(IS=7.075E-9 RS=0.78 N=1.95 TT=7.2E-9 CJO=4E-12
    + M=0.4 BV=100 IBV=0.0001 )
  4. Bob Eldred

    Bob Eldred Guest

    I don't think this thing is connected as a PUT. There shouldn't be a
    capacitor between the base of Q1 and the collector of Q2. In a PUT, the base
    of Q1, PNP goes to the collector of Q2, NPN and the base of Q2 goes to the
    collector of Q1. It's DC connected, no caps. The trigger voltage is set by a
    resistor voltage divider connected to the base of Q1 between +V and gnd.
    This is what is programmable. The LED is between the emitter of Q2 and gnd.
    The timing cap is connected between the emitter of Q1 and gnd. And, a timing
    resistor goes from the emitter of Q1 to +V.
    In operation, the capacitor charges up towards +V until it gets one Vbe
    drop higher than the trigger voltage set by the voltage divider on the base
    of Q1. At that time both transistors turn on and current dumps from the cap
    through the transistors and the LED pulsing it on while the cap discharges.
    When the cap is sufficiently discharged the circuit goes off and the cycle
  5. zero

    zero Guest

    When I run the simulation in PSPICE, when the 6V voltage is applied, V(C1:2)
    rises to 3.1693V and stays there. V(C1:1) rises to 1.9077V then
    exponentially decays to a small number about 1.1128mV.

    V(Q1:b) also rises to 1.1128mV and stays there.
    V(Q2:b) rises to 5.6401V.

    I(C1) peaks at -91.795uA then exponentially decays to a sine wave with a
    very small amplitude.

    IC(Q2) = I(D1) = 3.0831A.

    I think IC(Q2) should discharge V(C1:2) but because all the current flows
    through the diode the voltage is maintained at one diode drop below the

    IC(Q1) = -471.532mA, IB(Q1) = -505.800mA,
    IE(Q1) =977.332mA.

    IC(Q2) = 3.0833A, IB(Q2) = 471.561mA,
    IE(Q2) = -3.5549A.

    As for the oscillation, is this what is supposed to happen: when the voltage
    is applied, both ends of the capacitor jump to Vsupply - Vdiode, turning the
    LED on, then V(C1:1) slowly discharges, and when V(Q1:b) = Vsupply - Veb(Q1)
    both transistors turn on, discharging C1 and turning off the LED?

    If so, what turns the transistors off? Does Q1 turn off when there no longer
    is any base current available?
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