The app note mentioned in post #13 explains the voltage limitation of the control voltage pin for reliable operation when used in astable mode: 1.7v to Vcc.

eT

 

The app note mentioned in post #13 explains the voltage limitation of the control voltage pin for reliable operation when used in astable mode: 1.7v to Vcc.

eT

Hi,

Oh that's great. Would be nice to see some real life test results too.
Over that limited range it looks quite a bit linear too, nice.

 

Hello

There is a PWM circuit in app note AN-170 that uses the control pin with op amp.

Figure 22 Voltage to Pulse Duration Converter.

eT

Hi et, I was considering using an op amp voltage follower to provide voltage to the control pin. But it is still hard to figure out.
What I am trying to do is use the 555 to provide pwm charging from my solar panel to charge a 12 volt battery.
A solar panel can go as high as 22v. The battery specs allow a maximum of 15 volt to charge the battery. I want the pwm to provide an average voltage around 15v.
Thanks for the reference. Looks like the AN-170 has the answer. This would be an AN for the 555 I take it?

 

Are you looking for a duty cycle only V control, one that does not also
vary frequency ?

Curious, no specs for freq accuracy or range, duty cycle accuracy or range....

Regards, Dana.

Hi Dana, Duty cycle only. Frequency does not matter. I have applied on tina-ti software 1 to 6v on the control terminal for a 8v 555 supply. The low period is constant while the hi period changes thus the frequency changes. Also of course the duty cycle changes a lot to.

 

<wrong link> on using a 555 for a pulse width modulation circuit includes an equation for the component values. I just searched for it and this onecame up. My search criteria was “555 pulse width modulation calculator”.

Note: posted the wrong link in the original post

Hi Fanatic, Looked that up. does not use control voltage.

 

Hi,

Not that complex see post before yours.
Also it's somewhat linear already.
Note 1 is 100 percent, 0.1 is 10 percent.
Vcc=5v.
RA=RB/5.

View attachment 182218

 

Hi Hymie, That looks like the answer. A constant current source. I knew the non linear charging of the capacitor was going to be the problem. Thanks for that insight.

 

Hello again,


Here is the formula for the total period in plain text:
Tp=-log((2*(Vcont-Vcc))/(Vcont-2*Vcc))*C*RC-log((2*Vsat-Vcont)/(2*(Vsat-Vcont)))*C*RB
RC=RA+RB

[note we might be able to get away with the identity -log(x)=log(1/x) but no big deal]

Checking the formula with RA=200 and RB=1000 and C=0.1uf and Vcc=5v and Vcont=3v and Vsat=0.023v, the total period comes out to:
Tp=137us
and with a simulator it comes out to:
Tp=140us
so it should be pretty close in theory.

In LaTex:
\(
Tp = \[-\mathrm{log}\left( \frac{2\,\left( Vcont-Vcc\right) }{Vcont-2\,Vcc}\right) \,C\,RC-\mathrm{log}\left( \frac{2\,Vsat-Vcont}{2\,\left( Vsat-Vcont\right) }\right) \,C\,RB\]
\)
\(
RC=RA+RB
\)

As an image:




Frequency plot with RA=200, RB=1000, C=0.1uf, Vcont=0 to 5v, Vsat=0.023v, Vcc=5v
(Note the useful range of Vcont may be more limited due to comparator input limits, such as the part of the curve shown in blue)

 

Hi hymie, Can't get the same results as you. For example at vc=2 I get a 99% low. I used a 0.1uf capacitor but didn't use a constant current source. I presumed you didn't either in the graph you show. If you used a constant current source you wouldn't need the charging resistor, only the discharging resistor but you show RA=RB/5. I used RA=2k, RB=10k.
Can you give me more info on how you derived the graph please. What size capacitor did you use?
I use Tina-ti software.

 

Hi hymie, Can't get the same results as you. For example at vc=2 I get a 99% low. I used a 0.1uf capacitor but didn't use a constant current source. I presumed you didn't either in the graph you show. If you used a constant current source you wouldn't need the charging resistor, only the discharging resistor but you show RA=RB/5. I used RA=2k, RB=10k.
Can you give me more info on how you derived the graph please. What size capacitor did you use?
I use Tina-ti software.

Who is hymie? ha ha.

C=0.1uf, RA=200, RB=1k

If RA=2k and RB=10k, then C=0.01uf

 

Who is hymie? ha ha.

C=0.1uf, RA=200, RB=1k

If RA=2k and RB=10k, then C=0.01uf

 

Mr AI Sorry I thought I saw hymie somewhere. Found 2 problems relating to the 2 resistors and duty cycle. Your RA was my RB and RB was my RA. Turned the resistors around and I get your results.
I have always seen say 90% duty cycle as a 90% hi voltage and 10% low because the hi voltage usually does the duty or work. But I see that on your graph it means a 100% low voltage.
Anyway no problem. Should be able to go from here to get my average of 15 volt for the battery charging max from the average of the pulse modulation.
There is a simple equation for the average of pulse modulation. It is "duty cycle x peak voltage= average voltage.

 

Mr AI Sorry I thought I saw hymie somewhere. Found 2 problems relating to the 2 resistors and duty cycle. Your RA was my RB and RB was my RA. Turned the resistors around and I get your results.
I have always seen say 90% duty cycle as a 90% hi voltage and 10% low because the hi voltage usually does the duty or work. But I see that on your graph it means a 100% low voltage.
Anyway no problem. Should be able to go from here to get my average of 15 volt for the battery charging max from the average of the pulse modulation.
There is a simple equation for the average of pulse modulation. It is "duty cycle x peak voltage= average voltage.

Not sure what you mean about 90% and 10% being swapped i thought i checked that but ill check again later today or tonight.

LATER:
Yes it is backwards ill fix it.


Here we go...ill fix the others later.


Ok all images have been fixed, thanks.