The AM Forum
May 14, 2024, 05:07:37 PM *
Welcome, Guest. Please login or register.

Login with username, password and session length
 
   Home   Help Calendar Links Staff List Gallery Login Register  
Pages: [1]   Go Down
  Print  
Author Topic: Designing a Percentage Modulation Meter  (Read 3799 times)
0 Members and 1 Guest are viewing this topic.
ka1tdq
Member

Offline Offline

Posts: 1506


Red part turned in for a refund.


« on: March 21, 2015, 02:24:14 PM »

I just picked up this meter from the Scottsdale hamfest this morning for $5.  It's a percentage modulation meter that goes to 133%.  My transmitter will peg this meter at times, but I still want to build it. 

Firstly, I need to find out what full scale voltage on this meter is.  I'm assuming 50 mV, but I'll find out.

Secondly, what is the relationship between drain voltage and % power?  I have in my head 2 x Vdrain(carrier) = 100% modulation.

So then, I'd just need to build a circuit that would subtract the Vdrain(carrier) and at that point voltage would be in a 1:1 relationship with % modulation.  I would just need to sample that linearly down to 50 mV.

Is this correct?

Jon
KA1TDQ


* IMG_0355.JPG (1868.6 KB, 3264x2448 - viewed 357 times.)
Logged

It’s not just values, it’s business.
KA2DZT
Member

Offline Offline

Posts: 2190


« Reply #1 on: March 21, 2015, 02:46:12 PM »

Unlikely it's a 50mv meter movement.  Look at the very bottom of the face it may say what full scale is.  50mv movements are usually for use with DC current shunts.  The meter you have probably came out of some piece of equipment.  The full scale voltage or current could be anything.  There are ways to find what the full scale is.

Fred
Logged
ka1tdq
Member

Offline Offline

Posts: 1506


Red part turned in for a refund.


« Reply #2 on: March 21, 2015, 02:49:19 PM »

Ah, yes.  It's 400 uA DC. 
Logged

It’s not just values, it’s business.
ka1tdq
Member

Offline Offline

Posts: 1506


Red part turned in for a refund.


« Reply #3 on: March 21, 2015, 11:25:06 PM »

Ok, I've come up with this.

I already sample a small voltage off of the antenna output jack for my LED bar graph.  My sample voltage is 1.8 vdc at carrier.  I calculated that 100% modulation would equal 300uA on the meter since 400uA is full scale at 133%.

Doubling the sample voltage to represent 100% modulation would be 3.6vdc.  Using ohms law with the mod meter in series, a 12k resistor would cause 300uA to flow at 3.6vdc.

The opamp circuit has a gain of 1 and simply subtracts V1 from V2.  So at carrier, the output of the opamp is zero volts causing zero meter deflection.  The 1.8 reference voltage would be taken off of a potentiometer to reference the meter to zero with the carrier on.

...I think...

Jon

**Update:  Actually the series resistor with the mod meter would be 6k since the output of the opamp would be 1.8 volts at 100% modulation.


* IMG_0356.JPG (1769.19 KB, 3264x2448 - viewed 312 times.)
Logged

It’s not just values, it’s business.
MikeKE0ZUinkcmo
Member

Offline Offline

Posts: 443



« Reply #4 on: March 22, 2015, 12:38:42 AM »

If you want to do it the old fashion way, here is a schematic of an old Harris M-6659A meter.

Logged

Mike KE0ZU

Bold Text and PICS are usually links

https://mikeharrison.smugmug.com/
KA2DZT
Member

Offline Offline

Posts: 2190


« Reply #5 on: March 22, 2015, 02:54:52 AM »

12K would be the meter movement resistance plus the resistor.  You need to find the movement resistance.  Don't use an ohmmeter directly.  Put a series resistor in series with the movement.  Measure resistor and movement then subtract the exact resistor value.  May be very small compared to the 12K, may not need to worry about it.
Logged
Pages: [1]   Go Up
  Print  
 
Jump to:  

AMfone - Dedicated to Amplitude Modulation on the Amateur Radio Bands
 AMfone © 2001-2015
Powered by SMF 1.1.21 | SMF © 2015, Simple Machines
Page created in 0.066 seconds with 19 queries.