[GreenKeys] Model 33 Q1 transistor replacement

Tue Jul 22 12:52:46 EDT 2025

Erik --

If your magnets are running open, there are a couple of voltage tests that
might be helpful.

1.  Put your VOM negative lead on the Zero volt bus.

2.  Measure the voltage on the -20V bus. Is it really -20 volts?
 (anything from -16 to -24 is probably OK).

3.  With the board connected normally to the TTY, connect a 20 mA (or 60
mA) current loop to the right pins with the right polarity to the board.

4.  What is the voltage at the junction of ZD1 and CR4 (I'd expect it to be
about -4.7 volts

5.  What is the voltage at the cathode end of CR4 (which is also the
collector of Q1 and the base of Q2).  I'd expect it to be about -5.4 volts.

6.  What is the voltage at the base of Q1?  I'd expect it to be about +0.3
volts.

7.  What is the voltage at the emitter of Q2.  I'd expect it to be about
-4.7 volts.

8.  What is the voltage at the collector of Q2.  I'd expect it to be about
-10 to -14 volts.

9.  Disconnect the NEG end of your VOM from the zero volt bus and now
measure the voltage across R5 (8 ohms) - POS lead on the junction of R4 and
R5 - NEG lead on the collector of Q2.  The current through R5 is the
Voltage measured divided by 8.

10  Measure the voltage across R7 (14 ohms) - POS lead on the magnet side
of R7 and NEG lead on the -20 bus.  The current through R7 (and thus the
magnet current) is the Voltage measured divided by 14.

Not to cast aspersions about designs by other people, as I understand that
engineers are sometimes constrained (restricted, subdued, subverted and
crushed) by their management, but when I see heat scars like those visible
on the board, I think that more effort should have been put into heat
transfer and dissipation.  Heat and static are the destructors of
electronic components.

Paul

On Tue, Jul 22, 2025 at 11:59 AM Erik Bruchez <erik at bruchez.org> wrote:

