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Overdrive voltage relationship to the op-amp speed

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Junus2012

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Dear friends,

In the Jakob book he discussed about selecting the suitable value of the overdrive voltage of the transistors in the operational amplifier. for long channel he is using Vov = 250 mV (VDD= 5V).

My question how the overdrive voltage affect the speed of the operational amplifier ? consider please for example GBW = gm1,2/CL. in this equation only the input differential transistors are concinerned, and I can say yes higher overdrive voltage leads to higher gm1,2 and thus increases the gain. but what about other transistors? they are not included in the formula of GBW so why should they follow this value of overdrive voltage ?

Thank you in advance
 

Not just the gm. Everything past the front end sees a current
-difference- as the "driving force" for large signal swings. If
you think of the gain stage input as a simple capacitor and
call its slew rate a proxy for the output slew, what's the
difference between a diff pair driven to a (say) 99nA / 101nA
tail current pair, and a 0 / 200nA pair? 2nA vs 200nS difference
current, fighting over where the gain stage out to go... right.
 
Dear friends,

In the Jakob book he discussed about selecting the suitable value of the overdrive voltage of the transistors in the operational amplifier. for long channel he is using Vov = 250 mV (VDD= 5V).

My question how the overdrive voltage affect the speed of the operational amplifier ? consider please for example GBW = gm1,2/CL. in this equation only the input differential transistors are concinerned, and I can say yes higher overdrive voltage leads to higher gm1,2 and thus increases the gain. but what about other transistors? they are not included in the formula of GBW so why should they follow this value of overdrive voltage ?

Thank you in advance

GBW = gm/2piCl and it increases linearly with increasing overdrive only if you increase it by increasing the current and not changing the W/L. If you keep the current constant and then increase the W/L, your Vgs-Vth will come down (lesser overdrive for the same current), but gm (= 2Id/(Vgs-Vth)) will increase. But this will increase the drain to bulk capacitance of your input pair. This will add to your load capacitance and decrease the speed.

As for other transistors go, think about a differential pair loaded with a current mirror load. Say you want to increase the output swing as much as possible but don't want to change the current. Then you will have to reduce the Vdsat of the load devices and because the current is constant, you will have to increase their size. That will increase the drain to bulk capacitance which will add to the load capacitance and reduce the GBW of your differential pair.

Now you can add another stage and repeat the same for the load current source as well.

GBW=gm1,2/Cl is valid for a simple differential pair but it is assumed that the parasitics have been included in the load capacitance term.
 
Hi,

I wonder... I can't remember to ever have read "overdrive voltage" in combination with OPAMS.

But I've read it in combination with COMPARATORS.

OPAMS and COMPARATORS have different requirements, regarding input voltage, output voltage, feedback, stability ...

Klaus
 
Hi,

I wonder... I can't remember to ever have read "overdrive voltage" in combination with OPAMS.

But I've read it in combination with COMPARATORS.

OPAMS and COMPARATORS have different requirements, regarding input voltage, output voltage, feedback, stability ...

Klaus

Vgs-Vth is called the overdrive in many textbooks.. Vth is the minimum drive required and anything above that is "over"-drive. :D

P.S. Even Wikipedia has an article on it: https://en.wikipedia.org/wiki/Overdrive_voltage
 
Hi,

Yes, I knew this with MOSFETs... but I'm not into OPAMP design.

Klaus
 

Dear friends,

Thank you for your helpful discussion,

I can understand now that increasing gm by increasing the current is different from increasing it by increasing W/L. The first method increases both gm and overdrive voltage and hence speed while increasing it by the second method will increase the drain to capacitor which reduce the speed. So basically the two method are not linearly gkiving the same sloution of speed
 

Also increasing current increases speed in another way. The current available during slewinv increases.
 
Presumably the open loop gain vs frequency plots contain all the information.

If you want the opamp output to be X volts you need the inputs to be X/gain. In short periods of time bandwidth limits the gain, hence over-driving may get a quicker response. You should be able to work backwards from the gain plot to calculate response time versus overdrive.

Slew rate is the final consideration however.
 
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