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Messages - Lazarus

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1
Ham Radio Circuits and Repair / Re: President HR 2510 CPU/PLL issues
« on: March 16, 2018, 12:59:55 AM »
IC306-1/2 (actually the b half of the chip according to the pinout) is half of a dual op-amp acting as an active loop filter (active low pass filter), its job is to take the phase error pulse train signal coming from the phase detector (A) output (pin 13) of IC301 (pll chip), and turn it into a smooth dc control voltage for the vco tank made up of D306, D307, and L317 (best i can make out, its blurry),

i highly suspect that the reason for its output being so high (as high as possible i suspect), is because of the too low freq going into the Fin pin 9 of the IC301, its putting out max voltage in an effort to raise the freq going into that pin up to where it should be,

there are many things that can go wrong in a pll system, especially one as elaborate as this one, but a maxed out loop filter voltage is almost always due to an either missing or too low Fin freq,

NOTE: the other half of IC306 is also an active loop filter for the VCO tank made up of D303, D304, and L302 (best i can make out, its even more blurry), its associated PLL chip is IC302, its associated VCO output, after being divided by 10 while going through IC303, eventually
winds up going into IC304 to be mixed with the previous PLL/VCO, then this newly down mixed signal looks to be what feeds the Fin pin of IC301, so looks like you have your work cut out for you narrowing down where the fault is,

and i just noticed, looks like the first VCO goes through yet another mixer (IC305) before being mixed with that signal, wow there's sure alot of mixing going on in this PLL system,

anyway hope that helps a little, just holler some more if you need anything further, my hope is that my giving a little detail on what things are doing will help you to narrow the problem down,

:)


2
New member welcome / Re: Hello. I've just adopted a FT-101
« on: October 05, 2017, 09:57:40 AM »
Quote from: mael on October 04, 2017, 02:59:06 PM
My goal tonight is to replace at least one cap.

A couple of errors in my above posts bothers me.  One was I said the FT-101 weighed 35kg.  Well it's 35lbs.  The other was the location of a relay inside the enclosure for the output tubes.  It isn't.  Sorry.  I was a bit tired when I posted.  I felt that people who are familiar with the FT-101 would have immediately noticed I was wrong and wondered whether I was actually had an FT-101.

That's all.
you should be able to edit your own posts, by cliking the "modify" button at the top of the post,

:)

3
New member welcome / Re: Hello
« on: September 11, 2017, 03:15:42 PM »
welcome aboard Dean, good to see you here,

:)

4
New member welcome / Re: I've migrated from CBT
« on: September 09, 2017, 06:10:02 AM »
welcome aboard,

:)

5
Electronic connection / Re: AC 101: Voltage Doublers & Voltage Multipliers
« on: January 28, 2016, 04:30:48 AM »
Full Wave Voltage Doubler,


6
Electronic connection / Discussion thread for: AC 101: Rectifiers
« on: January 27, 2016, 07:35:25 AM »
lets talk about rectifiers:

:)

7
Electronic connection / AC 101: Rectifiers
« on: January 27, 2016, 06:55:00 AM »
this is a half wave rectifier, to make the output negative, just turn the diode and cap around reversed,

this is an animated graphic, it updates every 2 seconds,


8
Electronic connection / Discussion thread for: AC 101: Voltage Doublers & Voltage Multipliers
« on: January 27, 2016, 05:28:27 AM »
lets Talk about voltage doublers and voltage multipliers,

:)

9
Electronic connection / AC 101: Voltage Doublers & Voltage Multipliers
« on: January 27, 2016, 05:12:49 AM »
this is a half wave voltage doubler,

if you want the polarity to be reversed, all you gotta do is turn the diodes around in the opposite direction, then the output will be negative instead of positive, and of coarse if the caps happen to be polarized then they will need to be turned around in the opposite polariry,

other than that, i think the animated graphic pretty much says it all, it updates every 2 seconds,



Note: i'm only doing this in my spare time, so check back from time to time for updates, i plan to add some scope pics to this when i get time,

up next: full wave voltage doubler,

:)

10
CB Radio Repair Forum / Re: Galaxy DX2517 horrible spurious emissions and low power
« on: January 20, 2016, 12:40:23 AM »
you might check out C218 for good measure, or either C216, or C220, might have gone leaky, but i've got an all too familiar sneaky feeling that it will end up being either L40, L42, L43, or L44, been there too many times to think it'll be any different in this case, lol

see what the output from the TX mixer looks like on the scope, if the signal appears to be missing or way off, then check L43 and L44, or if the signal is there, then look at the outputs of L42 and L40, should be a nice clean healthy 27MHz signal, maybe you'll be able to nail down which one it is pretty easy that way,

you can pull the bias check jumper board during this test so that the final section wont be working and getting hot for nothing,

