Design of astable multivibrator

[Merlin3189]

I have used C= 2.1uF and R= 1k . . But the time period of the observed square wave response doesn't match with the calculated value.

Well I get that roughly to 250Hz, which is in fairness about 10^3 less than I said you might get. So as Steve says, reduce your capacitor to nF rather than uF and you should get a much higher frequency. (I'd try around 47nF aiming for 10kHz for starters, because I'm not so confident it will go up to the 100kHz+ range.)

Further to my previous post, my "rough" calc turned ot to be rather less reliable than I'd thought. Concentrating on the slew rate issue, I'd completely neglected the gain characteristics of the op amp. Although a Gain bandwidth product of 400,000 sounds a lot at low frequencies, it is only 4 at 100kHz. So I wonder about my chances of getting anything like the 250kHz I suggested!

But you've got me a bit hooked on this now! Unfortunately I don't have simulation software,scope or frequency counter to do a practical test and my maths is more than a bit rusty these days. So I'm reduced to creating a discrete simulation in Excel.
For what that may be worth, my observations are:
1- Slew rate affected the frequency quite predictably. But the effect is small at low frequencies and other issues arise before it can become the dominant factor at higher frequencies.
2 - The gain of the opamp has some effect on the frequency and this quickly became the dominant factor, if I set it to the low values given by Gain = GBW / operating frequency. As the gain is lowered, the frequency increases. In my simulations, varying the gain (once you get down to the order of 10) made much more difference than varying the slew rate.
3 - Looking at the graphs of the Voltages in the circuit, I decided the mechanism for this is:
Assuming an ideal opamp with infinite gain, the capacitor charges until it reaches the switching point; the opamp switches from one saturated state to the other; the capacitor starts charging the other way. Time is two simple charges to the switching point, driven by the saturated output Voltage.
But with a finite gain opamp, as the charging capacitor nears the nominal switching point, the Voltage between the two inputs is not enough to drive the (low gain) opamp to saturation; so the output Voltage starts to fall BEFORE the capacitor has charged to the nominal switching point; the feedback to the non--inverting input (+) immediately starts to lower the switching point, so the opamp switches early; the capacitor starts charging the other way, but has bit of a start, since it had not charged up as far as it should have done; and switching in the opposite direction similarly will occur early. Time is two shortened charges to less than the nominal switching point from a closer starting point.
*NB.* This is a result obtained from my model, which may not be a good one.
4 - I also tried modelling the frequency response by adding integration to the amplifier transfer function. If I kept the basic gain above 100, this allowed me to operate at higher frequencies.

I'd be very interested to see the trace of Vout (if possible alongside Vcap) from an actual circuit, to see if my simulations are anything like realistic.