Insane Posted July 14, 2004 Report Share Posted July 14, 2004 Hi there, I have a problem with my circuit. I suppose to get a square wave output with voltage levels 0-5V at point A, but when I test my circuit with the circuit maker simulation, it gives me a different answer. What should I need to do in order to get the desire output? Quote Link to comment Share on other sites More sharing options...
audioguru Posted July 14, 2004 Report Share Posted July 14, 2004 Hi KM,The 2N5484 has a broad range of gate-source cutoff voltage, from -0.3V to -3.0V. Maybe your simulator averages this voltage to -1.6V.Since D2 limits the gate voltage to about -0.7V, then the 2N5484 never cuts-off, and the output at point "A" never gets close to ground. The 2n5484 datasheet is here:http://www.fairchildsemi.com/ds/2N/2N5484.pdfAlso, the 2N3906 has a saturation voltage of about 0.1V, so the output never goes higher than about 4.9V.If you remove D1 and D2, then the output will be allowed to swing from 0V to 4.9V, but probably won't be symmetrical because it is not biased for idling at half-supply. Quote Link to comment Share on other sites More sharing options...
Kevin Weddle Posted July 14, 2004 Report Share Posted July 14, 2004 I think you have a voltage mismatch. You have a an emitter follower that shares the same voltage as the collector of the PNP. Suppose you have a large signal at the collector of the PNP. This is going to make the VGS behave strangely. I think the bias is right, but a signal is not going to operate to give you a change in VGS. Quote Link to comment Share on other sites More sharing options...
audioguru Posted July 15, 2004 Report Share Posted July 15, 2004 Kevin,If the collector of the PNP is swinging the full 4.9V peak-to-peak, then the voltage divider action of R4 and R5 will reduce that to only 158mV at the source of the FET. This reduction of Vgs (which is negative feedback) is negligible when compared to the 10V input swing. So Vgs will actually swing about 9.8V, which is plenty.KM,Now I notice that the voltage divider of R6 and R7 at the output will cause a voltage of about 0.25V across C4.So without the diodes, point "A" will actually swing from 0.25V to 5.15V. But since point "A" is such a high resistance, if you measure it with a 10Megohm 'scope probe, then it will show a swing of from 0.129V to 5.029V. Quote Link to comment Share on other sites More sharing options...
Kevin Weddle Posted July 15, 2004 Report Share Posted July 15, 2004 I think your PNP is biased at too low a current. It's supposed to be at 100mA. Quote Link to comment Share on other sites More sharing options...
audioguru Posted July 15, 2004 Report Share Posted July 15, 2004 Kevin,The collector load resistors for the PNP total 310 ohms. If 100mA was flowing through them then their voltage drop would be 31V.How can 31V be developed from a 5V supply?With the transistor fully saturated in this circuit, its collector current is 15.8mA, which is fine.Like all transistors, the 2N3906 operates very poorly at a high collector current that is near its absolute maximum rating. Look at how its gain drops off sharply above a collector current of only 20mA. Its datasheet is here:http://www.fairchildsemi.com/ds/2N/2N3906.pdf Quote Link to comment Share on other sites More sharing options...
Kevin Weddle Posted July 15, 2004 Report Share Posted July 15, 2004 I operate my transistors at mid current and the beta is fine. This transistor has a beta of about 113. Quote Link to comment Share on other sites More sharing options...
audioguru Posted July 15, 2004 Report Share Posted July 15, 2004 But the datasheet shows that the 2N3906's beta could be as low as only 30, at 100mA of collector current.In this circuit, beta doesn't matter since the FET supplies plenty of base current. Quote Link to comment Share on other sites More sharing options...
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