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After looking at the schematic for this particular darlington transistor (see picture pasted below) I had some questions about basic functionality.

First, the diode at the top right of the picture has its cathode leading towards the postivie Vcc voltage supply, wouldn't this reverse bias the diode and prevent any current from flowing even when current is applied to the base of the transistors?

Second, the output is not in series with the supply voltage so when the the transistors are switched on, what prevents the current from just following the path from directly from Vcc to ground (emitter)?? I guess I just don't understand how tying a load to the output terminal would do anything if the supply voltage has a direct path to ground like that.

Darlington Transistor

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  • \$\begingroup\$ Oooh! I call dibs on this one! \$\endgroup\$ – Jason_L_Bens Mar 30 '17 at 0:37
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There needs to be a tad bit of context for this circuit to make sense. The ULN200X series of chips are designed to drive relays and other inductive loads. VCC isn't really VCC like you think of it. That diode to "VCC" serves as a flyback diode to simplify relay connections. most datasheets refer to it as "common" since it can be very much different from VCC on your board.

This datasheet has a better idea on how that works, look at page 12 and 17.

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  • \$\begingroup\$ And if it's feeding a pulse transformer that would qualify as inductive load yes? The emitter pin 9 is connected to a +24 supply it looks like and the various outputs connect to the primaries of various transformers to provide current signals to fire SCRs. I was just looking at the board itself not a schematic so it was difficult to trace. I just don't see how the current path can flow given that the voltage supply is blocked by that diode, but I guess I must just be missing some connection that allows the electricity to flow given that I dont actually have the schematic. Does that make sense? \$\endgroup\$ – Jacob Mar 30 '17 at 4:57
  • \$\begingroup\$ And the chip I'm looking at specifically was ULN2004A. \$\endgroup\$ – Jacob Mar 30 '17 at 5:02
  • \$\begingroup\$ Flyback Diodes, how pin 9 is set up don't supply power, they stabilize the system. When inductors fire off, they push and pull the current as the magnetic field stabilizes. This causes some crazy negative voltages, the diode allows the "pull" to bypass the inductor so the field stabilizes faster and with a lot less voltage. It's a topic that you either take time to understand, or acknowledge that physics and really smart people exist and move on. I barely understand them, and I dissect circuits for a living. \$\endgroup\$ – Omagasohe Mar 31 '17 at 11:10
  • \$\begingroup\$ I misunderstood where you're having issues. Pin 9 isn't a supply pin. Transistors don't need their own source of power. Your original drawing mislabeled that. It's there to reduce component count and simplify connections. this is an open collector chip, which means the chip makes the ground connection(sinks current) not power(sources current). If it helps put your finger over pin 9 and that diode and look at the circuit. the diode is used for a millisecond when an inductive load is turned on. You can pretend it's not there and get a good grasp of the circuit. \$\endgroup\$ – Omagasohe Mar 31 '17 at 12:03
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Your ULN2003A darlington darling is configured as an open collector output. Open collector outputs are great, because they let you connect lots of devices to a bus without loading it heavily. I2C, for example, uses open-collector outputs. Normally, the output pin would be connected to VCC with a pullup resistor. When you drive the input high, the BJTs begin conducting, resulting in a low impedance path to ground. The output is then low. When you release the input, the output pin has a high impedance to ground, and the line is pulled to VCC. When many open-collector devices are sharing a bus and one device outputs a logic low while the rest are outputting a logic high, the low wins out and pulls everything low without carrying all of the current from every connected logic-high device to ground.

To answer your questions, the reverse-biased diode on your VCC pin is definitely supposed to be there. It's used to clamp your output pin voltage to no higher than VCC plus the diode drop. That'd be so you don't fry anything else on the bus. That chip's full of protection diodes. Input to GND and Output to GND both have diodes that protect against a reversed polarity on the pins. Is this device meant to be hot-swapped? I expect this is all about shunting transient voltage spikes, like what you'd see from ESD or disconnecting a cable while current is flowing through it, to ground and protecting the device and everything upstream of it.

For your second question, I guess I jumped the gun and partially answered it at the top. Current shouldn't flow into the VCC pin at all. You're right in that, if the diode were reversed, you'd either have an output high or a hard short to ground. With the pull-up on the output pin, though, that's no longer the case.

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  • \$\begingroup\$ I didn't mention it in the answer, but you don't necessarily need to use a pull-up. I took a look at the IC, and it can carry a lot of current. If you're trying to drive a motor, for example, this'd be a good chip to use. That also explains the diode from Output to GND. It'd act as a flyback diode, so when you turn the motor off, the sudden voltage surge is shunted to ground, rather than destructively through your BJTs. \$\endgroup\$ – Jason_L_Bens Mar 30 '17 at 0:55
  • \$\begingroup\$ Please see my reply on the other comment. I understand the need for flyback diodes with inductive elements, and I understand how once electricity was flowing it would create a low impedance path to ground, I just didn't see how the current could begin to flow due to the supply voltage being essentially blocked. But like I said maybe I'm just missing an external connection to voltage that allows electricity to flow once base current is fed into the input. Because I'm still not seeing how that path would flow, it must go from high to low potential somehow and I'm just not seeing it \$\endgroup\$ – Jacob Mar 30 '17 at 5:02
  • \$\begingroup\$ Thanks for the response and bare with me, I know it's difficult to work on vague generalities and my limited grasp on the subject \$\endgroup\$ – Jacob Mar 30 '17 at 5:02

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