I have a simple oscillator, as shown below. As far as I understand, a negative bais is induced on the base of either c1 or c2 which in turn pulls the base of the other transistor below 0 shutting it off. However, I am wondering how a difference in size of the capacitor affects the circuit. It appears that the current can flow to charge the capacitor as well as charge the base of the transistor so how does this work? 
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\$\begingroup\$ Do you understand how a capacitor works in say a standrard relaxation oscillator based around a schmitt trigger? Link: google.com/… \$\endgroup\$– Andy akaJul 31, 2016 at 19:29
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2 Answers
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As far as I understand, a negative bais is induced on the base of either c1 or c2
No. C1 and C2 are capacitors, which don't have bases.
which in turn pulls the base of the other transistor below 0 shutting it off.
One transistor turning on lowers the base voltage of the other. It might or might not make it go below 0, depending on other parameters of the circuit, such as the power supply voltage.
However, I am wondering how a difference in size of the capacitor affects the circuit.
First see how the whole thing works. For whatever reason, one transistor turns on, let's say T1. This causes its collector voltage to decrease. Instantaneously, the voltage on a capacitor stays the same. The same decrease in collector voltage of T1 therefore shows up as a decrease in base voltage of T2. This turns off T2. The collector of T2 stays high, so T1 stays on.
However, eventually the right side of C1 charges up due to R2. This will eventually get to a point where it turns on T2 at least just a little. That causes its collector to go lower, which causes the base of T1 to go lower, which at least slightly turns off T1. T1 turning off even slightly drives T2 harder, which turns off T1 even more, etc. The net effect is that T1 and T2 flip states.
The same thing happens regardless of the sizes of the capacitors over a wide range. However, the time to flip to the other state depends on how long it takes to charge up the base side of a cap. Both a larger capacitance and a larger resistor increase this time. The time to flip in one direction is related to the product of C1 and R2, for the other direction the product of C2 and R3.
answered Jul 31, 2016 at 20:17
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This circuit is astable. As long as power is available, any small fluctuation of voltage will build up an oscillation, until the amplitude (say, at the collector of T1) of the voltage repeatedly goes up to the power supply rail on the positive swings, then down to the saturation minimum (about 0.1V) of a 2N3904 transistor.
As for the capacitor size, THAT is what sets the oscillation period. That period, which has units of time, cannot be computed from voltages, transistor gains, resistor values, or ratios. None of those quantities has any time-dependence, only by combining resistance and capacitance can a time scale be established by these components. If there were only ONE capacitor, there would be no stable oscillation, because a single capacitor does not determine an irreversible condition (oscillating means a positive peak followed by a negative trough, not by another positive peak). It takes two capacitors to determine both a voltage level for the output AND define the future (by holding charge, and in a way, representing the past history).
