0 Members and 1 Guest are viewing this topic.
Here's some pictures. Went to electronics store & they unsoldered POT for track #2 & came up with a reading for a 1K ohm POT. I have installed a new 1K ohm POT for track #1. Had a 0-15 output volt reading at track terminals when turned on. Put a loco on the track, went 3', new POT smoked when it had a load. Not the loco because it runs perfect on track #2. There has to be another problem. Also terminal with orange wire (pencil pointing to) has broken off from part that pencil is pointing to. This happened when I unsoldered the second smoked POT. Looks to be in a heat sink. That part is now bad. I don't know what it is. Track #2 works great, will power 3 KATO F3's fine. AC & DC acc. terminals work fine also. The white plastic POT shaft has #145 molded on it. I don't see any other #'s or letters anywhere. There has to be another shorted part somewhere.
Also I would like to THANK "glakedylan" for letting me acquire another Troller from him. For some reason this sight will not let me contact him.
While I agree with your evaluation that the pot's power rating was not sufficient, looking at where the damage occurred on the pot, how severe it was, and how it failed quickly only after a load was placed on the output, the destruction was caused by current feeding back from the transistor's base terminal to the pot's wiper. When the throttle knob was adjusted to towards either slow or maximum speed, that placed the wiper on a very low resistance path inside the pot, causing high current and temperature. Poof! If I designed this circuit I would have added a series resistor between the pot's wiper and the transistor's base.
Given the Troller's notorious reputation for blowing those power transistors, if it were mine, I would replace them both with newer ones just to avoid future melt-downs.So, this thing has only a single Darlington for all the current amplification, apparently. I agree with Peteski on the series resistor that should have been put between the pot and the transistor base. There is no current limiting to the base except the pot, so the whole thing blows up if the transistor fails. It was designed with minimum components, assuming nothing would fail.
As a retrofit that the OP could apply, what value (and power rating) of resistor would you recommend be installed between the wiper of the potentiometer and the base of transistor?
2.2k sounds reasonable to me. I'm thinking if the max current output is 2A, and the gain is 1000, then the base current would never need to be more than 2mA. At 12v, that would be 12/.002 = 6000 ohm. But we need to allow for a transistor that doesn't really have a gain of 1000, certainly not uniformly over its whole operating range, and for other variability in the components, so a lower value that would let more like 5-6 mA in the base current wouldn't be a bad idea. So yeah, 2.2k seems good.If it doesn't seem to reach full output (12v or so) under load, then you can always lower it a bit.6ma of current at 12v is only 72 mW, so the pot will not be stressed in the least. A 1000 ohm would push you up to 144mw, still way under 1/4 watt, so a 1000 ohm would be fine, and give you more power from the throttle if the replacement transistor's gain is only like 100-200 instead of 1000).
Without doing any calculations, I would add a 2.2k 1/4W resistor.I'm also a bit puzzled about the specs on the back of the pack. It shows 18 VDC and 16VAC for accessory outputs. That makes no sense. There are no visible filter caps in the circuit. So the AC from the transformer (16V) is present at the AC accessory terminals and also fed into a rectifier. The rectifier produces a pulsing DC voltage which is lower than the AC being fed into the circuit by however much voltage is dropped by the rectifier diodes (I assume that will be 1.4V). The rectified voltage is the present on the DC accessory terminals and also is fed into the variable throttle circuit. How the got 18VDC from 16VAC is a mystery to me. It should be more like 14.6VDC.
Won't a value of 2.2K limit the output current too much at relatively low output voltages, when nearly the full resistance of the potentiometer is in series with the 2.2K resistor?
How are the two transformers connected relative to the AC and DC accessory outlets and the two throttles?
I don't think so. Think of the speed control pot as a source of variable voltage (a voltage divider). The base of a properly functioning transistor just drains a very small amount of current from that voltage divider.I don't know. I never owned that throttle or seen complete schematics. I suspect that the AC acc. voltage comes from one transformer and the DC acc. voltage from the other transformer. Just speculation....
Now that you bring this up, that speed pot could probably be a wide range of values and still work, if it is just connected between the + and - rails and forms a divider to drive the transistor. What I'm saying is that a 2k or 5k pot would probably work just as well as a 1k, no?
Yes, with a Darlington transistor used in that circuit (with a gain of 1000 or more) a higher resistance voltage divider (pot) would most likely work. But at some point when the pot's resistance get s high enough, the current draw of the base will visibly affect the voltage supplied out of the voltage divider. But (without calculating it) I think that a 2k or maybe even 5k pot would work.
Given that the throttle should be able to supply up to about 2.5A, and assuming a gain of 1000 for the transistor, you wouldn't want to go above 5K for the pot to ensure full current output at low output voltage - when most of the potentiometer's resistance is in series with the base of the transistor. Keeping those parameters in mind, let's return for a moment to the 2.2K value of the added resistor proposed to be connected between the wiper of the potentiometer and the base of transistor. If there's 2.5mA flowing through that resisitor (and into the base of the transistor), then there will be a voltage drop of 5.5V across the resistor. Also, since the transistor is a darlington type, the voltage drop between base and emitter will be at least 1.4V. So, the total voltage drop from the wiper on the potentiometer to the output of the throttle (at a 2.5A load) will be on the order of 7V.Given the throttle's nominal 15V maximum output, doesn't this imply that for a high current load, the throttle knob would have to be turned up to nearly the 50% mark before the train would start to move?Also, if the load current were to vary substantially - as in the case of a train ascending versus descending a steep grade - won't the variation in transistor base current (and thus the voltage drop across the 2.2K resistor) result in a significant change in the output voltage (as much as 5.5V), even with no change in throttle setting?