In my opinion Dave you should clearify better that the differential gain of the differential amplifier is the same, regardless if the input is single ended or differential. (Vo1-Vo2)/Vin = Avd. In your table it suggests that the differential gain of the differential input version is twice the one of the single ended version (-2Rc/re vs -Rc/re). I actually know what you meant (which is not the differential gain) but it still a bit confusing.
Why isn't the connection between the two emitters also connected to ground in the Small Signal Model (AC) since in they are connected to constant DC (at least for common mode)? In which case, wouldn't the two emitter currents just flow to ground?
Andrew Hughes make sure that constant voltage and constant current is different. for Constant voltage is short circuit while constant current is open for removing source. Btw, i read some demonstration, Iee is replaced by a Re resistance and connected to ground. for the best performance , Re is as large as possible so that a current source is replaced as it is seen as resistance infinity .
So is this the simplest operational amplifier that can be made? Does it meet all the definitions of an op amp? The inverting and non-inverting inputs of an op amp are a a differential amplifier, which is why I ask.
This is old but... I think one of the inherent characteristics of an operational amplifier is the absurdly high voltage gain. So, I think that the simplest operational amplifier can be constructed with this differential amplifier and a block that give an ultra-high voltage gain in cascade.
hi there, really great video, thanks for putting it up. i am just a little unclear about when you introduce the differential input. you say define the input of Vi2 as 180 degrees out of phase. its not out of phase though, right? like the input should be just inverted polarity, but in phase. thats how balanced signals work i thought.
Ib is like a gate for Ic to Ie. If Ib > 0: current can flow from Ic to Ie and Ie = bèta * Ic. If Ib = 0: the gate is 'closed' and no current flows from Ic to Ie.
@@ElectronXLab Thank you so much,I have seen some questions about this top,c but they didn't have exact symmetry and I thought i this way,but now it's clear,thanks :)
You're simply superb! Love your explanation style. Makes the most hard to crack circuits appear so easy to solve
serious comment. this is very clear and simple video to understand. top 1 tutor! I have watched few videos from him and most of them are fantastic.
In my opinion Dave you should clearify better that the differential gain of the differential amplifier is the same, regardless if the input is single ended or differential. (Vo1-Vo2)/Vin = Avd. In your table it suggests that the differential gain of the differential input version is twice the one of the single ended version (-2Rc/re vs -Rc/re). I actually know what you meant (which is not the differential gain) but it still a bit confusing.
Remind me how to compute re. Suppose I don't have a base resistor how do I compute the base current?
great video on differential amp
Thanks for the great work! I may have missed something, but what is the common emitter between Q1 and Q2 usually connected to? ground?
Why isn't the connection between the two emitters also connected to ground in the Small Signal Model (AC) since in they are connected to constant DC (at least for common mode)? In which case, wouldn't the two emitter currents just flow to ground?
Because there is current source which is open for ac...disconnecting the vee to ground
Andrew Hughes make sure that constant voltage and constant current is different. for Constant voltage is short circuit while constant current is open for removing source. Btw, i read some demonstration, Iee is replaced by a Re resistance and connected to ground. for the best performance , Re is as large as possible so that a current source is replaced as it is seen as resistance infinity .
So is this the simplest operational amplifier that can be made? Does it meet all the definitions of an op amp? The inverting and non-inverting inputs of an op amp are a a differential amplifier, which is why I ask.
This is old but... I think one of the inherent characteristics of an operational amplifier is the absurdly high voltage gain. So, I think that the simplest operational amplifier can be constructed with this differential amplifier and a block that give an ultra-high voltage gain in cascade.
hi there, really great video, thanks for putting it up. i am just a little unclear about when you introduce the differential input. you say define the input of Vi2 as 180 degrees out of phase. its not out of phase though, right? like the input should be just inverted polarity, but in phase. thats how balanced signals work i thought.
plz what is re in 4:00 ? is it the entire resistor of the transistor ?
It's the equivalent resistance in the emitter at the particular operating point (DC bias point) of the transistor
How do you calculate v1 and v2 in a differential amp with no resistors specifically a bjt with a mirror
4:24
Why does Ie = 0 automatically mean that Ic and Ib = 0?
Ib is like a gate for Ic to Ie. If Ib > 0: current can flow from Ic to Ie and Ie = bèta * Ic. If Ib = 0: the gate is 'closed' and no current flows from Ic to Ie.
normally, ie = ib + ic .. It means that if ie = 0, the sum of the two is 0, meaning both are at 0A, assuming there are no complex plane currents
you are the Best
Hope All Good for You Bro
This just hurts my brain trying to understand all these equations...It mega hurtz.
I hope I didn't amplify your pain
What about input and output impedance's for this amplifier?
input: 2re while ouput :2Rc
how to calculate re from r_pi or rb.
WHAT IS VD1?
Everything is awesome except at the last part Add should not be multiplied by 2
It should be just -Rc/re
how do you find re though?
re is determined by the DC bias(VCC to be exact), usually use VT/IB(mA) this formula to get re, VT is just a constant normally around 25
Mutant Llama this resistance only affect AC signal only.
@@fanpeter-z3c Hi, in this case there is no DC bias, so IB is 0 (I assume?), then isn't VT/IB close to infinity?
If there's not Rc in one of the branch can we still say both Ic's are equal in DC analysis?
Rc does not affect the Ic, it will only affect the point at which the transistor enters saturation.
@@ElectronXLab Thank you so much,I have seen some questions about this top,c but they didn't have exact symmetry and I thought i this way,but now it's clear,thanks :)
with a current mirror sorry meant to say with my first comment