Simple Oscillator (non ideal)

P: 9
 Quote by Averagesupernova So after C1 has been charged up by the positive feedback the base current in Q1 starts to sag again. As I said before, there is not enough base current to turn the transistors on hard without aid from C1. So, when the base current sags, the collector of Q2 drops which in turn reduces base current in Q1 even farther. Cycle starts over.
Which direction is the base current in the PNP going? I'm still confused on that. Also, does the base current in Q1 sag because of the discharge versus charging rate? Man, I think I'm confusing myself way too much here.
 P: 1,072 Roughly speaking, when there is a voltage change on one side of a capacitor, the same change appears on the other side. Just before the output switches off the Q2 side of the capacitor is at 4.5V (assuming large signal multivibrator action, which may not be what is actually happening). When Q2 switches off, that causes a transition to 0 volts, which is a -4.5V change. On the other side, the base of Q1 was in the region of 0.7V so it would transition by -4.5V to -3.8V. But, again, all that was predicated on the voltage at the Q2 - C1 junction making 4.5v swings. What does it actually do?
P: 9
 Quote by meBigGuy But, again, all that was predicated on the voltage at Q2 and C1 making 4.5v swings. What does it actually do?
I dont have an oscope on me (poor college student), and I havent simulated it yet. But wait if Q2 switches off wouldn't there still be the constant +4.5 volts on the other side of the capacitor from the voltage source canceling out the -4.5volt change?
 P: 1,072 If Q2 switches off completely (hypothetical at this point) the speaker takes the node to zero. Since there is nothing on the other side that can supply the charge, that side will transition by -4.5V (again, assuming such a transition occurred at the Q2/C1 junction) and then charge through the resistors.
 P: 2,497 I had a quick look last night in an old radio shack book called: Getting Started in Electronics by Forrest Mims. Same circuit except a bigger cap and probably different resistor values. It is described as a metronome.
 HW Helper Thanks P: 5,142 This is a relaxation oscillator, the transitions are fast because the transistor switching uses positive feedback. Waveforms here are nothing like sinusoidal. Generally speaking, the term "oscillator" as in your title without qualification is used for sinusoidal oscillators, using LC elements or RC staged filters, and where transistors operate as linear amplifiers/buffers.
 P: 405 This simply multivibrator is nothing more then two stage CE amplifier with positive feedback add by C1 capacitor. When we first start-up the circuit C1 start to charge until T1 start to conduct. The current that is flow through T1 also opens T2 transistor. T2 collector voltage reach positive saturation voltage (4.5V). And C1 capacitor will quickly charge in the circuit: +4.5V ---> emitter-collector T2---> C1 --->base-emitter T1--->GND The capacitor is rapidly charge to V_C1 = Vcc - Vbe - Vce(sat) ≈ 3.8V. And when Capacitor is full charged. The T1 base current is provided by R1 resistor. And in order to circuit work as a multivibrator we need to select R1 so that R1 current is so small, that T2 immediately after C1 is full charging comes out from saturation to the linear region. So voltage on T2 collector is start to drop (T2 comes out from saturation). This change in VecT2 voltage is "feedback" by C1 capacitor to T1 base. This will decrease the voltage at T1 base. So T1 and T2 will decrease his currents. This will increase VecT2 and Q1 base voltage drop even faster (for example if Vec = 0.3V then Vb = Vcc - Vec - Vc1 = 4.5V - 0.3V - 3.8V = 0.4V). So C1 that was previously charged to 3.8V "pulls-down" the voltage on the T1 base below GND (-3.8V) So we have a very strong positive feedback in this circuit. So T1 and T2 will be immediately go into cut-off. And C1 immediately start to discharge in the circuit: right C1 plate ---> R2 --->battery--->R1----> left C1 plate. The voltage on the "left" plate of a C1 is rising during discharging phase, at some point becomes equal to zero volts (end of discharge), and voltage at C1 "left" plate (at T1 base) will continue to rising (C1 is now charging). When voltage on C1 will increase to 0.6V the T1 is start to open. And we buck to the beginning. Attached Thumbnails
P: 2,497
 Quote by Tactified Which direction is the base current in the PNP going? I'm still confused on that. Also, does the base current in Q1 sag because of the discharge versus charging rate? Man, I think I'm confusing myself way too much here.
The base current of Q2 (conventional flow) comes out of the base. It can't go the other way. Base current in Q1 sags because once the cap is charged only the resistors provide base current. This lack of current in the base is reflected on the output which of course pulls the base low.
-
I would bet those of you who question this circuit would not having any problem seeing how an op-amp configured with positive feedback through an RC network would oscillate.
 Engineering Sci Advisor HW Helper Thanks P: 6,936 @mebigguy: I think you are trying to understand the behavior of a very nonlinear circuit using a linear model of how an oscillator "ought" to work (frequency-dependent feedback etc). This circuit is a switch, with some positive hysteresis - like an op-amp comparator designed to avoid "flutter" if the input is changing slowly. When the switch is flipped one way, the cap charges up, slowly. When it's flipped the other way, it discharges quickly. The "duty cycle" might be 100:1 or even 1000:1, hence you get a "pop" from the "discharge" part of each cycle.
 P: 2,497 An oscillator is generally defined as a device that outputs an AC signal with no AC input. I would say the circuit in question meets that requirement. It may be crude, but sometimes it is all that is necessary.
P: 9
 Quote by Averagesupernova I think I see how it works. When the transistors start to turn on C1 gets an extra 'kick' from the collector of Q2. I suspect that without this extra 'kick' there is probably not enough base current in Q1 to turn it on hard enough to latch up like it appears that it would.
Earlier I didnt know what you meant by turning it on hard. But thinking about it a little further, did you mean that the current is actually flowing through the cap from +4.5 through Q2 (Emitter base) and through Q1 (base emitter), but there is only a very small amount of current becuse none of the diodes have hit the linear region yet? And then once it builds up it discharges all its built up charge through Q1?

