Transistor analog switch to stitch out of phase signal

Thread Starter

rfpd

Joined Jul 6, 2016
101
I was googling around, and this concept gave me an idea, which in paper works, the thing is it wasn't working in real life with a guitar signal, since the output pretty much sounds like the input.
upload_2017-10-22_16-15-4.png
OUTPUT is at the emitter.

Imagine something like this, not exatly like this because the square wave has a 180 degree phase, but I think the result would be the same. In the collector it's the out of phase signal, in the emitter the original signal. I amplify the signal by a factor of 5.45, I would get, tops, a signal oscilating between 7.225V and 1.775V, being the bias 4.5V. Most likely, it's much less than that.

I mean I tried a lot of things, low pass filter after that, voltage follower, transistor buffer at input, everytime the result is the same. It seems like the signal is ignoring the transistor. I think it works well (the transistors), I tried with many transistors, and the result is the same for all of them. I tested all of them with a multimeter. I also tested R1 and R2 values from 10k to 150k.

The only time I got something working, was yesterday, when I connected by mistake a pnp transistor and the resistor of the emitter to + of an op amp, like below (basically 4.5V).

upload_2017-10-22_16-39-21.png

This sould be the output:
upload_2017-10-22_16-39-56.png
Also, everything in the schematics is what I'm using. I also use a flip flop, an inverter, an amplifier, a transistor buffer, and a voltage follower.

Thanks in advance.
 

Thread Starter

rfpd

Joined Jul 6, 2016
101
How do you think that inverting the phase will change the sound?
Not inverting the phase, but stitching together the inverted signal with the original signal, the point is dividing the frequency in a 'clean' way.
 
Last edited:

Thread Starter

rfpd

Joined Jul 6, 2016
101
I don't know why you think this would work. AFAIK, a transistor needs DC BIAS in order to function
This is working between cutoff and saturation, like I said the square signal oscilates between 0 and 9 V, and the sinusoidal signal, oscilates between 1.775V and 7.225V (tops). So, when it has 0V, it's in cutoff and the emitter gets the voltage from V1. When it has 9V, it activates and it also enters saturation mode, therefore, I would get at the emitter V2.
 

Thread Starter

rfpd

Joined Jul 6, 2016
101
The signal would have to be 90° out of phase to get anything close to that.
Why? I mean the one controlling the switch has half the frequency from the rest. Therefore when it's deactivated, the signal goes through, one it's activated the out of phase one goes.
 

Papabravo

Joined Feb 24, 2006
12,514
This is working between cutoff and saturation, like I said the square signal oscilates between 0 and 9 V, and the sinusoidal signal, oscilates between 1.775V and 7.225V (tops). So, when it has 0V, it's in cutoff and the emitter gets the voltage from V1. When it has 9V, it activates and it also enters saturation mode, therefore, I would get at the emitter V2.
You're the one that said it wasn't working -- now you say it is. Which is it?
 

kubeek

Joined Sep 20, 2005
5,625
Why? I mean the one controlling the switch has half the frequency from the rest. Therefore when it's deactivated, the signal goes through, one it's activated the out of phase one goes.
Ok looking at it closer I think the 90 is wrong and 180 is correct, but you are definitely not going to get that waveform from a sine wave using a single transistor or opamp.
 

Thread Starter

rfpd

Joined Jul 6, 2016
101
Ok looking at it closer I think the 90 is wrong and 180 is correct, but you are definitely not going to get that waveform from a sine wave using a single transistor or opamp.
upload_2017-10-22_19-13-34.png

I based on this sketch, and I since I don't have that mosfet (J112), I tried to do it with a transistor.

Why not, though?
 

kubeek

Joined Sep 20, 2005
5,625
J112 isa JFET, not a mosfet. It maybe could be possible to do with a bipolar transistor, and it would have to be connected pretty much the same way, i.e. common base.
 

crutschow

Joined Mar 14, 2008
23,521
Here's an op amp circuit that inverts the analog signal under digital control, if that helps.
The circuit uses two back-to-back MOSFETS to control the analog switching so that it can block both the plus and minus swing of the analog input voltage when off.

When the MOSFETs are off, the signal goes to the plus input, which causes the op amp to act as a follower (non-inverting).
The minus input simply follows the plus input so has no effect on the output.

When the MOSFETs are on, the signal to the plus input is shorted to ground, and the op amp now acts as an inverter with a gain of -1.

Note that the op amp requires plus and minus supply voltages since the input signal is centered around zero volts.

upload_2017-10-23_11-50-11.png
 
Last edited:
Top