Hi all,

I want to build an LED driver using the LTC3454 buck/boost IC.

However, I want to get tricky and include multiple different LEDs, driven by the same circuit. The plan is to parallel the LEDs (4 in total) between the Vout and LED pins of the IC. However each LED will be in series with a p-channel MOSFET connected between the Vout pin and the LED+ contact. The theory here is that I can select which LED to power by turning the MOSFETs on or off as needed. No more than 1 LED will be turned on at any time.

I have successfully worked with and built a circuit using an n-channel MOSFET in the past, but p-channels are totally alien to me.

What I want to confirm is how the mosfets should be connected. I am assuming it is not much different to an n-channel, but flipped. So in the case of the LTC3454 and this application, Vout will be connected to the source of the p-channel, then the drain to LED+. The gate will be connected to an output from my MCU (to turn the FET on and off) via a 100ohm resistor. A large vlaue resistor (100k?) between the gate and Vout could (should) also be used.

Yay? Nay?

Any advice would be appreciated.

Thanks,

- Matt

 

Your source and drain connections are correct. The P-MOSFET gate must go between ground (to turn it on) and the source voltage (to turn it off). If the source voltage is greater than the MCU logic-high output voltage then you will need a buffer circuit to control the gate voltage. This can be either an NPN transistor or a small logic-level N-MOSFET. In either case you also need that resistor (say 10k ohm) from the source to the gate to provide the turn-off voltage.

 

The P-MOSFET gate must go between ground (to turn it on) and the source voltage (to turn it off).

OK, I did not realise that zero voltage at the gate turned on a p-channel mosfet, while a positive voltage turned it off. This is the opposite of n-channels. Told you p-channels were alien to me lol.

If the source voltage is greater than the MCU logic-high output voltage then you will need a buffer circuit to control the gate voltage. This can be either an NPN transistor or a small logic-level N-MOSFET.

I'm not entirely sure I'm following you here, but the MCU logic-high output is, as far as I can tell, the same as the entire circuits VIN value (voltage supplied by the battery pack). The voltage present on the Vout pin of the LTC3454 chip will however be whatever the forward voltage of the LED is at that particular drive current. Given the LTC chip is a buck/boost driver, VIN (and hence ADC output of the MCU) can vary from 5.5V down to 2.7V. The forward voltage of the LED will, depending on the mode chosen, only shift between approximately 2.7V and 3.5V.

So the source voltage should never be higher than the MCU ADC output (what the p-channel gate sees), but the voltage the p-channel mosfet will see on the source pin will be different. Sometimes higher, sometimes lower.

I'm not understanding something I think?

 

The P-MOSFET works like an N-MOSFET except the voltage polarities are reversed. In normal operation the drain-source voltage is negative and the required gate-source voltage (Vgs) is also negative to turn it ON. For either type of MOSFET Vgs=0V turns it off.

The thing you may be forgetting is that it's the relative MOSFET voltages that are important, not the absolute voltages.

So when the P-MOSFET source is connected to a positive voltage then connecting the gate to ground means Vgs equals the gate voltage (0V) minus the source voltage (Vs) or -Vs which turns it ON. It's the relative gate-source voltage that determines the state of a MOSFET, whether N-channel or P-channel.

Similarly when the gate ground connection is removed, then Rgs pulls the gate to the source voltage, giving Vgs=0 and shutting off the MOSFET.

So whatever the source voltage is, the gate must be equal to that to turn OFF the MOSFET. That's why, if the source voltage is higher than the MCU output, you need to add a buffer/driver transistor.

 

Hmmm, I'm struggling here though I can't quite figure out why:

• Pull the P-Channel MOSFET gate to ground, and that will turn it on.
• Pull the P-Channel MOSFET gate to Vs, and that will turn it off.

So why not connect the gate to ground using a 10k resistor, and then the gate to Vs using a transistor. The transistor is switched using the MCU output. Turn the transistor on and the gate sees Vs, turning the P-Channel MOSFET off. Turn the transistor off, and the P-Channel MOSFET gate sees ground, turning the MOSFET on.

I'm confused because you said:

...you also need that resistor (say 10k ohm) from the source to the gate to provide the turn-off voltage.

But everything else suggests my set-up above is how it should be implemented.

Sorry if I'm being a bit slow on the up take. I spent a lot of time thinking about this before I replied I promise!

 

You really need to think in terms of Vgs (voltage on the gate terminal, as referenced to the source terminal) instead of anything else; as the MOSFET will be destroyed if the maximum limits of Vgs are exceeded, even for an instant. Most frequently, the maximum Vgs is specified as ±20v, but very frequently it is ±15v; as always, you must consult the manufacturer's datasheet to make certain. Also, you need to reference the Rds(on) specification to determine the Vgs required to turn the MOSFET ON, and look at the Threshold voltage specification "Vgs(th)" to determine how close to Vs you need to have the gate in order to have the MOSFET considered turned off.

One disadvantage of P-ch MOSFETs is that the physical size of the gate is about 2.5 times greater than an N-ch MOSFET for a rough equivalent due to electrons moving easier than holes; so a P-ch MOSFET's gate charge is roughly 2.5 times larger than an N-ch, making it slower to turn on and off. This may not be of concern for your circuit.

You also need to start posting schematics of your proposed circuit so that many questions are answered up front, and to enable us to help catch possible errors or omissions before they cause you great vexation.

 

I will work on the schematic tonight when I am home from work. At this stage I'm trying to work out the best way to approach it; P-Channel or N-Channel. I'm thinking I should switch to N-Channel purely because they seem more common. Plus for reasons unknown to me I seem to understand (though obviously I use the word 'understand' very loosely) how to use them better

Food for thought though - the voltage source supplying the entire circuit will range from 2.7V to 4.2V (single cell Li-Ion). Doing a quick search I have found this N-Channel MOSFET:

http://www.diodes.com/datasheets/ZXMN3B01F.pdf

Full turn on is guaranteed below the lowest input voltage, and it will happily function well above the maximum input voltage. It can also comfortably handle the continuous drain current I require (1A).

The only problem I am seeing here is that in the LTC3454 circuit, the LED return path goes to the LED pin on the LTC3454 chip, not ground via a set resistor as per normal constant current regulators. That being said the block diagram on page 6 of the LTC datasheet indicates that the LED return pin does connect to ground as well as an error amplifier.

Will this cause problems?

To be clear - I think I will go N-Channel instead of P-Channel. I will supply a schematic tonight.

 

...........................

• Pull the P-Channel MOSFET gate to ground, and that will turn it on.
• Pull the P-Channel MOSFET gate to Vs, and that will turn it off.

So why not connect the gate to ground using a 10k resistor, and then the gate to Vs using a transistor. The transistor is switched using the MCU output. Turn the transistor on and the gate sees Vs, turning the P-Channel MOSFET off. Turn the transistor off, and the P-Channel MOSFET gate sees ground, turning the MOSFET on.

I'm confused because you said:

"...you also need that resistor (say 10k ohm) from the source to the gate to provide the turn-off voltage."

But everything else suggests my set-up above is how it should be implemented.
..................

If you try to connect up the transistor to do the switching between the source and gate as you suggested, then the transistor will be floating at Vs and cannot readily be controlled by the MCU. If you draw the circuit, then the problem should be apparent. Again, note that all transistors operate on the relative voltages between their terminals, not absolute voltages measured to ground. Do you understand the concept of relative voltages?

That's why a grounded transistor is commonly used to switch the gate of a high-side P-MOSFET to ground to turn it on.

Edit: If you use an N-MOSFET then the source must be connected to ground with the LED in the drain circuit for the MCU to be able to fully switch it ON.