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how convert squarewave to sawtooth

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mohamadzi

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hello
i convert squarewave to sawtoothwave with this circuit
figure_01.gif

it worked well with squarewave 100khz as input , of course by change c1 ,c2 to properly value.
but i try to use this ciruit in 1MHz but dose not worked,please help me!
 

Seems as if this OpAmp could be not suited to work at such frequency, once exhibits a Gain-Bandwidth Product of 1-1.5MHz, which fits on the same magnitude of the working frequency. What exactly didn´t work ?




+++
 
Look at the datasheet for the opamp:
1) The opamp is supposed to be an integrator. It isn't at 1MHz because at that frequency it has no voltage gain.
2) The low slew rate of the opamp begins to cut its output above only 30kHz.
 
thanks , but i used LT1361 for integrator that work in 70MHz and slew rate 1000v/us but again this circuit not good work and sawtoothwave don't shaped well in the output.
i
i request that guide me about how choose c1 and c2 R1 R2 to work cicruit better.
 

The LT1361 should be fast enough to generate the 1MHz sawtooth. You need approximate values for R1 = 2.13kΩ and C1 = 150pF.

Don't know about R2 and C2 since it's not clear to me what they do.

Do you have a good high-speed layout with a 0.1µF ceramic cap directly from the op amp power pin to a good ground (preferably copper ground plane) and short connections for all the high speed circuit paths?
 
i think that 74hc4053 don't work well in this frequency, in datasheet this frequency response presented
Untitled.png
so is it reason for circuit dosn't work well????
please help me to convert squarewave to sawtoothwave in 1MHz from different way.
 

hello
i convert squarewave to sawtoothwave with this circuit
View attachment 107530

it worked well with squarewave 100khz as input , of course by change c1 ,c2 to properly value.
but i try to use this ciruit in 1MHz but dose not worked,please help me!

Do you have any oscilloscope waveforms you can share?

Because the statement "it doesn't work" is ambiguous.
 

Do you have any oscilloscope waveforms you can share?

Because the statement "it doesn't work" is ambiguous.

this is waveforms in 100khz that worked well
100khz.png
but when input is 1Mhz the waveform
1meg.png
worked weakly.
of course that i changed c1 c2 r1 but not useful.
 

The circuit shown in post #1 is designed for a frequency of 1 kHz, respectively the 15 us time constant of the charge pump circuit isn't a problem. It sounds unlikely that the 100 kHz waveforms are achievec with the original circuit component values.

When modifying the circuit to work at a higher frequency, several component values have to be changed, not only the OP bandwidth. I also believe that there are more effective circuits for high frequency sawtooth generators.
 
The circuit shown in post #1 is designed for a frequency of 1 kHz, respectively the 15 us time constant of the charge pump circuit isn't a problem. It sounds unlikely that the 100 kHz waveforms are achievec with the original circuit component values.

When modifying the circuit to work at a higher frequency, several component values have to be changed, not only the OP bandwidth. I also believe that there are more effective circuits for high frequency sawtooth generators.

yes for 100khz i changed component values and c1 and c2 changed to 560p and 820p
tell more about effective circuit for high frequency sawtooth generators and consider that i need a square converter to sawtooth and my input is squarewave.
 

Operation at 100kHz is no-good. Rounding at bottom and overshoot at top. Unless you are satisfied with that.
 
Below is the simulation of a simple transistor sawtooth circuit using a current-mirror current source to charge the capacitor and a short pulse to discharge it. Q4 and Q5 are complementary emitter followers (for cancellation of the emitter follower offset) to buffer the output.

Sawtooth.gif
 
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Operation at 100kHz is no-good. Rounding at bottom and overshoot at top. Unless you are satisfied with that.

yes,your right. for my work that's good enough
 
Last edited by a moderator:

Below is the simulation of a simple transistor sawtooth circuit using a current-mirror current source to charge the capacitor and a short pulse to discharge it. Q4 and Q5 are complementary emitter followers (for cancellation of the emitter follower offset) to buffer the output.

View attachment 107556
very very very thanks for your simple soultion
excuse me do you know any replacement or equivalent transistor can replace instead of 2n2369 and 2n3906?
is bc557 and bsx20 good repalcement for that purpose??
 

.................... do you know any replacement or equivalent transistor can replace instead of 2n2369 and 2n3906?
is bc557 and bsx20 good repalcement for that purpose??
Those two appear to be reasonably close to work okay in the circuit but you may have to tweak the values some to get the desired output (which is true in any case).
The value of R3 controls the current-mirror current and thus the amplitude of the sawtooth (along with the value of C1 of course). R3 can be a 10k pot in series with a 2k resistor for adjustment of this.
The value of R1 affects the sawtooth fall-time.
 
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