What's the job of a voltage follower? Low current, high impedance input --> low impedance output. This means it stabilizes the outpout voltage.I know the opamp is a voltage follower,
Oh, why is that?I does not seem a well designed circuit at all ....
I understand, but on the graph you provided X1-out climbs to 7v and not 4v as I expected because of the 10k voltage dividers. This is what puzzles me.Note the R divider keeps the inputs, more or less, inside the CM range
of the OpAmp inputs as the power to the opamp is ramped up.
Regards, Dana.
I understand, but on the graph you provided X1-out climbs to 7v and not 4v as I expected because of the 10k voltage dividers. This is what puzzles me.
Out of curiosity, what tool do you use?
Thanx Brian. I'm very new to the gyrator. I'm reading up on this. Very interesting stuff. The setup in this diagram is different from the other schemas I see on the net. Most of them are build around an op amp, and the ones with transistors have a different setup. Like this one for instance.What confuses this is the pulse input. The circuit is basically a gyrator followed by a voltage divider. Normally it would be powered by a battery or similar steady voltage but here it seems to be driven by a pulse.
As I understand it the transistor stage is the gyrator, it serves to roughly pre-regulate the voltage. If its emitter voltage drops under load, C2 holds the base voltage up hence Vb-e tries to increase and the transistor conducts more. Effectively it makes the capacitor seem to have a much larger value but at the expense of losing about 0.6V in the transistor B-E junction.
The op-amp is a unity-gain buffer stage to isolate the load. The voltage is set by R3/R2 which being equal value makes it half the supply. Without the amp, any load in parallel with R2 would affect the division and the voltage would drop. C4 is, I think, there to filter noise from the potential divider but with a pulsed input it will be slow to charge, hence the slow rise in output voltage.
Brian.
Thanx DanaI also noticed that, but got distracted and did not re-address the issue. My apologies.
The supply ground of the OpAmp was not connected. DUH !
Note SIM is a bit incomplete in that the input diode and C1 100 uF cap are essentially
not in circuit since I am driving that node with a V source. I should put some generator
R in the circuit to get a more complete sim. I tried it with 10 ohms, that did not alter the
results in any significant way.
Here is updated sim (note the Vcesat of Q1 curve that affects max V sent to OpAmp) -
View attachment 167044
SIM is SIMetrix. Most folks are using LTC Spice these days. SIMetrix was default SIM
used at Analog Devices I gather, and when they bought out LTC they settled on
their spice. I find SIMetrix much more user friendly than LTC in general.
Regards, Dana.
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