Is that what you were saying?
P: 2,497
I don't know what you mean by this:
 ...did you mean that the current is actually flowing through the cap from +4.5 through Q2 (Emitter base) and through Q1 (base emitter), ...
That really doesn't make sense.
-
At power up, both transistors are turned off. At first, the voltage across the base emitter junction is zero because of the cap. The voltage on the collector of Q2 is at zero and the low impedance of the speaker does not allow it to float up. The cap needs time to charge up. The voltage starts to ramp up at the base of Q1 due to the cap charging. Eventually it gets high enough so the transistor Q1 starts to conduct which pulls the base of Q2 lower. This starts Q2 into conduction. The voltage on the collector starts to rise. Up to this point, we have not gone through even half a cycle of oscillation. The voltage across the cap is a little less than .7 volts with the left side of the cap being more positive than the right side. Here is where it gets interesting. The collector of Q2 is tied to the cap. So the voltage on the right side of the cap starts to rise which makes the voltage on the left side of the cap attempt to follow it. This is the start of the discharge. The base voltage of Q1 can only get as high as .7 volts because of the diode in the base-emitter junction. There is some serious positive feedback going on at this point. The higher the collector voltage at Q2 gets, the harder it drives Q1 which in turns makes the collector voltage of Q2 go higher. Eventually it tops out close to supply voltage. At this point the cap is discharged and is now recharged in the opposite polarity. Now the right side of the cap is more positive than the left. The cap is now fully charged in this polarity (we'll call it reverse) so it no longer conducts current. At this point there is not enough base current on Q1 to keep the collector of Q2 close to supply voltage. The collector voltage of Q2 starts to fall which in turn lowers the voltage on base of Q1. Again we have some serious positive feedback. The more the collector voltage of Q2 falls, the less Q1 is driven which lowers the collector voltage of Q2 even farther. The collector voltage of Q2 is now at zero and the voltage across the cap is still what we called reverse with the right side more positive than the left. But, there are resistors tied to the left side of the cap and they will start the discharging of the cap and then eventually recharge it in the other direction again with the left side more positive than the right. That is one complete cycle. I don't think I can explain it any more clear than that. If you want to know more about it try building it and tweak a few things. Try it without the cap. Or try bringing up the power supply voltage REALLY slowly. You may find it won't oscillate.
 P: 1,072 I fully understand a bistable mult-vibrator like "oscillator" and was attempting to analyze with that in mind, especially since that was the description. I had a problem with the off cycle. " At this point there is not enough base current on Q1 to keep the collector of Q2 close to supply voltage. " That's where I had a problem. My thoughts were that if R could supply enough current to turn on Q1, the current into Q1 Base would never drop enough to turn it off. I still have a small problem with that since it take very little current on Q1 to keep Q2 saturated. But, then again, it requires very little voltage change at Q2 to get Q1 fully off - maybe even noise is enough since gm is high. The other thing I don't like is that the base of Q2 is open circuit when Q1 is off, meaning leakage current (Q2 Icb and Q1 Icb and Ice) will tend to keep it on a bit. Feels sloppy. Looks like starving Q1 Ib with high R is what makes it work.

 Related Discussions Introductory Physics Homework 3 Introductory Physics Homework 1 Electrical Engineering 0 Electrical Engineering 2 Introductory Physics Homework 0