#### pager48

Joined Nov 25, 2018
161
Is there a simple way to adjust fall time of a simple ne555 square wave oscillator? Preferably with a pot.

#### ericgibbs

Joined Jan 29, 2010
10,213
hi 48,
What are the details of frequency and fall times you have in mind.?
E

#### pager48

Joined Nov 25, 2018
161
hi 48,
What are the details of frequency and fall times you have in mind.?
E
That would be anywhere from 100hz to 1.25Khz. The output frequency is going to be varied by a pot but the falling edge should be constant ~545uS +/-125uS long.

#### ericgibbs

Joined Jan 29, 2010
10,213
Is there a simple way to adjust fall time of a simple ne555 square wave oscillator? Preferably with a pot.
hi,
You said in your opening post that the falling edge would adjusted by using pot. but now you say it should be constant at ~545uS +/-125uS long.

If an exponential fall time was acceptable, you could use a diode on the 555 output that charges a capacitor, which is discharged by a parallel resistor.
E

#### pager48

Joined Nov 25, 2018
161
hi,
You said in your opening post that the falling edge would adjusted by using pot. but now you say it should be constant at ~545uS +/-125uS long.

If an exponential fall time was acceptable, you could use a diode on the 555 output that charges a capacitor, which is discharged by a parallel resistor.
E
The pot was for fine adjustment.

Is there a linear method?

#### ericgibbs

Joined Jan 29, 2010
10,213
hi,
Linear could be done by using a Constant Current discharge of the cap.?
Do you want the slope to be from Vmax to 0V.?

E

#### pager48

Joined Nov 25, 2018
161
hi,
Linear could be done by using a Constant Current discharge of the cap.?
Do you want the slope to be from Vmax to 0V.?

E
Need it to slope from max to base line.

Maybe an MMBT3904 constant current source?

#### ericgibbs

Joined Jan 29, 2010
10,213
That MMBT3904 on Google is a transistor type, is that what you mean.?

#### pager48

Joined Nov 25, 2018
161
The MMBT3904 is for a constant current source for the capacitor to discharge make the linear falling edge.

Would a gate driver correctly output the falling edge into a MOSFET gate?

#### ericgibbs

Joined Jan 29, 2010
10,213
OK, please post a datasheet for MMBT3904 and I can look thru it.

#### pager48

Joined Nov 25, 2018
161
OK, please post a datasheet for MMBT3904 and I can look thru it.
Its only 2n3904 in SMT but was a constant current source what you meant by linear falling edge?

#### AnalogKid

Joined Aug 1, 2013
8,493
If the question is - can a 3904 be used to construct a constant current source that will discharge a capacitor for form a linear voltage ramp, the answer is yes - BUT -

A constant current course (CCS) circuit needs an operating headroom of at least one base-emitter voltage drop (Vbe). With an NPN transistor and the emitter tied to GND, the linear part of the output waveform would stop at approx +0.6 V.

AND, because it is a current source circuit, the output impedance will be high. A load of even 1 mA will distort the linearity of the waveform.

A cure for both of these issues is bipolar power supplies powering an output buffer.

ak

#### pager48

Joined Nov 25, 2018
161
There is no distortion on the output. Is it just a spice artifact?

Last edited:

#### TeeKay6

Joined Apr 20, 2019
572
There is no distortion on the output. Is it just a spice artifact?

What load (resistance, capacitance, inductance) must Vout drive while maintaining the rise/fall times you spec? For example, in your schematic, Vout will fall more slowly as the load capacitance rises.

#### Audioguru

Joined Dec 20, 2007
11,249
The 100 ohms input resistor quickly charges the 220nF capacitor but the 7mA current source slowly discharges it producing a falling ramp voltage.

#### TeeKay6

Joined Apr 20, 2019
572
The 100 ohms input resistor quickly charges the 220nF capacitor but the 7mA current source slowly discharges it producing a falling ramp voltage.
@Audioguru
If you were responding to @TeeKay6:
That is indeed what is expected at the base of the final 2N3904. It is not what will be seen at the emitter of that transistor IF there is any significant capacitive load. There is no active pull-down at Vout, only a 1Meg resistor.

#### Audioguru

Joined Dec 20, 2007
11,249
Of course the 1M emitter resistor has a very low discharge current. I was even going to mention the capacitance of the transistor plus stray capacitance.