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This is one of the circuits in the Electronics 9/10 class.

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The package did not explain how transistor combination worked and why the capacitor will reverse its current.

Can you explain this circuit? Why I can't make a transistor combination with only two NPNs?

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  • \$\begingroup\$ I'm not sure what you are talking about in the circuit. The link provided just brings up an RLC circuit example. \$\endgroup\$ – mjcarroll Oct 30 '10 at 5:59
  • \$\begingroup\$ This looks like a non-symmetrical relaxation oscillator, but it is drawn unusually. I've seen them drawn like this before. \$\endgroup\$ – Thomas O Oct 30 '10 at 10:07
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The cap charges through all three resistors, until it reaches a voltage where the NPN base-emitter junction starts to conduct. At that point, the NPN allows current to flow into its collector, which means it begins to draw current from the base of the PNP, turning it on, too. Now, the base of the NPN will be somewhere in the range of 0.6 to 0.8 volts while that transistor is on, and that's one leg of the cap. The other leg of the cap experiences the I-R drop that occurs on the 10 ohm resistor. The closer the PNP gets to saturation, the closer the top of that 10 ohm will get to being a voltage divider formed with the 47 ohm resistor. Now I don't see the battery voltage here, but I'd bet its at least 5V. At 5V, the voltage at the top of the 10 ohm will approach 5*(10/(10+47)) = 0.88 V (discounting Vce-sat of the PNP). This means that the capacitor now sees a voltage higher on the right side, so current begins to flow the other direction. While charging up, the cap built up a charge with (+) on the left and (-) on the right, so the capacitor now looks like a tiny battery in series with the 0.88V on the 10 ohm, so it basically discharges into the base of the NPN, helping keep that transistor on. Once the cap has charged up in this opposite direction, it ceases to conduct, and behaves again like a DC open circuit, leaving the NPN biased only by the 120K, which isn't enough to keep the transistor active. So the NPN shuts off, this kills the PNP, and we're back to the start .. only you can see that the capacitor not only has to charge 'up', it has to actually discharge the reverse-polarity charge first.

As for trying to use just NPN - you might be able to do it with three NPN's, but two seems unlikely. Essentially, the second transistor needs to be something that passes current under control of a node that is either very high impedance or connected to the negative rail, which just shouts PNP. To go NPN-only, you'd need to move the 2nd transistor (call it Q3) to between the 10 ohm and the negative rail, and you'd need an intermediate stage (call it Q2) to act like an inverter, delivering current into the base of Q3 only when the original NPN (call it Q1) is conducting. There would have to be a pull-up resistor on the collector of Q1, to supply current to the base of Q2 when Q1 is off. You'd need another pull up resistor on the collector of Q2, which would feed current into the base of Q3 when Q2 is off.

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  • \$\begingroup\$ "Now I don't see the battery voltage here" -- if you hover over any of the components in the circuit, additional details appear in the lower right corner. The battery is 9v. \$\endgroup\$ – tcrosley Oct 30 '10 at 17:21
  • \$\begingroup\$ @tcrosley - ah. i normally run my browser with a boat load of script blockers, that must be why when i hover the mouse over a component it only just turns blue. \$\endgroup\$ – JustJeff Oct 30 '10 at 22:24
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It looks to me (with quite limited electronic circuit analysis skills) like the 100nF capacitor charges until it reaches a voltage high enough to forward bias the BE junction of the NPN transistor. When this happens, it discharges its energy through the BE junction to ground until its voltage drops below the minimum BE voltage, thus switching off the NPN. The capacitor reverses its current because that's what they do when faced with a falling DC voltage.

Additionally, because the -ve leg of the capacitor is tied to the Collector of the PNP, the capacitor voltage receives a positive offset equal to the voltage at the Collector / the voltage across the 10 ohm resistor. This is also why there is no current flowing from the voltage source through the Base of the NPN.

Unfortunately I'm not really sure where to start for your second question so I'll have to defer to someone more experienced (and hope they don't have too many corrections to make about my original circuit analysis). :]

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