<20% of 5V is technically 1V >

 

Please let me add a word of caution on the pwm issue here. Although the 
statement seems intuitively true, it falls apart when current (and therefore 
power) is taken in consideration. In order to calculate the required pulse 
width ratio the following formula must be used:

 

Using the following definitions:

d = t/T (ratio = on time / period time),

R = filament resistance

V = Voltage

I = current

P = power (I*V)

Vdc = rated voltage of the filament

Vpwm = supply Voltage

 

First statement: Pdc = Ppwm

==> Vdc*Idc = d*Vpwm*

==>Vdc*(Vdc/R) = Vpwm*(Vpwm/R)

cancel out R ==>Vdc^2 = d*Vpwm^2

than solving for d: d = (Vdc)^2 / (Vpwm)^2

 

Soooo.... In order to drive a filament rated for 1V dc from a 5V dc supply d 
will equal (1^2/5^2)*100% = 4%. A 20% duty cycle WILL burn the filament in very 
short order.

 

Bill

 

From: neonixie-l@googlegroups.com [mailto:neonixie-l@googlegroups.com] On 
Behalf Of taylorjpt
Sent: Tuesday, March 07, 2017 12:30 PM
To: neonixie-l
Subject: Re: [neonixie-l] Rookie question about driving VFD filament

 

An important aspect of the filament has not been discussed in this thread:  
Yes, the filament needs to be heated so that it emits electrons efficiently but 
it also sets the potential difference between the cathode (Itself) and the 
anodes (the segments).  For the later reason you can't simply calculate the 
number of watts and then use a PWM to get to that number.  20% of 5V is 
technically 1V but that means that one end of the cathode is at 5V potential 
(for 25% of the time) and the other end is at 0V.  If the anodes are at 30V 
then the end of the display nearest the PWM will get 28V (anode to cathode) and 
the end nearest the ground side will get 30V (anode to cathode) or 7% brighter. 
 This is why filaments, especially for wide multi digit displays, are operated 
with an AC drive to minimize the end-to-end voltage differential.  For a single 
digit like the IV-6 this will result in uneven illumination between the left 
and right side of the display.

Another consideration is the grid to filament bias.  The grid for an "Off" 
digit is typically held at a negative potential relative to the filament to 
repel the emitted electrons thus preventing them from hitting the anodes.  A 
grid that is not properly biased this way results in segments that are dimly 
lit when they should be off because of electrons that sneak through the grid.  
This is why the filament is usually biased so that ground is below the lowest 
voltage of either end of the filament so that grids at ground potential are 
negative with respect to the filament... or else the grids are operated at a 
negative voltage for filaments at zero volts.  An IV-6 operated by itself does 
not need the grid to be used but for a group of them in a multiplexed display 
this is critical.  If 5 filaments are connected in series to 5V, they will all 
get 1V but the cathode to anode potential (For a 30V anode voltage) for the one 
nearest the 5V will be 25.5V (Average) and the one nearest ground will be 0.5V 
(Average) or a 20% difference in brightness.





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