Weired voltages for Magnus Dallas Rangemaster

[Leonka]

Hi, I've just finished a Magnus mod RM, negative ground. I used a Mullard OC44 tested at 106hFE. I also replaced R5 with a 3k trimpot to fit the led light.
I triple checked all the wirings and these are the voltages i get:

Using a battery at +8.7V:
Collector +1.02V
Base +3.99V
Emitter + 4.29V

and with a power supply at +9.14V
Collector +1.1V
Base +4.23V
Emitter + 4.55V

The sound with the battery is weak with almost no sustain but, if you want, acceptable, while with the power supply horrible with a lot of terrible noises...

Please help...

[PRR]

Welcome.

Show the schematic you followed. There are way too many mods, polarities, and part designations (what is "R5"?) of the RM.

But at a glance, either emitter or collector resistor is not the value it should be. Wrong color-code, bad connection, hard to know.

[Leonka]

This is the schematic i followed...



Thanks

[R.G.]

Hi, I've just finished a Magnus mod RM, negative ground.

I type this in over and over. Using a PNP germanium transistor with signal ground on the collector is a Bad Idea. I realize you found this somewhere and are now puzzled at why it doesn't work. This scheme works often enough to trick people into thinking it always works. It doesn't.

But that's not why you called. 

I used a Mullard OC44 tested at 106hFE. I also replaced R5 with a 3k trimpot to fit the led light.
I triple checked all the wirings and these are the voltages i get:
Using a battery at +8.7V:
Collector +1.02V
Base +3.99V
Emitter + 4.29V

And yes, something is rotten in the state of Denmark.
First, have you checked your battery voltage while it's powering the circuit? Is it still 8.87 or does it sag?
Next: What is the DC voltage across C7? The object of these two questions is to find out if your power supply is collapsing under load.

Next: If you keep doing this, you're going to have to learn Ohm's Law. That says that the voltage across a resistor >>>MUST<<< be equal to the resistor times the current through it. You've reported that in your circuit, the collector sits at 1.02V. The collector resistor is that 10K pot for DC purposes. So the voltage across that 10K resistor is 1.02V, and the current >>>MUST<<< be 1.02V/10,000 = ~100uA. On the other side of the transistor, we have an 8.87V battery and a 3.9K resistor in series with 100 ohms: call it 4K. the emitter measures 4.29V, and that's 8.87-4.29 = 4.58V less than the battery voltage. So that 3900+100 ohms has 4.58V across it, and must have a current of 4.58V/4000 = 1.145ma flowing through it. That's 1,145uA.

1,145 microamps flowing in, 100uA flowing out through the collector. Transistors can't store current. Where it it going?? Right - out the base.

So there's 1000uA flowing out the base: where does it go? there are three choices other than these schematic).  It can't be flowing out R2. R2 is making this balance of payments problem worse by contributing 8.87-3.99V/68000 = 71ua of its own to the base. So the current has to be flowing out either the input capacitor, which is impossible as capacitors are complete DC insulators, or it has to flow out R3. So 1,071uA through 470000 ohms is the voltage across R3. That's 0.001071 * 470000 = 503V.

Wait a minute??? over 500 V across R3? Not possible. And this is correct - it's not possible. The voltages you show us must either be mistaken, or the resistor values are not as you've shown them, or the circuit is not connected up as you show it in the schematic.

So you have some re-checking to do. The circuit and components are not as you think they are. Sorry - back to checking it again.  Wait - there's another possibility: your meter could be giving your wrong answers for some reason. This happens so often that all experienced techs are always questioning what their meter told them, because meters have problems too - and measurements must constantly be checked for sanity.

The sound [...] with the power supply horrible with a lot of terrible noises...

This is one of those things I referred to in the comment at the top. The inverted power supply connection is particularly prone to power supply problems, noise and oscillations.

[Rob Strand]

Sure you have C and E around the right way?  Check pin-outs and make sure you view it from the correct perspective: underneath or through the top of the device..

How did you test Hfe, with a multimeter or a special circuit for to measure hfe and leakage for germaniums?

[Leonka]

Mmmm...i didnt know Hamlet was a skilled electronic engineer, cool....anyway thanks for the calculations that say there s something wrong but i still dont understand if the scheme is correct or not or in other words if i can follow it; because if i cant we are talking about nothing.

To C7 cathode i mesure 4.3V
To C5 8.8V

C is the red dot and i let it go to the 3 pin pot.

I use a multimeter to test the transistor.

Heeeelp

[R.G.]

He wasn't skilled at electronics - but the bard knew when something was rotten.

Let's take a look at that schematic and see if it's correct.

