You can use a single PS. Ensure proper de-coupling of the OP Amps power lines.I'm working on a VCO circuit. It is powered by a 9V power supply and uses three op amps. I wonder if I need to use separate split power supplies for each or one can do the trick? Is there much crosstalk or noise with a single supply?
You can use a single PS. Ensure proper de-coupling of the OP Amps power lines.
Ramesh
Yes, this may be a problem because - I assume - you want to bias the opamps all at Vcc/2.By the way, does it make a difference if the op amp feedback networks differ, i.e. there are both inverting and non-inverting configuration in a quad op amp?
The power supply in the right hand corner provides the virtual ground, 4.5V. The lowest op amp is a comparator - that's why it doesn't have any feedback. Not sure about the capacitor in the feedback. Can you explain that?Alkopop79, your circuit does not show the dc supply concept you have chosen (and this was your main question, was it not?).
For example, what is the dc level at the input of both most right opamps?
And - the lowest opamp has no feedback and cannot work as a linear amplifer.
More than that, please note that - in case of single supply - a bias dc voltage at the pos. opamp input (normally chosen as Vcc/2) is transferred to the opamp output only if the dc gain of the opamp is unity (a capacitor is required in the feedback path).
I stand corrected! Yes, I wonder if any of the op amps need them.Are you sure you're not confusing this with a series DC-block capacitor?
Excellent, thank you!If T1 is meant to be a voltage-controlled constant-current source it would be better to add an emitter resistor to reduce the effects of transistor gain variation and ambient temperature changes.
If T1 is meant to be a voltage-controlled constant-current source it would be better to add an emitter resistor to reduce the effects of transistor gain variation and ambient temperature changes.
Ok. The best I can figure out you would need to use this formula:I've spent a day trying to figure this out. What resistor value? How does that change the range and the linearity of the transistor?
Let's suppose the base voltage Vb is at 3V and you have an emitter resistor Re of value 10k. The voltage Ve at the emitter will be 3V - Vbe = 3 - 0.7 = 2.3V. So the emitter current Ie will be 2.3/10k = 0.23mA. You don't need to know the transistor beta value. Even if Vbe changes with temperature (at ~ 2mV per degree) the emitter current will not change much, since Vbe <<3V. Emitter current is thus effectively constant and controlled by (a) the base voltage and (b) the value of the emitter resistor.I've spent a day trying to figure this out. What resistor value? How does that change the range and the linearity of the transistor?
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