I know I can use an inverter and a power transistor on the output of a 555, but that adds parts. In the last 30 years, somebody must have developed an improved 555 without going digital with counters.
John Nagle
25% high or 25% low, output? ...Jim Thompson
-- | James E.Thompson | mens | | Analog Innovations | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | STV, Queen Creek, AZ 85142 Skype: skypeanalog | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at http://www.analog-innovations.com | 1962 |
Hello,
You can use a 556 or 1/2 558 to build a circuit which offers a pulse rate and width totally independent. The first timer gives the clock, the second one is configurated as a one-shot.
There are alternatives for example with cmos/ttl digital circuits
Best regards,
alpibucky
Hello,
Above is a possible solution with one 555 I think.
Best regards,
alpibucky
You might be able to use a schmitt trigger oscillator with an HC14 sort of gate.
There are tons of switcher controller chips around that do all sorts of extra tricks.
-- John Larkin Highland Technology, Inc lunatic fringe electronics
...Jim Thompson
-- | James E.Thompson | mens | | Analog Innovations | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | STV, Queen Creek, AZ 85142 Skype: skypeanalog | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at http://www.analog-innovations.com | 1962 |
Or use a CMOS 555, and feed back the charge and discharge current independently from the output pin. You just add one extra resistor and a BAV99 dual diode, and you can have any duty cycle you want. It does make the frequency depend on input voltage and temperature a bit.
That's only one extra part compared with the two-resistor astable connection.
Cheers
Phil Hobbs
-- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 hobbs at electrooptical dot net http://electrooptical.net
A 555 works like a precise schmitt, if you ignore all that trigger/discharge stuff. One RC from output to input makes it a square wave oscillator. One added resistor to Vcc or to ground could make it
25% duty cycle. The OP didn't say if the 25% was constant or might need to vary according to some feedback or something.-- John Larkin Highland Technology, Inc lunatic fringe electronics
Good point, but that requires algebra. ;)
Cheers
Phil Hobbs
-- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 hobbs at electrooptical dot net http://electrooptical.net
Just add a diode between the DIS terminal and TRIG/THR pair with anode to DIS. That's not so many parts.
More than Algebra... gets _very_ unreliable (in the real, noisy, world) for duty-cycle < 20% ...Jim Thompson
-- | James E.Thompson | mens | | Analog Innovations | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | STV, Queen Creek, AZ 85142 Skype: skypeanalog | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at http://www.analog-innovations.com | 1962 |
Or Spice.
-- John Larkin Highland Technology, Inc lunatic fringe electronics
you're going to have to add parts.
eg:
Put a diode parallel with the discharge resistor, cathode to the capacitor. Alternativley pull the control-voltage pin up with a resistor (perhaps 2.2K - a guess: I have not done any math)
-- This email has not been checked by half-arsed antivirus software
Why can't the duty cycle be less than 50%? There are plenty of options for how to do that, but to me the simplest is to invert the output.
-- Rick C
That looks promising, so I modeled it in LTSpice IV. Here's the file:
But the math in the article makes no sense. It says
"The duty cycle (n) is dependent on P1 and P2 in the following manner:
n = 1 + P2/P1
If P2 = 0 (n = 100%) then the frequency can be approximately calculated with the following formula:
f = 0.69/((2*P1 + P2 + 4.7kO)*C1)"
So, for a 25% duty cycle,
0.25 = 1 + P2/P1 -0.75 = P2/P1 No way.Actually, it seems that the setting of pot P2, from 0 to 1, directly changes the duty cycle without affecting the frequency. In our circuit, pot P2 is represented by R3 and R4, and the duty cycle seems to be simply
n = R3 / (R3 +R4)
Changing R6 and C1 changes the frequency without affecting the duty cycle, so this is convenient.
Here it is, set to 100KHz, 25% duty cycle.
Thanks.
I'm working on a switching power supply design for old Teletype machines. These need a data signal at 120VDC (!) current limited to 60mA to operate the selector magnet coil. The traditional solution is to use a 120VDC supply with a 2K ballast resistor to drive the
55 ohm 4H (yes, 4 henries) selector coil. 95% of the energy goes into heating the ballast resistor, and you need too much power supply. The selector magnet only needs 120VDC for about the first 2ms after turn-on; the high voltage is just to overcome the huge inductance quickly. So an alternative is to charge up a 1uf capacitor to 120V, and on turn-on, dump that into the selector magnet. Then follow up with a low sustain voltage, about 4V, to maintain 60mA.Here's the current circuit:
V3 is simulating an input data stream at 45.45 baud, turning on and off ever 22ms. L3 is the antique Teletype selector magnet. (See
This seems to work in LTSpice, but I want to see if I can do it without surface mount parts. Those are easier for hobbyists to build. With this 555 circuit, I can get rid of the inverting LTC1693-5 MOSFET driver. The 555's power transistor has enough drive to trigger the MOSFET in the switcher. Now to find a suitable through-hole transformer and MOSFET.
This is my first switcher. Any comments?
John Nagle
On 9/17/2016 3:03 PM, John Nagle wrote: Correction:
[snip]
Here is the simulation for your ttydriver16 circuit.
As you can see, there is a lot of high voltage ringing (3.5 kV) when the switch turns off. You really need a flyback diode across the coil to eliminate this, but it will tend to keep current flowing and thus delay the deactivation of the coil.
You could produce the 120 VDC by using the transformer of a 12V PSU, running it backwards so 12V will produce at least 120 VDC and likely more like 350 VDC.
If you already have 120 VAC available, you could just rectify that to about
180 VDC and then use a 3k 15W resistor for the 60 mA hold current. Put a 10-20 uF capacitor across the resistor to get an initial high voltage pulse for about 30-60 mSec to pull in the solenoid, and then it will settle down to 60 mA. Of course, when the switch is opened, there will be a similar TC to discharge the capacitor so it's ready for the next pulse.It may be better to use a PIC and an optically isolated driver to a MOSFET in the secondary circuit (or the 180 VDC rectified line voltage). You can apply the full voltage for about 30-60 mSec and then set up a PWM of about
10-20% to get the holding current you need.Paul
Adding the commutating or "flyback" diode gets rid of the transients, but delays the current drop-off. I don't think this will work for what you want:
BTW, don't worry about using SMT components for prototypes. If you use somewhat larger packages such as SOIC and 0805 they can be soldered using ordinary tools, and you can get adapters pretty cheap:
Paul
Right. Dumping the energy from a Teletype selector magnet fast is difficult. The RC snubber shown is known to work on real-world machines, despite the ringing. A diode makes it worse. The selector magnet needs to dump in about 2-3ms or the mechanism won't work.
We need enough snubbing to keep from blowing out the drive components. I use 400V opto-isolators. Actual ring voltage measured with a similar circuit using a more traditional power supply is about 200V peak. The SPICE parameters for L3 may need work. 4H and 55 ohm series resistance are valid, but the parallel resistance is just a guess.
Pre-WWII Teletype machines often had visible arcs at the keyboard contacts.
John Nagle
I put a 200 ohm resistor in series with the diode, and it dumps most of the energy before the next pulse. Here is the simulation:
A higher value resistor might help. You might be able to use another opto-isolator that turns on when the series switch turns off, and dump the energy that way. The visible arcs probably helped to dissipate the energy.
Here is the simulation with 500 ohms:
There might be a way to store the energy from the coil into a capacitor rather than wasting it as heat. It would be basically a tuned LC circuit where you could "break" it at the current zero crossing where the capacitor would be fully charged.
Here it is with 1 uF across the 500 ohms:
Changed the 500 ohms to 5k, and it looks just about perfect:
Paul
ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.