I built a phase shift oscillator and now I want to turn the output of the oscillator on and off using a pulse, generated with the NE555 IC.

I am trying to do this using a BJT. I connected the oscillator output as Vcc and the pulse to Vb.

enter image description here

I got more or less the short sinusoidal pulse that I wanted but as you can see the output only follows Vcc for values larger than -Vce. I have only used BJT's with DC values for VCC so I am not really sure how to redesign this circuit in order to get the output to follow Vcc for larger negative values.

enter image description here

So my questions are:

1) Is it possible to get what I want using a BJT?

2) If it is possible how should I redesign my circuit?

UPDATE: This is the new circuit I am using: enter image description here


2 Answers 2


The sinewave must always be positive relative to the transistor emitter.
So, you must either

  • "Move" the sine wave up to be always above ground or

  • Move the transistor "down" to have its emitter at sine wave minimum or lower or

  • Some combination.

You can move the centre of the sine wave "up" by AC coupling it with a capacitor and then providing a high impedance resistive divider using two resistors, one from capacitor output to V+ and another to ground. Sine wave amplitude can then be a maximum of < (V+-ground) peak to peak.

You can move the clamping transistor (call it Q1) "down" by connecting the transistor emitter to DC supply V-.
You must now provide a means of driving the base high relative to V-.
This would usually take 2 transistors to do but a "clever" circuit. is to use a PNP transistor Q2. Connect Q1 base to V- with say 100k. Connect Q2 collector to Q1 basee with say 10k. Connect Q2 base TO GROUND by say 10k. Drive Q2 EMITTER with the pulse signal.
Alter resistor values to suit.

  • \$\begingroup\$ I used the second method with the PNP transistor and it worked perfectly. Vout is -6 V instead of 0 V though but that doesn't matter in my application. \$\endgroup\$
    – Ace
    Aug 22, 2015 at 14:47
  • \$\begingroup\$ @Ace How did you drive the base of Q1? - with a second transistor as above or other? \$\endgroup\$
    – Russell McMahon
    Aug 22, 2015 at 14:54
  • \$\begingroup\$ Base of Q1 is connected to -6 V with 100k resistor as well as to Collector of Q2 through 100k resistor. See updated circuit \$\endgroup\$
    – Ace
    Aug 22, 2015 at 15:36

A common and simple approach is to use a transistor such as a 2SD2704K in a circuit such as this one (as from this previous answer):

enter image description here

The transistor I mention is notable in this application for having high symmetry (large emitter-base breakdown voltage),so it does not have to be used upside down like the above example, and the hFE is very high (maybe 30x better than an ordinary BJT in reverse) so it will mute better.


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