function generatoroperational-amplifierwave
So I'm trying to design a sawtooth wave generator that goes from 2.5V to 3.25V. I'm using this design:
I'm getting a voltage range from around -4V to 4V. Is there any way for me to get 2.5 to 3.25 by editing the values in this circuit? If not is there any other configuration that will help me achieve this, preferably using op amps and/or mosfets.
Edit:
Thank you for the quick responses I was able to work something out.
500kHz is a ways past the maximum specified frequency for square waves for this generator: 100kHz. It appears to have a single-pole RC of about 3.5µs, which would work great for a 0.35/(3.5µs)=100kHz square wave. The output may have a LPF for slew limiting. Also, it is a 50Ω source, so it should be terminated properly to avoid ringing. Try using the TTL and CMOS waveforms, too. B&K have put together this document: Function & Arbitrary Waveform Generator Guidebook .
The external CMOS inverter is not a 50Ω source -- it's source impedance is only a few ohms at most (for low currents) due to VCC and ground impedances and FET RON equivalent resistances. Notice that the output duty cycle isn't 50%, and the edges are ringing.
There a few problems with your circuit:
1) The second stage will generate an exponential rather than a triangle. 2) The resistor values for feedback on the second stage are much too low - the amplitude may be limited by the output current capability of the 741. You can use something in the range 10k-100k. 3) The actual amplitude of the output will be limited by the opamp. The 741 os a very ancient opamp and is not as good as more modern ones - as Peter mentioned the output cannot get closer than ~2.5v to the rail.
This configuration should give better results:
It still uses two opamps and only needs one capacitor to set the timing.
The first stage is arranged as a schmitt trigger to detect when the output of the second stage reaches voltages determined by resistors R1 and R2. The second stage is configured as an integrator and will produce a triangular wave with linear slopes.
The square wave output is still determined by the saturation of the opamp as before, an opamp with rail to rail output will get very close to the supply rails.
Best Answer
Your desired hysteresis is 0.75V but your observed is 8V (implying a +/-10V swing at the output of the comparator U1A) so decrease R2 by a factor of 0.75/8 to 3.75K.
You want it centered at +2.875V so add a resistor from the non-inverting input of U1A to the negative supply. Value should be 12/2.875 * 3.75K or 15.65K
That gets you the desired swing and the duty cycle is unaffected, but the frequency will be different (about 10x higher) because the hysteresis is less, so increase the capacitor by the inverse ratio to R2 or 8/0.75 to give you 2.1uF or leave the cap the same and increase both R3 and R4 by that ratio.
And better to replace the 1N4xxx with 1N4148 or similar switching diodes.