VCO

Thread Starter

Zeeus

Joined Apr 17, 2019
616
hi Z,
Tried that, removed R9.
LTSpice cannot solve the equations so goes into a continuous loop.

So no waveforms are created.

Why do you want to remove R9.?
E
I see

Trying to understand
Can current flow from the source of the JFET to the drain?
 

Jony130

Joined Feb 17, 2009
5,598
D1 and D2 is just for input protection?
No, these diodes together with Zener diodes and R7 and R8 form the inverting Schmitt trigger.
Without these diodes, this Schmitt trigger cannot work. So you would have to use a traditional Schmitt trigger.
Trying to understand
Can current flow from the source of the JFET to the drain?
Yes, it can. In this circuit, the JFET is working as a switch.
When the voltage at A3 output is "Low" the JFET switch is "OFF" (open circuit, no current flow). On the other hand, when the voltage is "High" (A3 output) JFET switch is "ON" because the D5 diode is reverse biased thus JFET gate is pull-down via R10 resistor to GND so, JFET is "ON" (conducting current).


Also, notice that when the JFET is "ON" the current is equal to:

I_sd = 2*I_inV = 2*1V/20kΩ = 100μA
. (if we ignore the JFET Rds_on)

Iin = Vin/20kΩ = 50μA


Thus the capacitor current is I_cap = I_sd - I_in = 50μA


12d.PNG
Therefore the voltage at the A2 output will rise with the rate of I/C = 27.78mV/μs

So all you need to know to determine the output frequency equation it to know the Schmitt trigger threshold voltage.
 

Thread Starter

Zeeus

Joined Apr 17, 2019
616
No, these diodes together with Zener diodes and R7 and R8 form the inverting Schmitt trigger.
Without these diodes, this Schmitt trigger cannot work. So you would have to use a traditional Schmitt trigger.

Yes, it can. In this circuit, the JFET is working as a switch.
When the voltage at A3 output is "Low" the JFET switch is "OFF" (open circuit, no current flow). On the other hand, when the voltage is "High" (A3 output) JFET switch is "ON" because the D5 diode is reverse biased thus JFET gate is pull-down via R10 resistor to GND so, JFET is "ON" (conducting current).


Also, notice that when the JFET is "ON" the current is equal to:

I_sd = 2*I_inV = 2*1V/20kΩ = 100μA. (if we ignore the JFET Rds_on)

Iin = Vin/20kΩ = 50μA

Thus the capacitor current is I_cap = I_sd - I_in = 50μA


View attachment 212058
Therefore the voltage at the A2 output will rise with the rate of I/C = 27.78mV/μs

So all you need to know to determine the output frequency equation it to know the Schmitt trigger threshold voltage.
Thanks

Understood the frequency and R5 has to be 1/2 of R4
Thanks

Please what's use of R6?

Also, TS doesn't think the diodes and zener are necessary for operation. Can just do normal by using maybe 10k from output to "+" input and 10k from "+" input to ground

Correct? (SPice not working for me)

Most importantly, please how JFET is operating as switch?
Though it is as analog switch when the signal is at the drain or source? confused
 

Jony130

Joined Feb 17, 2009
5,598
Please what's use of R6?
It looks like the only purpose of R6 resistor is to limit the VDS so the JFET can work in triode region (ohmic region) or for safety reason (Vgd_max)
Also, TS doesn't think the diodes and zener are necessary for operation. Can just do normal by using maybe 10k from output to "+" input and 10k from "+" input to ground
It won't work. Try to calculate the threshold voltage in such a case when R1 = R2. And dos do not forget about the "phase" (inverting Vs non-inverting Schmitt trigger )
https://en.wikipedia.org/wiki/Schmitt_trigger#Inverting_Schmitt_trigger

Most importantly, please how JFET is operating as switch?
Though it is as analog switch when the signal is at the drain or source? confused
The JFET is a bidirectional device thus source and drain can be interchanged. And it will behave as a switch because JFET is "ON" when the Vgs = 0V and the actual drain current (Id) is smaller than the drain saturation current (Idss).
 

Thread Starter

Zeeus

Joined Apr 17, 2019
616
It won't work. Try to calculate the threshold voltage in such a case when R1 = R2. And dos do not forget about the "phase" (inverting Vs non-inverting Schmitt trigger )
https://en.wikipedia.org/wiki/Schmitt_trigger#Inverting_Schmitt_trigger
Thanks
So dc current can flow from source to drain of NJfet?

Strange, will test it

Really, it won't work?
Using 1k from output to '+' and 1k from '+' to ground (remove all the diodes)
like in the image

When output of A3 is 15V, "+" is 7.5V
when it is -15V "+" is -7.5V

Missing something?
Screen Shot 2020-07-13 at 1.12.28 PM.png
 

Jony130

Joined Feb 17, 2009
5,598
Really, it won't work?
Using 1k from output to '+' and 1k from '+' to ground (remove all the diodes)
like in the image

When output of A3 is 15V, "+" is 7.5V
when it is -15V "+" is -7.5V

Missing something?
Ah yes, with this configuration it will work. But I was thinking about this configuration (shorting the diodes)
https://en.wikipedia.org/wiki/Schmitt_trigger#/media/File:Op-Amp_Schmitt_Trigger.svg

Thanks
So dc current can flow from source to drain of NJfet?
How can a gate tell which terminal is a Source and which one is a Drain?
 

Thread Starter

Zeeus

Joined Apr 17, 2019
616
Last edited:

Thread Starter

Zeeus

Joined Apr 17, 2019
616
Thanks..
PDF has a sample and hold was trying to understand

Please someone kindly make Spice in #28 work (when you can)
 
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