3 MHz Sawtooth Generator Needed with Discrete Parts Only!

[crutschow]

It may be obvious, but it should be noted that any load on the C capacitor node will affect the waveform. You will likely need to add a buffer, such as an emitter follower, to that node before you connect it to any other external circuitry.

 

[Roff]

It may be obvious, but it should be noted that any load on the C capacitor node will affect the waveform. You will likely need to add a buffer, such as an emitter follower, to that node before you connect it to any other external circuitry.

Good point. A PNP emitter follower is probably the best choice, for a couple of reasons:
1. There is a lttle headroom for a current source to bias it, which can't be said or an NPN.
2. A PNP can better handle the high current demand of the fast negative transition acting on a capacitive load.

 

[Teketron357]

Good point. A PNP emitter follower is probably the best choice, for a couple of reasons:
1. There is a lttle headroom for a current source to bias it, which can't be said or an NPN.
2. A PNP can better handle the high current demand of the fast negative transition acting on a capacitive load.

Roff, can you please explain why you chose to use that configuration of emitter follower? I mean the two PNP BJTs (Q1 & Q2) configured as shown on the image.
Q2 is being directly biased by the R10 10K resistor, also I see that C3 100nF has been added. Also, you are simulating the Rload and Cload as if they were actual loads, correct?

Regards,
-Steve.

 

[Teketron357]

If you can't get a 2N5771 try a 2N3906. Sim shows it works nearly as well.

Thanks, I was thinking the same thing. I really was asking on the 2N5771 because it seems as it is the only PNP high speed saturated switching in existence <--- This is odd, normally manufacturers will provide an equivalent.

Thank you,
-Steve.

 

[Roff]

Thanks, I was thinking the same thing. I really was asking on the 2N5771 because it seems as it is the only PNP high speed saturated switching in existence <--- This is odd, normally manufacturers will provide an equivalent.

Thank you,
-Steve.

I think you can deduce from this that there is little demand for a high speed, low current saturating PNP.

EDIT: You will probably find that base-emitter breakdown of the PNP will screw up the linearity of your ramp. Base-emitter breakdown is not modeled in spice on any transistor that I have seen.

 

[alec_t]

Base-emitter breakdown is not modeled in spice on any transistor that I have seen.

Which is a shame. I wonder why? Presumably you can slap a zener across the b-e junction to give a similar effect in a sim, though how valid that would be over a wide frequency range is uncertain.

 

[Roff]

Which is a shame. I wonder why? Presumably you can slap a zener across the b-e junction to give a similar effect in a sim, though how valid that would be over a wide frequency range is uncertain.

Zeners have too much capacitance.
Here is a subcircuit I made last year for a 2N3906 with base-emitter breakdown included. I'm sure it is not a completely accurate model, but I'm pretty sure it's better than a zener.
Note that the breakdown device is actually a diode with 6 volts forward voltage drop, in parallel with the BE junction.

*
.subckt 2N3906bd 1 2 3
Q1 1 2 3 0 2N3906
D1 2 3 IdealZener6V
.model D D
.lib C:\Program Files\LTC\SwCADIII\lib\cmp\standard.dio
.model NPN NPN
.model PNP PNP
.lib C:\Program Files\LTC\SwCADIII\lib\cmp\standard.bjt
.model IdealZener6V D (Ron=10 Roff=1e12 Vfwd=6)
.ends

 

[crutschow]

...........................

EDIT: You will probably find that base-emitter breakdown of the PNP will screw up the linearity of your ramp. Base-emitter breakdown is not modeled in spice on any transistor that I have seen.

Which transistor and when is the reverse base-emitter voltage high enough to cause breakdown?

 

[Roff]

Which transistor and when is the reverse base-emitter voltage high enough to cause breakdown?

In the simulation, the Q6 base voltage rises faster than the emitter. The breakdown occurs for about 60nS, beginning right after the ramp starts.
It can be avoided by increasing the time constant on the base of Q6, by changing the PNP to a higher capacitance device and/or increasing the Thevenin resistance of R6/R7.
The fix I like best is to bootstrap the base of Q6 to the emitter by adding a 1N4148 diode from the output of the emitter follower to the base of Q6, anode to the base of Q6. This prevents the base of Q6 from rising faster than the emitter. It holds reverse Vbe to about 1.5V.

