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In this sawtooth oscillator circuit below, I understand that the values of Rs2 and C1 detertmine the rate at which the output ramp charges. As long as this value is the same (220u), it should charge at the same rate. I have experimented with different values and found that simulating at Rs2 = 60K, C1 = 3.8nF, the output begins to have an error (this is the lowest RS2 can go). What would be the cause of this? In the other direction where RS2 is increases, RS2 = 1000kohm, C1 = 220pF there is still no error at all. However I imagine there is some limiting factor eventually to the minimum input current ?

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The lower Rs2 is the higher the gain, until at some point you saturate the op-amp. This would cause severe distortion. Your V6 battery is backwards. Quality op-amps with JFET or CMOS inputs should have the most range.

Since Rs1 is dynamic, Rs2 sets the overall gain. The lower it is the higher the gain, up to the op-amp limits.

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  • \$\begingroup\$ thank you, what is the gain equation for an integrator? I have read somewhere that it is k = 1/RC? so the op amp has reached its maximum gain capabilities ? \$\endgroup\$ – konobyBYnight Apr 8 '18 at 13:37
  • \$\begingroup\$ also, if the RC value stays the same, and only the variation of resistor, capacitor value changes. then surely the gain will stay the same? as the integrator is still charging at the same rate? is this wrong ? \$\endgroup\$ – konobyBYnight Apr 8 '18 at 14:10

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