Hi
I'm hoping someone could help me out?
See the 12v to 9v DC circuit design below, not being very technical myself, could any one tell me the purpose of the 2 resistors R51 and R52.
I have been playing around with the value of R51 and @100ohms I get a 10.2v output and @50ohms it drops to 8.9v.
Just wondering if this is legitimate way to adjust the output voltage or do they serve another purpose?
Many thanks

 

Welcome to AAC!

It's a gate driver circuit for a half bridge. What makes you believe it's a voltage regulator?

 

Thanks for the info Dennis, all I know is was designed to provide a 9v rail.
Sometimes on power on it fails, dropping R51 to 80ohms seems to cure the problem, but I'm not cleverer enough to know why?

Thanks again

 

Sometimes on power on it fails, dropping R51 to 80ohms seems to cure the problem, but I'm not cleverer enough to know why?

Messing with R51 is not a legitimate way to control the output voltage.
The duty-cycle of the PWM input (pin 2) controls the output voltage.

What do you mean it "fails"?

 

Thanks for the info Dennis, all I know is was designed to provide a 9v rail.
Sometimes on power on it fails, dropping R51 to 80ohms seems to cure the problem, but I'm not cleverer enough to know why?

Thanks again

It's not a voltage regulator. If it was, the voltage on the right side of the inductor would be fed back to a compare circuit. To get 9V, the load would need to be constant and a specific PWM signal would be needed.

If you want a 9V voltage regulator, look for a 9V LDO voltage regulator. Or look for a real switching regulator (like MC34063 that can be configured for step-up or step-down).

 

Hello,

In the NCP5111 datasheet there are some examples for DC-DC converters given:



As you can see there are no gate resistors and there is an output transformer used.

Bertus

 

Many thanks for the reply.

When it fails on power up the attached leds flash rapidly. Adding more leds seems to help. So with my limited knowledge it is as though it needs a certain load to function correctly? Cheers

 

The circuit implements a buck converter with synchronous rectification. The upper FET is the switch and the lower FET is the rectifier.

Provided that the current through the inductor is never allowed to reach zero during the course of a switching cycle, the output voltage is the product of the input voltage and the duty cycle. Ignoring losses, for 9 V output with 12 V input, the required duty cycle would be 0.75. In reality, it will need to be a bit higher due to conduction and switching losses. If the inductor current goes "discontinuous" - drops to zero during a switching cycle - the duty cycle must be reduced to maintain voltage regulation.

Due to the large deadtime (neither FET on) enforced by the IC, the circuit would be improved by adding a Schottky diode across the rectifier FET (cathode to junction of the FETs). As it is with the long dead time, the intrinsic body diode of the lower FET will be acting as the freewheeling diode. This isn't a terrible thing, but certainly isn't ideal. In reality, with the high duty cycle required, the gain in efficiency with sync rectification versus a simple Schottky diode won't be great. I have to look at the FET specs to quantify the difference.
[EDIT] - had a brief look at FET data; it's inexpensive and would save enough power versus a simple Schottky to make it worth using

100 ohms is a very high for a gate current limiting resistance (the resistors inquired about). I'd have to look carefully at the whole circuit to recommend an "optimum" value, but something in the range of 10 to 20 ohms is probably suitable. 1% precision certainly isn't required. The changes in output voltage observed due to changes in the resistor values are because the losses in the circuit have changed and the effective duty cycle has been altered by changing the rise and fall times of the currents through the FETs.

The capacitance of C31 is probably far too small. If the circuit is getting input power via a "barrel connector" it is probably getting it down a long piece of wire that has relatively high resistance. Buck converters "chop" the input current and place high demands on the input capacitor, especially in a case as described. The amount of capacitance required depends on the output current of the circuit, but capacitance in the range of several tens to hundreds of microfarads would likely be required. Ceramic-only at the end of a long connecting cable can result in very high amplitude ringing as the capacitance resonates with the cable inductance in response to the step changes in current.

I dislike circuits like this because there is no protection for anything in the event of output overload.

 

Thanks ebp, great response and very helpful.
Not quite getting what you mean with sync rectification, is this a different type of FET?

The other 2 comments make a lot of sense, so I will investigate making the changes

Thanks again

 

In an ordinary buck converter there is a diode between the "input" side of the inductor and ground, cathode to the inductor. When the switch (usually a FET in modern circuits) between the input supply and the inductor turns off, the energy stored in the inductor must so "somewhere." An inductor "tries" to keep the current through itself constant. The diode forms the path that allows this current to flow. The energy stored in the inductor flows through the diode and hence into the output capacitor and load circuit. The voltage on the capacitor doesn't change much during a single cycle, so the current ramps down more or less linearly, starting at exactly the same magnitude as it was at the instant the FET turned off. If it gets to zero before the FET turns on again, it is said to be operating in discontinuous inductor current mode. If current is still flowing when the FET turns on again, it is said to be operating in continuous inductor current mode. These are sometimes abbreviated DCM and CCM. Ignoring switched capacitor types that are normally used only for low power, all switch mode converters rely on two energy storage elements - an inductor and a capacitor, and high-efficiency energy transfer between them.

The diode can be replaced with a FET that is turned on when the main FET is turned off. This FET then acts as the path for discharge of the inductor current and is referred to as a "synchronous rectifier" because it is controlled synchronously with the "requirement" for it. With modern FETs with very low ON resistance, this can result in a path with only a small fraction of a volt across the "rectifier" element instead of the 0.5 to 1 V or so that is common with conventional or Schottky diodes. You can quite easily get the voltage drop down to 0.1 V or even less with a suitable FET. This can produce a significant improvement in efficiency, especially when the input voltage is much higher than the output voltage so the rectifier is conducting most of the time. The complexity is a little higher since you need to actively control the FET. One consideration is that the sync rectifier FET can conduct in reverse so the inductor will operate in forced CCM - after it has discharged, it will begin to charge with current drawn from the output capacitor. This may be unacceptable, so many dedicated power supply control ICs detect it and turn off the FET to prevent it.

Synchronous rectification can be used in other switcher topologies - boost and inverting (flyback) and their derivatives.

 

R51 and R52 "typically" are to control gate ringing in the drive circuit due
to stray L. Normally one maximizes gate C charge current so that FET
can switch fast and to minimize gate drive power.

http://www.ti.com/lit/wp/snva595a/snva595a.pdf

Gate resistors and switching characteristics Generally, a resistor is connected to the gate terminal of a MOSFET. The purposes of the gate resistor include suppression of inrush current and a reduction in output ringing. A large gate resistor decreases the switching speed of a MOSFET. This results in an increase in power loss, a reduction in performance and potential heat issues. Conversely, a small gate resistor increases the switching speed of a MOSFET, which makes it susceptible to voltage surge and oscillation and therefore to device failure and damage. It is therefore important to optimize the MOSFET switching speed by adjusting the gate resistor value

https://toshiba.semicon-storage.com/info/docget.jsp?did=59460


Regards, Dana.

 

Thanks Dana
Would R51 and R52 always be the same value?

Cheers, Fred