Electronics > Repair
(FIXED) Keithley 225 (current source) repair
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cncjerry:
So r105 is 180ohms.  At 55V my ohm's law says it is handling 16.8W  (55^2)/180 = 16.8W.  So that means the bias is wrong on Q105 which is determined by Q104, those diodes, R104, the cap I can't see and what looks like a ladder network of some type.  I assume you checked Q104 and those diodes as well as the cap.  It could be just a gain issue with the transistor you put in.  But Q105 probably won't last long with all that base current.  Without digging in to far, I would go back and do some ohm checks.  If it get tricky to figure out, you can throw the circuit into LTSpice and try to figure out what would cause the 55V across r105.

I've been thinking about a 225 to complement my 228.
motocoder:

--- Quote from: cncjerry on June 05, 2015, 10:34:07 pm ---did r105 pop again?  Sometimes the burn out and don't flame.  I don't remember the value, but do the power math at 55v to be sure it isn't too much which would point you somewhere.  when I had a resistor popping in my keithley 228 VI source, one cool product by the way, I put 1/4w in to make it easier to blow.

I'm going to go look at the schematic again.

--- End quote ---

It's actually a little toasted, but I've measured it a few times with the ohmmeter and it's still measuring 179 ohms.
motocoder:

--- Quote from: cncjerry on June 05, 2015, 10:45:08 pm ---So r105 is 180ohms.  At 55V my ohm's law says it is handling 16.8W  (55^2)/180 = 16.8W.  So that means the bias is wrong on Q105 which is determined by Q104, those diodes, R104, the cap I can't see and what looks like a ladder network of some type.  I assume you checked Q104 and those diodes as well as the cap.  It could be just a gain issue with the transistor you put in.  But Q105 probably won't last long with all that base current.  Without digging in to far, I would go back and do some ohm checks.  If it get tricky to figure out, you can throw the circuit into LTSpice and try to figure out what would cause the 55V across r105.

I've been thinking about a 225 to complement my 228.

--- End quote ---

I can't reproduce the 55V across R105. Maybe I just misread a "mV" on the meter instead of "V". Probably not, though, given it burned out already once.

I've been staring at the schematic for a couple of weeks now, and I think I have most parts of it figured out. I think the diodes you're referring to are D108, D109, and D110. These, along with R121 bias the voltage across Q104 and Q102 such that they are separated by a couple of volts. This last stage is basically a push-pull amplifier, so they're biased apart a little so there isn't a dead zone near zero. A lot of times people use a "Vbe multiplier" to do this, but here they just used 3 diodes and a potentiometer (an inferior choice, IMHO). The bias point difference, however, means both complementary stages will be conducting some current at 0 output. The current can go from +130V, through Q101, out through R102 or R103 (depending on range switch setting), and then back through R107 or R108, and then through Q105 to -130V. All of that can happen without changing anything in the feedback loop, so without some limiting mechanism, it could easily run away. That's what Q103 and Q106 are for. Using Q103 as an example, and assume range switch has seletected R102. Q103's base is on one side of R102, and its emitter is on the other. R102 is 3.9 ohms. Q103 will be conducting pretty good when it's Vbe is around 0.6V. That happens when the current through R102 is 0.6 / 3.9  = 153mA.

Now I think the beta of Q101, the replacement I put in, was around 48. So if ~153mA is flowing through Q101 emitter, that means Ib is around 153/(1+48) = 3.2mA. Let's assume most of this current is coming from Q102. At 0 current output, it's dropping about 130V minus a couple of Vbe drops from collector to emitter on this transmitter. So that's about 0.4W being dissipated. So I did the calculation for the wrong side :), but let's assume they're similar for the negative side. I replaced Q104 over there with a 2N5416S, which looks to have a slightly lower power dissipation rating. Maybe it's just overheating?

But either Q104 is now burned out, or there is some other part that went out because the negative polarity is not working at all - even when I start with everything cold. It does seem to work on positive polarity just fine, although as I turn up the current, Q104 seems to be heating up faster.

I did check all the transistors and diodes in this part of the circuit with the diode check function on my meter (in circuit), and they all showed up fine. However, I've had that mislead me on a couple of parts already. I hate to desolder anything, though, as the traces on this board are in rough shape after all the rework...

motocoder:
Yes, Q104 was bad. My transistor tester said it was a "common cathode diode network", so it still looked like valid PN junctions, but just wasn't operating like a transistor. This is why I didn't detect the issue in circuit.

I've replaced it with a NTE397, which NTE lists as an equivalent part for the MM3003. I am going to go very slowly and make sure nothing is overheating this time.
motocoder:
Ok, it's *mostly* working, but have a few problems:

- I am not able to set the bias for transistors Q101 and Q102 as described in the manual, table 6.2d on page 19. I was able to set this before, so something is definitely still not right. The bias voltage is way too high, which means there is too much current flowing at equilibrium. I think the problem might be that the gain for the two polarities is too imbalanced.
- There is definitely some oscillation that was not there before when it was briefly working perfectly.
- I can't set the positive output voltage past 30V. I can set the negative polarity, but just not positive. I think this is related to the gain problem as well.
- I noticed that the package for the power transistor I am using (NTE124) says maximum Ic is 0.1A. Now the NTE data sheet says 1A, but I'm not sure I trust that. I did find a source for the Central Semiconductor parts, mentioned earlier on the thread, that are a very close match. I may try those.
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