driver compression logic question

Thread Starter

yef smith

Joined Aug 2, 2020
805
Hello, I have the following circuit where the heart is the BJT.when i put the resistor to be 50Ohms i get a good sine current on the inductor L1 shown below.
But when i put 25 ohms i get compression .
In the compressed version i see that the current going threw the inductor is much higher and it seems like for input voltage of 3.8V the current reaches its maximum and going threw a compression.

does the current compression situation happens because the PNP NPN switch threw other states?
What is the logic of the cause the current threw L1 gets compressed?
Thanks.

1702657882404.png

1702657898990.png
 

Thread Starter

yef smith

Joined Aug 2, 2020
805
Hello ,even when i flipped the PNP i got the same compression plot.
hi guess that Q1 and Q2 are not opening simulatiosly.I assume that Q1 and Q2 enter the saturation state at some point Correct?
why saturation state in Q1 and Q2 causes current threw L1 to stop increasing?
Thanks.

1702661921009.png
 

bertus

Joined Apr 5, 2008
22,289
Hello,

Reduce the values of R1 and R3 to 10 Ohms and see what happens.
It can be that those resistors limit the current.

Bertus
 

Thread Starter

yef smith

Joined Aug 2, 2020
805
Hello Bertus, do you think that R1 and R3 plays a role with the state of Q1 and Q2.
So when we change R1 and R3 our Q1 and Q2 not changing state during voltage input swing?
Thanks.
 

bertus

Joined Apr 5, 2008
22,289
Hello,

Can you post the .asc file?
When the transistor is in saturation, R1 limits the top current and R3 the bottem current.
Put probes on the basis of both transistors.
You will see that only one conducts.

Bertus
 

bertus

Joined Apr 5, 2008
22,289
Hello,

I have drawn the current source myself:
Current source LTspice.png
I only used some other transistors, as I could not find the ones you used.

I even reduced the sens resistor to 10 Ohms. I did reduce the emittor resistors to 3.3 Ohms for no clipping.
Current source LTspice 10 Ohm.png

In the following picture the base voltages are given:
Current source LTspice 10 Ohm with base voltages.png

Bertus
 

Ian0

Joined Aug 7, 2020
9,981
Hello,

I have drawn the current source myself:
View attachment 310072
I only used some other transistors, as I could not find the ones you used.

I even reduced the sens resistor to 10 Ohms. I did reduce the emittor resistors to 3.3 Ohms for no clipping.
View attachment 310073

In the following picture the base voltages are given:
View attachment 310074

Bertus
That's more than a little bit unstable!
I think a Bode plot and some compensation capacitors is called for.
With that arrangement of inductance and feedback the voltage gain increase at 6dB per octave, and at some point reaches the op-amp open-loop gain which reduces at 6dB/octave. At some point there is phase-shift of greater than 180° at a point where the gain is more than unity!
 

bertus

Joined Apr 5, 2008
22,289
Hello,
That's more than a little bit unstable!
I think a Bode plot and some compensation capacitors is called for.
With that arrangement of inductance and feedback the voltage gain increase at 6dB per octave, and at some point reaches the op-amp open-loop gain which reduces at 6dB/octave. At some point there is phase-shift of greater than 180° at a point where the gain is more than unity!
How do you implement the compensation?

Bertus
 

Ian0

Joined Aug 7, 2020
9,981
Hello,

How do you implement the compensation?

Bertus
First I might try a zobel network on the output. Then, if that doesn't work I would insert a resistor between the current-sense resistor and the inverting input, then add a capacitor from output to inverting input. Not sure without trying it whether the capacitor should go from the op-amp output pin or the circuit output on the collectors of the transistors.
The op-amp will be compensated for standalone operation, but this circuit changes its open-loop gain and phase response, and then adds a load which requires the gain to rise with frequency. It doesn't bode well (apologies for the pun).
 

Alec_t

Joined Sep 17, 2013
14,368
According to the datasheet , for a +-15V supply the LT1028's output swing decreases with load and is only about +-12V for a load of 600Ω, so likely a lot less for the load being driven here. Hence the clipping that is occurring.
 

Ian0

Joined Aug 7, 2020
9,981
According to the datasheet , for a +-15V supply the LT1028's output swing decreases with load and is only about +-12V for a load of 600Ω, so likely a lot less for the load being driven here. Hence the clipping that is occurring.
But the LT1028 is not driving the load. The transistors do. It was clipping because the PNP transistor was reversed.
The circuit is a standard application note circuit to boost the output current of an op-amp.
 

Ian0

Joined Aug 7, 2020
9,981
I don't understand how that is supposed to work. Why are the transistor bases connected to the supply?
As the op-amp tries to supply power to the resistor on the output it draws current from the supply, which results in a voltage drop across R2 and R4. When that voltage drop exceeds 0.6V the transistors start to switch on supplying power to the load.
There are versions where R5 is connected to ground and versions where it is connected to the final output.
In this case, it is connected to the final output.
http://www.t-es-t.hu/download/analog/an106.pdf
 

Ian0

Joined Aug 7, 2020
9,981
In my 10+ years of seeing electronic circuits, I've never seen one like this where you intentionally drop voltage to supply the op amp, and it's not because its voltage should be lower than the supply voltage. IMHO you should redesign your circuit paying attention to this detail too. I didn't delve too much into it because it seems a wrong circuit to start with, but imagine that downward current depends on how much the op amp consumes, this dictates how much voltage drop you have for each transistor to operate. Seems wrong to me.
Can't be that wrong if it's in a Analog Devices Application note!
 

kiroma

Joined Apr 30, 2014
66
Can't be that wrong if it's in a Analog Devices Application note!
I deleted my comment because when I was typing that comment then another post with the application note came up.
Even though, there's the 100 ohms resistor in series with the output. I don't see an use for that.
 

kiroma

Joined Apr 30, 2014
66
I see that if the 100 ohm resistor (initially R5) is there, then you must have a lot of voltage to push those transistors into conduction (to help the op amp). In the application note there isn't a circuit where you do that.
So if you have a "free output", then there isn't a need for high voltage to conduct, thus conducting and not saturating the output.
 
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