 

[Roff]

Roff, can you please explain why you chose to use that configuration of emitter follower? I mean the two PNP BJTs (Q1 & Q2) configured as shown on the image.
Q2 is being directly biased by the R10 10K resistor, also I see that C3 100nF has been added. Also, you are simulating the Rload and Cload as if they were actual loads, correct?

Regards,
-Steve.

Q1 is the emitter follower. Q2 is a current source, providing a total of 4mA to the emitter follower and the load. Note that the current through the load capacitor is high (I=C*dv/dt) during the fast ramp reset. The PNP emitter follower sinks this current.
Yes, Rload and Cload are as labeled - loads.

 

[alec_t]

@Roff
Thanks for the breakdown model.

 

[Roff]

@Roff
Thanks for the breakdown model.

You're welcome.
Obviously, you can add it to any transistor, either internally (by creating a subckt) or externally.

 

[unclejed613]

using a current source for the load on Q1 makes the output more linear. this is the same "design trick" used inside op amps to improve linearity (before feedback is applied). with just a resistor as a load, the current in the output circuit would change, and make the output nonlinear. not as badly as if it were a common emitter amp. current sources are more often used as active loads for common emitter amps.

 

[Roff]

using a current source for the load on Q1 makes the output more linear. this is the same "design trick" used inside op amps to improve linearity (before feedback is applied). with just a resistor as a load, the current in the output circuit would change, and make the output nonlinear. not as badly as if it were a common emitter amp. current sources are more often used as active loads for common emitter amps.

It is even more important to linearity when the base voltage is a ramp generated by a constant current charging a capacitor. A resistive load on the emitter follower contributes to ramp nonlinearity on the capacitor. A resistor used for emitter bias would make that even worse.

 

[hiboux]

Hi,

I know this is an old post, but I tested this schematic (https://www.electro-tech-online.com/attachments/sawtooth-gen-3mhz-png.69403/) and it works pretty well. But I want it to work with a 5V supply.
I tried to find the equations for the resistances but it's a little bit difficult...

Can somebody help me? It would be very nice!

Thank you for your help!

 

[crutschow]

Try changing R1 to 3k ohm, C1 to 370pF, R3 to 2k ohm, and R4 to 169 ohms.

 

[hiboux]

Thank you! It works well! But, how did you know the values of the components?
Because if I want to change the frequency (for example 270kHz), when I change only the capacitor value the generator doesn't work anymore... I tried to change R1 and R3 but nothing happened.

I know how this schematic works, but I dont't understand how to simply change the values correctly...

 

[crutschow]

I'm lazy so I experiment with changing the values and simulating the circuit to see what happens. I know what each section of the circuit does so I can make informed guesses about what part values to change.

For example:

C1 along with the value of the constant current through Q3 is the primary factor that determines the oscillation frequency so changing C1 changes the frequency.

The constant charging current to C1 is determined by Q3 and Q4 along with the value of R4 so you can adjust R4 to adjust the frequency (R4 could be a pot if you want to fine tune the frequency in the actual circuit).

Q1 and Q2 perform a bistable function similar to an SCR. The upper trigger point of that (which discharges C1) is determined by the voltage at the base of Q2 which is generated by the voltage divider of R1 and R2. If that voltage is too high or too low the circuit may not oscillate. The voltage determines the peak value of the output. If the voltage is too high the peak will become rounded and compressed.

R3 affects the sharpness of the trigger point for C1's discharge and the linearity at the top of the sawtooth peak.

Using the circuit values in my previous post while changing C1 to 5.1nF gives an output frequency of about 271kHz. See below.

 

[hiboux]

Ok, I know why I couldn't change the value... It's because I was precharging C1 with 15V.... Now I can change them like I want!

Thank you verry much for your help and for your explanations!