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enter image description here

The figure shows my circuit.

The transistor I am currently using is a MJH6287 (PNP) as opposed to whats shown in the figure (MJH6284 - NPN)

The circuit with the PNP transistor worked just once in turning the LED on and off, Would it make a difference if I replace the NPN transistor with a PNP, if the circuit is designed for an NPN?

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PNP and NPN transistor are diametrically different: NPN transistor conducts current / switched on when saturated (specified current applied to its base) while PNP transistor inhibits current / switched off when saturated current applied to its base (cut-off condition).

The NPN transistors on your circuit diagram is arranged in Darlington pair, a simple voltage polarity swap will NOT work on this configuration.

Anyway, you can use PNP transistors by changing the position of the resistors (see: PNP Darlington Pair) and put an inverter (you can built it with a transistor and two resistors, or simply use inverter IC such 7404) right after the opAmp output. But why would you do it (adding complexity and more components to your circuit), while the NPN transistor is as available as it's PNP counterpart?

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  • \$\begingroup\$ I am wondering however, the PNP transistor does light 1 single LED (Max characteristics of the LED (BXRA-C0402) are 10V and 1000mA) but as the number of LED's increase the brightness of individual LED's decreases (few LED's do not get turned on) Yes I agree using the NPN would be a logical choice. \$\endgroup\$ – ShP Feb 20 '14 at 0:49
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    \$\begingroup\$ PNP transistors do NOT inhibit current when saturated. \$\endgroup\$ – Brian Drummond Feb 20 '14 at 11:57
  • \$\begingroup\$ @Brian Drummond: you're right, I was mistakenly used term 'saturated'. What I mean is PNP transistor inhibits current / switched off when positive current applied to it's base (which in case of NPN transistor it will be saturated). Correct term for this condition is 'cut-off'. \$\endgroup\$ – excalibur Feb 21 '14 at 14:04
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Even the NPN transistor is going to have problems driving a regular LED. The LM741 is capable of supplying up to about +8V from a +10V supply and this will create a voltage of about +7V on the anode of the LED. The cathode of the LED is connected to -10V and this will burn and fry a regular LED. You probably need a current limit resistor unless the LED you are using already has one BUT, you haven't told us and it is perfectly reasonable to assume it hasn't (given that most LEDs will fry with more than 3V across them).

Now to the PNP version and I can only assume you had the emitter connected to +10V BUT even if you didn't and it was the collector connected to +10V you might easily pull too much current through the base (via collector or emitter connected to +10V) and destroy the transistor. You said it worked once but you haven't said that it never worked again even after a power cycle so I'm assuming you have killed it with too much base current.

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  • \$\begingroup\$ I tested the transistors individually later as i figured they might have burnt up, but they seem to work fine. The LED's (BXRA-C0402) need a minimum of 7V just to turn on and can handle upto a max of 10V. I am wondering how the PNP transistor is still fine when I connected the emitter to -10V and collector to +10V. The LED does glow, as I increase the voltage on the collector the LED grows brighter, Why is this happening when its not supposed to? (The circuit is connected EXACTLY as shown) \$\endgroup\$ – ShP Feb 20 '14 at 0:41

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