removing offset voltage from charging capacitor

crutschow

Joined Mar 14, 2008
34,472
For minimum parts count, below is a linear sawtooth generator using a dual rail-rail op amp as a comparator and an integrator, whose output goes close to 0V and the supply rail.
Pot U3 controls the frequency.

(It's basically one of the circuits in Bertus's post # 10 reference).

upload_2019-8-21_19-34-15.png
 
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Thread Starter

pager48

Joined Nov 25, 2018
161
Thats a very decent oscillator. Thanks for getting it to work. What frequency can an opamp with a 0.5V/uS slewrate going to reach at the same 12v rail?

The square waveform already appears to distort but it might be better to just drop the rail to 3.3v-5V and amp it with a transistor instead.



How close to 0V do you need the bottom of the waveform to be?
It should be less than a typical power to220 mosfet Vgs(th) about 2V-4.5V.
 
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crutschow

Joined Mar 14, 2008
34,472
What frequency can an opamp with a 0.5V/uS slewrate going to reach at the same 12v rail?
It's a simple calculation.
With a 0.5V/μs slew-rate it would take 12V * 0.5μs/V = 6μs for 1/2 cycle or 12μs per cycle.
That is a frequency of 83kHz.

Edit: Opps, faulty equation. See Post #27.
it might be better to just drop the rail to 3.3v-5V and amp it with a transistor instead.
Depends upon the highest frequency you need.
 
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Thread Starter

pager48

Joined Nov 25, 2018
161
It's a simple calculation.
With a 0.5V/μs slew-rate it would take 12V * 0.5μs/V = 6μs for 1/2 cycle or 12μs per cycle.
That is a frequency of 83kHz.
Depends upon the highest frequency you need.
For a TLV271CS at 2V/uS its slower

0.5v/uS
((12*0.5e-6)*2)^-1 = 83,333Hz

2v/uS
((12*2e-6)*2)^-1 = 20,833Hz
 

crutschow

Joined Mar 14, 2008
34,472
Sorry my math was faulty. :oops:
Obviously a faster slew-rate should give a higher frequency.

So the correct formula should be the voltage divided by the slew-rate or, for the first op amp, 12V / (0.5V/μs) = 24μs for 1/2 cycle, giving a frequency of 20.8kHz.
For 2V/μs the time for 1/2 cycle is 12V/(2V/μs) = 6μs for 1/2 cycle, giving a frequency of 83.3kHz.
 

Thread Starter

pager48

Joined Nov 25, 2018
161
Whats keeping a single non inverting amplifier from raising the output voltage of the BJT pair? The non inverting needs a buffer amp or it won't amplify.

 
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crutschow

Joined Mar 14, 2008
34,472
Which are the resistors which change the rising and falling edges of the output wave in the opamp only generator?
Pot U3 and C1 determine the U1 integrator time-constant and thus the triangle-wave rise and fall time, along with the frequency.
 
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crutschow

Joined Mar 14, 2008
34,472
Does this oscillator need a rail to rail opamp to remove offset?
I dunno.
Is the offset shown in the simulation tolerable for your application?
What's wrong with using rail-rail op amps?

First you show a triangle-wave oscillator and now you show a sawtooth-wave oscillator.
Which do you want?
(Or don't you care which it is)?
 

iimagine

Joined Dec 20, 2010
511
Thats design is inefficient. Wasting too much current on threshold voltage references. Anyway a simple emitter follower will removes the offset down to .~250mv
 

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Thread Starter

pager48

Joined Nov 25, 2018
161
I dunno.
Is the offset shown in the simulation tolerable for your application?
What's wrong with using rail-rail op amps?

First you show a triangle-wave oscillator and now you show a sawtooth-wave oscillator.
Which do you want?
(Or don't you care which it is)?
The circuit works best with a sawtooth which charges an large inductance to prevent damage to it by charging it slowly. Theres nothing wrong with rail to rail opamps the other circuits posted here are useful.
 

Alec_t

Joined Sep 17, 2013
14,338
Why are you so concerned about removing the offset? You're always going to end up with at least a few mV of offset. For what following stage is the offset critical?
 
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