> Paul,
>
> Amazing, thanks for the work on the schematics! Got it regarding earth. I
> had added the symbol myself to indicate "common" to myself, but it is just
> clumsiness on my part.
>
> -Erik
>
>
> On Tue, Jul 22, 2025 at 8:02 AM Paul Newland, ad7i <ad7i at ad7i.net> wrote:
>
>> Oh, gosh.  That schematic reminds me of the good old days at AT&T.  I
>> used to have mild arguments with drafting as to how a schematic should be
>> shown on paper.  Their approach was often that the schematic was just a
>> graphical netlist and the goal was to use as few pages as possible.  My
>> position was that a schematic told a story, in graphic form, a story about
>> how a circuit worked, so that the reader could easily understand its
>> operation.
>>
>> Anyway, here's a redrawn version of the schematic with some commentary
>> below that.  Please let me know if you spot any transcription bugs in the
>> schematic.
>>
>> [image: image.png]
>>
>>
>> A couple of things to note here.  First, I don't see any reference to
>> earth ground in the original schematic.  To me, that means that this
>> circuit "floats" with respect to earth ground, and the relationship with
>> ground is determined by the applied 20mA (or 60mA) current loop (one side
>> of which is likely grounded directly or through some finite resistance).
>> When I call out a voltage on the schematic or text, that voltage is with
>> respect to the Zero volt bus.
>>
>> Second, the selector current is BIG, about 0.5 A.  And when Q2 is
>> conducting,  Q2 is operating in its linear range, so it will get HOT.
>>
>> Let's look at what happens in the MARKing condition.  In the marking
>> condition the 20 mA (or 60 mA) of positive current is shoved toward the
>> base of Q1.  In this condition the loop current flows through CR5 and the
>> R3 (0.8 ohms) as well as through R1 into the diode network.   Also in this
>> condition the voltage at the base of Q1 is clamped to about +1.1V, and the
>> BE junction of Q1 is cut off, so Q1 is out of the picture in the MARKing
>> state.  The base of Q2 is held at -5.4 volts by the diode network and the
>> voltage the the emitter of Q2 is held at 4.7 volts (due the diode voltage
>> at the base, less the diode drop across the Q2 BE junction).  So with a
>> constant voltage at the emitter of Q2, and a constant resistance between
>> the emitter and the 0V bus, Q2 forms a constant current driver.  If Q2 has
>> a decent gain value, then Ic = Ie = 4.7V / (0.8 +1.5 + 8) = 456 mA.  So
>> that's the marking condition.
>>
>> In the case of the spacing condition, the 20 mA loop current is disabled
>> and Q1 is biased on via R1 and the diode network.  This causes Q1 to turn
>> on and current is sent from the collector of Q1 toward R8, which tends to
>> rob current from the diode network, sending the voltage at the base of Q2
>> toward zero volts.  It won't get to zero but it will be close (maybe -1
>> volts).  With lower voltage at the base of Q2 (and 0.7 to 0.5V less than at
>> Q2 emitter), that will reduce the selector magnet current significantly
>> (but I don't think it will go to zero).  In a working circuit the actual
>> MARKing and SPACEing selector current can be determined by measuring the
>> voltage across R7 and dividing by 14.
>>
>> Paul, ad7i
>>
>>
>>
>> On Mon, Jul 21, 2025 at 5:04 PM RODNEY HOGG <radio_man at wbsnet.org> wrote:
>>
>>> Paul,
>>> Do you have a type number or p/n for the transistor.  I have 1000s of
>>> transistors, civilian and military types, be glad to help you out.
>>> Rod
>>> K0EQH
>>>
>>> ------------------------------
>>> *From: *"Erik Bruchez" <erik at bruchez.org>
>>> *To: *"Paul Newland, ad7i" <ad7i at ad7i.net>
>>> *Cc: *"Greenkeys" <greenkeys at mailman.qth.net>
>>> *Sent: *Monday, July 21, 2025 1:58:36 PM
>>> *Subject: *Re: [GreenKeys] Model 33 Q1 transistor replacement
>>>
>>> Paul,
>>>
>>>> It's probably not all that difficult to find a working replacement, but
>>>> in a different form factor.   Can you show a larger portion of the
>>>> schematic?  I'd like to see if this transistor is operating as an on/off
>>>> switch, or if it's trying to work some magic to regulate the voltage on the
>>>> magnet for fast pull in and then low current idle.
>>>>
>>>
>>> I attach the whole thing. There are explanations in the schematics.
>>>
>>> I measure about 0 V on the base of Q1, but also about 0 V on the base of
>>> Q2. My understanding (which could be wrong) is that when one is on, the
>>> other one should be off, and vice versa, which means that 0 V on Q1 should
>>> lead to about 4.7 V on the base of Q2. In any case, I am not getting any
>>> voltage to the selector magnet through Q2 (the large transistor on the heat
>>> sink, separate from the board), which is the original problem I am having.
>>>
>>>
>>>> Some basic DC measurements might give you some insight into the
>>>> health of the transistor without removing it from the circuit board.
>>>>
>>>
>>> This is what is leading me to Q1: it seems that there is continuity
>>> between emitter and collector, which I don't think there should be! I have
>>> another instance of the board where this is not the case. (However, that
>>> second board also doesn't seem to work, but there might be something else
>>> wrong with it.) I could desolder the transistor on that one and transplant
>>> it to see if that gets my first board working, but I was trying to avoid
>>> doing that immediately.
>>>
>>>
>>>> I'd be curious to know the voltage from base to emitter when the
>>>> transistor is "ON" (or supposed to be on).   Also (can't see that part of
>>>> the schematic) but if there's a resistor in series with the base I'd be
>>>> curious to know the voltage across the resistor (and the resistance of the
>>>> resistor), and then use ohms law to determine the base current.  Same thing
>>>> for the selector magnet current, if that's possible.
>>>>
>>>
>>> -Erik
>>>
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