:)

11
Electronic connection / Re: Digital 101, digital numbering systems,
« on: April 14, 2015, 05:24:29 PM »
here is an animated table showing various numbering systems and their relationship, it updates every 2.5 seconds,



:)

12
Electronic connection / Re: Digital 101, digital numbering systems,
« on: April 12, 2015, 05:56:34 AM »
Someone wanted me to include Octal, it is an obsolete numbering system, as hex handles the task of representing binary much more efficiently, but here it is none the less,

Octal, or base 8,

in Octal, each digit has 8 possible values, beginning with 0 as the lowest possible value, and ending with 7 as the highest possible value, each digit is used to represent 3 bits of binary code, here is an example of 2 digit Octal used to represent 6 bits of binary code, counting from 00 to 20,

000,000 = 00
000,001 = 01
000,010 = 02
000,011 = 03
000,100 = 04
000,101 = 05
000,110 = 06
000,111 = 07
001,000 = 10
001,001 = 11
001,010 = 12
001,011 = 13
001,100 = 14
001,101 = 15
001,110 = 16
001,111 = 17
010,000 = 20

:)

13
Electronic connection / Discussion thread for: Bipolar Junction Transistors "BJT"
« on: January 03, 2015, 06:28:09 AM »
lets talk about transistors,

:)

14
Electronic connection / Re: Bipolar Junction Transistors "BJT"
« on: January 03, 2015, 05:35:47 AM »
A Practical Application:

to start off with we would need to select a transistor that has a high enough voltage and current rating for the job, so then we would need to know a few details about the load that the transistor would be powering, so lets assume that our load needs 10 volts at 1 amp, we would need to select a transistor with a Vce rating of at least 10v, and a collector current rating of at least 1 amp right? wrong! it is good design practice to pick a transistor rated at least 30% higher than we will ever need, and by doing so the transistor is likely to never fail, so a transistor with a Vce rating of 15v or higher, and a max collector current of 1.5 amps or higher, would do nicely,

ok now lets assume that our transistor has a current gain of 100, so to get started lets start with the collector current, we already know that our load is going to have 10 volts dropped across it, and a current of 1 amp flowing through it, so our collector current is obviously going to be 1 amp,

now lets calculate the load resistance using this formula:

Vcc / Ic = RL

where:
Vcc = collector supply voltage
Ic = collector current
RL = load resistance

so the voltage dropped across the load divided by the current flowing through the load equals the load resistance, so:

10v Vcc / 1A Ic = 10Ω RL

so RL resistance is 10 ohms,

now lets calculate the value for or base bias resistor, since our transistor has a current gain of 100 then we already know that our base current should be 10 milliamps, because 1 amp or, 1000 milliamps collector current divided by the current gain of 100 = 10, here is the formula:

Ic / β = Ib

where:
Ic = collector current
β = current gain
Ib = base current

now lets assume that our base bias supply is also 10 volts, and remember that it takes approx 0.6 volts on the base to turn the transistor on, so that 0.6 volts will be subtracted from the base bias supply voltage, now we can find the value needed for our base bias resistor by using the formula:

(Vbb - Vbe) / Ib = Rb

where:
Vbb = base bias supply voltage
Vbe = base emitter voltage
Ib = base current
Rb = base bias resistor value

so:

(10v Vbb - 0.6v Vbe) / .01A Ib = 940Ω Rb

so our base bias resistor Rb comes out at 940 ohms,

ok i think that just about does it for this little project, you could add a switch in series with the base bias supply to turn it on and off if you want, and it doesn't have to be a big switch either, in fact it would only need to be able to conduct 10 milliamps of current, the transistor is taking care of the heavy 1 amp of current that the load is drawing, which was really the purpose of using the transistor in the first place, so that a very small amount of current flowing through the base can be used to control a much larger amount of current flowing through the collector,

:)

15
Electronic connection / Re: Bipolar Junction Transistors "BJT"
« on: January 03, 2015, 12:19:14 AM »
Here is a PNP type, a PNP works exactly the same as an NPN in every respect except all the polarities are reversed, therefore it is an exact polar opposite of an NPN,



by now some of you have probably noticed the arrows showing the direction of current flow and are saying hey wait a minute, what's going on here, well contrary to popular belief, current flows from negative to positive, after all electrical current is the flow of electrons, and electrons are negatively charged, therefore they flow toward a positive charge, so therefore electrical current flows from negative to positive, and it just chaps my hide when i see tutorials plastered all over the internet with illustrations showing it backwards, in fact it really kind of ticks me off,

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