Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Frequency multiplier

Status
Not open for further replies.

Teakthecat

Newbie level 1
Newbie level 1
Joined
Jan 21, 2015
Messages
1
Helped
0
Reputation
0
Reaction score
0
Trophy points
1
Visit site
Activity points
14
Hi everyone,

I'm in a crunch with too much on my plate to figure this one out on my own, so I would appreciate any info with this design.

I need a small(ish) unit that we can use on a couple different machines here. The signal pulses count number of turns of a shaft, which is then translated to speed and distance travel. We have two different variants of machines right now. The first uses the pulses as they are and fed to the microcontrollers, and other systems in the machine. We now have a second variant which unfortunately uses a different gearbox, therefor the pulse counts don't translate properly.

If we can have a unit that with a control signal will allow either the frequency to pass through unchanged or "modified" then we can use same controllers, wiring, etc.

Requirements:
Supply voltage 10 to 32 Vdc
Current consumption 15mA max
Fin and Fout signal duty cycle 50 to 60%
Fin range DC to 2.1 KHz
Fin max 14Vp-p
Fout square wave 0 to 5 V +/- 0.5V
Temperature range -40 to +120*C​

Control Signal:
If Control signal is equal to Supply voltage than Fout=Fin/0.860
If Control signal is equal to 0V or open than Fout=Fin​


Thanks in advance for any help here.

Pat
 

It's rather difficult to multiply a frequency by such a non-integral value of 1/0.86 = 1.1628.

What accuracy is required for the multiplication?

A phase-lock loop comes to mind but that wouldn't work down to DC.

The best I can think of is to use a micro to count the pulses and then output 11,628 pulses for every 10,000 pulses in, but that may be unacceptable due to the time delay in the output.

For less delay with a slight loss in accuracy (0.02%) it could output 1,163 pulses for every 1,000 input pulses.

Or you could output 116 pulses for every 100 input pulses with a 0.24% loss in accuracy.

- - - Updated - - -

A further though on doing the multiplication with a micro:

If you output 116 pulses for every 100 input pulses and then 1 extra pulse every 357 counts, you would have 11,628.01 pulses for every 10,000 input pulses.

Taking that one step further you could output 11 pulses for every 10 input pulses with one extra pulse every 16 input pulses. This would give 11,625 output pulses for every 10,000 input pulses or within 0.03% of the desired 11,628.
 

Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top