I'm trying to figure out an issue I have with a Darlington Transistor. Seen in the bottom picture is the schematic I am testing.The op amp is powered with +/- 15 volts and the collector of the Darlington is powered by 15 volts also. What the expected behavior of the circuit is suppose to be is that when i apply a control voltage(Vc) to pin 12, the collector to base voltage(Vcb) will drop the required amount of voltage so that the voltage at the emitter(Ve) will be the same as the control voltage.
For example if i applied 12 volts to VC, i should expect that the Vcb will drop 3 volts so that VE will equal 12 volts. When there is no Load attached to the emitter, this is exactly what happens. However, when i connect the VE to alternator coils, which i measured to be about 4 ohms. The voltage drops from 12 volts all the way down to 1.4 volts. I believe that there must be something in the data sheet of the Darlington i must not understand that is giving me such a huge voltage drop. Is there any method to increase this voltage up to about 12 volts? I have supplied the datasheet of the transistor, and a copy of the image
It sounds like you are not fully saturating the transistor, perhaps because the op-amp doesn't have enough output drive? Unlikely though given the Darlington.
I'd start with an intermediate load and work your way up (down) to 4 ohms. Begin with 1k, then 100 ohms, etc. and see when things start going south.
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The Aussie Shield: breakout all 28 pins to quick-connect terminals
NPN Darlingtons are generally set up to sink current, vs sourcing it.
Go here http://www.kpsec.freeuk.com/trancirc.htm
~1/3 of the way down, try hooking it up like that.
Hey So i followed Rugged Circuits Method of dropping the resistance until i see a a drop in Voltage at the emitter
I applied a control voltage to the op-amp non-inverting terminal, then i measured the Voltage at the Emitter while varying this resistance. I started with 44 ohms and worked my way down to 23 ohms. here are the results.
Vc= 4.6 volts
Ve =4.6 volts
Re = 44 ohms
Then [ Here is when the voltage drops]
Vc = 4.6 volts
Ve = 4.5 volts {drops .1 volt}
Re = 23 ohms, So this is when the darling ton begins to fail.
IE = .1956 A
since IB max = .02 A
The gain must be .1956/.02 = 9.78 A/A I'm assuming this is when the gain starts failing because of the limitations of the op-amp to output a max of 20 mA, because we expect about 600 to be the gain when operating correctly.
Should i include a priming BJT to the Darlington and then retest the circuit?
additional thoughts: The Additional BJT will be powered by 15 volts at the collector. Would a Heat-sink be needed for this BJT also, since it is passing a big gain in current?
Cross Roads, which one would your recommend for the MOSFET, because in my school we have readily availlible 2n2222A BJT's. Would the circuit still be hooked up the same? Do you think that by placing this bjt there it should resolve the whole issue of the dropped voltage?
Hey CrossRoads i was about to implement the circuit you were going to use, I see how you state that it is easier to pull the current into the coils rather than push the current into it by placing it at the emitter.
However, before i tried your circuit, I placed a BJT and ANOTHER Darlington in series with the Darlington of the original circuit.
so it looks like
Op-amp ---> BJT ----> Darlington ----> Darlington ----> Field Coils.
The behavior I noticed when i apply the control voltage to the op-amp now is that, the voltage seen at the field coils is greater than Vc {good!}.
For example if Vc = 1.6 V
then Ve = 1.9 V also base current is .7 SO it does the amplification, But as I'm reading the output AC voltage from my alternator i notice a lot of noise and not a clean sine wave, could it be because of the amplification that there is noise introduced to the field coils.
Is it because i included these Bjt and Darlington, it just increases the current gain by a thousand fold? Why is the voltage holding at a higher voltage than the control voltage now?
