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N-Channel DC/DC Converter

Thread starter #1
Hi,

I am working on a schematic which I am unfamiliar with and after staring at the datasheet for hours I figured I would reach out here and see if someone can help lead me down the right path.

Attached is the schematic right from the datasheet I am trying to understand. My biggest concern is how the two PNP transistors are working with the LTC3875.

Please give me a "seat of the pants" view on how this thing works...and if I haven't spent enough time staring at it feel free to tell me to RTFM :)

The circuit I actually want to use it for is a -9V to -5v buck.

Thanks all.
 

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dknguyen

Well-Known Member
Most Helpful Member
#2
The two PNPs is a current mirror. It looks like it's output is inserted straight into the middle of the feedback loop and can influence it.

I think it's supposed to be taking the output voltage (referenced to power supply GND) and floating it down to to be referenced to the negative power supply rail since that is ground as far as the IC is concerned, and the IC is expecting Vfb to receive a feedback voltage referenced the GND pin.
 

ronsimpson

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Most Helpful Member
#3
The IC looks at pin 3 to 4 to see what the output voltage is. If V 3-4 is less than 1.2V the duty cycle in increased. If V3-4 is more than 1.2V the duty cycle is decreased.
In this circuit the IC is being used upside down. So it can not easily see the output voltage.
Look at my right hand circuit. When the "-5V" is at -5 volts the diode causes the base of the transistor to be at -5.6 volts. This puts the emitter at -5V. This puts 5V across the 5k resistor. So 1mA is flowing through the emitter of the transistor. This 1mA must flow through R2 causing 1.2 volts across R2. (5V on R4 makes 1.2V on R2) This way the IC can see the voltage on the output.
upload_2018-6-21_15-48-52.png
The left hand circuit is the same but used a transistor to act like a diode. In the right circuit the voltage drop in D1 might not equal to the voltage B-E on the transistor. In the left side we are using the same type of transistor so the voltage drop on Q1=Q2.
 
Thread starter #4
ronsimpson,

Thank you for such a great explanation. After digesting it a little I fully understand what is going on. Due to the fact that there is no easy way for the LTC3873 to sense the voltage output it is creating by pulsing Q1 on and off they have to come up with a feedback to the chip. They did this elegantly by using a current mirror to create the 1.2volts across pin 3 and 4. As the -5v output goes too high or goes too low due to load change, this will change the 1.2 voltage to go higher or lower, which then tells the chip to adjust it's duty cycle to the FET accordingly to maintain the constant -5v.

Again, thank you. The drawings you made helped a ton. Much respect.
 

ronsimpson

Well-Known Member
Most Helpful Member
#5
Post #2 used that word but I disagree.
Post #3 right side. The transistor is an amplifier. The gain is R2/R4. 1.2/5 so less than one. Normally a amplifier has a gain of more than one but they can have a gain of 0.24.
The input is injected into the base of the transistor. In this case -5 volts.
5 X 0.24 = 1.2 Input X gain = output.

The transistor or diode is just to make the amplifier temperature stable.
 

ronsimpson

Well-Known Member
Most Helpful Member
#7
If no diode then the voltage across the 5k resistor is 5V-D=4.35 volts. Where D is one diode drop. The voltage across the Base to Emitter of a transistor is about the same as a diode. 0.6, 0.65 or 0.7 depending on temperature. We start out with the 5 volts to measure. Then drop down 1D then back up 1D. So the voltage across the 5K resistor is 5V-1D+1D=5V If the -1D and the +1D are exactly the same then all the "diode drops" subtract out. The voltage across B-E of a transistor changes with temperature. By having a -1D and a +1D the "thing" that changes with temperature is removed from the formula.
 

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