Multiple op amps with single split supply?

Thread Starter

alkopop79

Joined Jul 30, 2012
12
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?
 

Ramussons

Joined May 3, 2013
940
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
 

Thread Starter

alkopop79

Joined Jul 30, 2012
12
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?
 

LvW

Joined Jun 13, 2013
1,229
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?
Yes, this may be a problem because - I assume - you want to bias the opamps all at Vcc/2.
It really depends on the circuitry of the gain stages and how they are coupled.
Show us the circuit and we can be more specific.
 

MrChips

Joined Oct 2, 2009
23,234
Why do you want to put decoupling capacitors on the signal inputs?

Doing so will affect the frequency response of the circuit.
 

LvW

Joined Jun 13, 2013
1,229
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).
 

Thread Starter

alkopop79

Joined Jul 30, 2012
12
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).
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?
 

MrChips

Joined Oct 2, 2009
23,234
An AC coupling capacitor aka DC blocking capacitor is used between stages for one of three reasons which all constitute the same reason:

1) when the DC offset of the signal from the preceding stage does not match the DC bias requirements of the next stage,

2) when the DC offset inherent in the input or output stage of the amplifier may drift significantly for various reasons, particularly with temperature, and thus alter the operating conditions and characteristics of the circuit,

3) when it is desirable to limit the low frequency response of the circuit.
 

Alec_t

Joined Sep 17, 2013
12,001
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.
 

Thread Starter

alkopop79

Joined Jul 30, 2012
12
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.

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?
 

shteii01

Joined Feb 19, 2010
4,644
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?
Ok. The best I can figure out you would need to use this formula:
100k at base is approximately equal to 0.1*(1+beta)*R of emitter

So. If you know beta of the transistor, you can calculate R of emitter.
 

ronv

Joined Nov 12, 2008
3,770
I'm a little confused by your schematic. I see 0V and GND with the same symbol. Are they the same. Are you trying to power everything thru the 100K resistors at the top right? What is the input voltage there? What does the digital logic run from?
 

Alec_t

Joined Sep 17, 2013
12,001
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.
If you want that expressed as a formula, Ie = (Vb - Vbe)/Re
 
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