The transistor is quoted at a current gain of ~100. If it's working correctly in the circuit, that means that the base current is only 1/100 of the current that flows through the emitter and on to the collector. So for a first approximation, we will say it is zero compared to the emitter and collector currents. We will also make a first approximation that the voltage across that 100R resistor R4 is zero compared to the voltages on R1, R6, etc. this may turn out not to be true, but we'll check it later.

If the circuit is workable (notice I did not say it's correct - correct implies that there's only one way, and there are many more) then the voltage on the base is set by R2 and R3. The battery voltage is - let's call it 8.87 for the example here - across R2 and R3. The base voltage is approximately 8.87V * 470K/(470K+68K) = 7.75V.

For bipolar transistors, the emitter and base >>must<< be only the Vbe voltage for the material apart. Let's guess this transistor is 0.2V, a common value for low-signal germanium. So the emitter should be sitting at 7.75V + 0.2V = 7.9V. That means that there is the same 7.9V across  R1, and the current through it must be (8.87V-7.9V)/3.9K = 249 uA. 1% of that goes into the base, so the other 99% goes into the collector. Then this 246ua times the 10K collector resistor (the pot) is 246uA times 10,000 = 2.46V.

And that is what the collector voltage should be with these values of resistors and with a properly functioning transistor. The illustrated values of resistors then can work and make an amplifier. Something is preventing them from doing so.

We know the base voltage is consistent with the emitter's actual voltage, but not the resistors that should be setting it. Just picking one possibiliity, what would happen if the 68k base resistor was actually open - one of its solder joints was open? That would mean that the base current was limited only by the 470K to ground, so the current in the base would be 3.99v/470k = 8.49uA. We know that at these voltages, R2 would be pouring 75uA into the base node, and only 8.49 can go out thorugh R3. Where is it going?

It can only go two places (unless there is a sneaky shorted contact to somewhere else under the board, to the case, a solder thread to another part, etc,) and those are into the collector and into the capacitor arrays. The collector shows a voltage of 1V, so the base-collector junction is still reverse-biased, and that means that it can't be going into there. Well, actually, it may be doing so. We're talking germanium, and germanium leaks. How much current is in the collector, anyway?

Well, 1.02V across 10k ohms is 102uA. This is remarkably close to the 75ua from the 68K R2. Could be here, because 102uA is about par for good germanium leakage. But that would mean that no current could be coming in from the emitter. So maybe the >emitter< is open. What if the 3.9K resistor has a bad solder joint?

This reasoning process just goes on and on until one has re-analyzed the circuit for all possible things being open, shorted, leaky, etc. and will probably bore you to tears if it hasn't already. The short way to find out the problem is to deliberately remove current paths. Temporarily (and deliberately) lift one lead of C1, C2,  C3, and C4. Now no current can possibly get out those paths. Did the voltages change? if no, the caps were not the problem.Similarly lift one lead of C6. Did voltages change? if not, C6 was not the problem.

Now lift the base lead. If the voltages do not change, the base bias string was open and the voltages you're reading are all emitter current. Now read the voltage at the junction of R2 and R3. It should be 7.75V if the base was not hosing it up. If it now does, and removing the base "fixed" it, something is wrong with the transistor.

And do measure the battery voltage under load. Batteries do sag under load, and yours may be doing so.

[PRR]

> Ohm's Law.

+1

> The short way to find out the problem

There's multiple "short ways".

I have learned (ouch) to expect certain errors to be done over and over again and again. Red? Orange? Oh-Oh? Oh-Oh-Oh?

But first: re-draw that thing to dummy-level. My dummy likes + on top and all +to- paths to run down. I could draw it on a matchbook, but not scan as fast as I can twist the image. (Either way, we risk introducing new errors...) Also skip caps because they "should not" affect DC levels (hmmm....).


It does look plausible; for *some* set of values.

Insert Voltages. Apply Ohm's Law to get currents.

You do want to "know" some things. Emitter and Collector currents are normally the same (almost). So go for those numbers.

Here we seem to have Ie 10X bigger than Ic.

Is the transistor wrong? Too much work to prove it is unlikely (though possible).

Is the meter wrong? Sometimes, but you got these numbers all in one session.

Are the resistor values wrong? Is R1 maybe Org-Wht-Org instead of Org-Wht-Red? (Maybe fluorescent light?) Or the "dot in 3.9K" is just a fly-spec on the resistor?

My 2 cents is on R1 being 39K or R6 being 1K. Such OoM goofs are high on my Top Ten list of favorite mistakes.

Bright light and good eye/glasses will find a builder mistake. If that seems OK, in this case an OhmMeter (power off) will read the actual values near-enough, to see if it is a factory mis-mark (rare but happens).