currentlampraspberry pitransistorsusb
I posted this question in the Meta Stack Exchange but David Tweed said that the idea was flawed so any alternative circuit suggestion would be appreciated.
I have this Ikea lamp. It runs off 5V usb power however it draws 0.6A which according to this is more than the GPIO pins can supply. So I need to use a transistor but I don't know how to find one that will work.
While using a one-shot timer circuit will work, I think an easier solution can be used. Take a look at this circuit.
For clarification, "VBAT" is a 12V source that is always on as long as the battery is connected. However, "ACC" is a 12V source that is only on when the ignition is on or the key is set to "accessory." Rather than using a 5V relay just to control the power to the RPi, why not use a standard 12V auto relay as shown. This way, there is no wasted power (except for the coil current while the power is on) because everything will be disconnected from the battery.
One side of the coil is always connected to 12V. The opposite side is connected to ground (chassis) through an N-Channel FET (Q1). While a MOSFET is used in the diagram, any FET capable of sinking the coil current can be used. When "ACC" is powered ON, Q1 will switch ON, connecting the coil to ground and actuating the switch. This will in turn power whatever 5V regulation circuit you plan to use (a simple 7805 regulator with heat sink, a switching DC-DC converter, the USB supplies mentioned, etc).
The diode D2 is there to ensure the capacitor can only discharge into Q1 and can be regular or Shottky. Other methods should probably be used for over voltage and current protection from the battery.
The "ACC" voltage can be put through a voltage divider to create a 3.3V signal for the RPi. Be careful with this voltage level, considering a 12V auto battery can really be more like 14V DC. As long as this signal is HI, the RPi knows that the power is on. Obviously, this GPIO pin should be set as an input with any internal pullups disabled. When "ACC" is turned off, the RPi should see the LO signal on the pin and begin its shutdown.
When the "ACC" voltage is turned off, the capacitor C1 will retain the charge for so long, discharging through the resistor R1. Once the capacitor voltage drops below the gate threshold of Q1, it will switch OFF, disconnecting the relay coil from ground and removing power from the peripheral circuit. If a "logic level MOSFET" is used for Q1, it will remain switched ON until C1 voltage is fairly low. I tested this circuit using an NTD4960 (Datasheet), and it remained on for around 15 seconds - until C1 was around 2V. To increase the time, increase the capacitance value.
I would use a relay for this. A relay provides electrical isolation between the controlling circuit and the controlled circuit. You would not need to know anything about the controlled circuit: whether either side is grounded, polarity, voltage, AC/DC, etc., as long as the relay contacts are rated for the voltage and current they will need to handle.
Best Answer
You've drawn an NPN transistor as a high-side switch. This won't work since you need to drive the base at least 0.7V higher than the emitter, and your RPi GPIO outputs only push 3.3 volts.
You should move the transistor into the ground path to operate as a low-side switch, and add a current limiting resistor to the base. Even then, you will probably run into trouble depending how much current the lamp needs compared to the h_fe of the transistor, necessitating two BJTs in a Darlington configuration.
I would recommend ditching BJTs altogether and using an N-channel MOSFET as a low-side switch. MOSFETs are voltage controlled devices, so you just need to supply a logic level to the gate to get it to operate as a switch. Add a pulldown (say 10k) from the gate to the ground so the input doesn't float when the RPi is off or Hi-Z. Something like this:
simulate this circuit – Schematic created using CircuitLab
When choosing a MOSFET, pick one which can handle the current without a heatsink (say 1A), and ensure that it can switch using a 3.3V gate drive.
Finally, 600 mA is more current than most standard USB ports can handle. A USB charger can do this, but a standard USB 2.0 port is limited to 500 mA. So, be careful what you plug this into.