Buck Converter Using TIP122

Thread Starter

ro169

Joined Oct 10, 2014
62
Hi guyz,

I'm building a Buck Converter to step down a 12V input to a 5V output using the TIP122 darlington pair as the switch.

I'm having a problem with the output stage. The tip switches smoothly with just VCC and signal to it. Once the output stage is connected. i.e the diode, the inductor and capacitor, the output reads 12V >> seems like it isn't switching anymore.

Here is a circuit diagram:
Please note the inductor value used is actually 220uH, and capacitor 22uF. The diode is a Fast Recovery diode. Currently I'm testing at open loop, no load conditions. i.e no feedback to the micro, no voltage divider and no load.
Untitled1.png
Please Help. I've tried using a variety of different Inductor and Cap sizes and this didn't change anything.
 

crutschow

Joined Mar 14, 2008
34,282
If you have no load and there's no feedback, then the output will go to maximum voltage.
Try adding some load resistance.

You do realize that a Darlington switch will have much lower efficiency than a MOSFET(?).
 

ian field

Joined Oct 27, 2012
6,536
If you have no load and there's no feedback, then the output will go to maximum voltage.
Try adding some load resistance.

You do realize that a Darlington switch will have much lower efficiency than a MOSFET(?).
The Darlington is also a lot slower - may even make that SB catch diode superfluous in terms of its speed capability.
 

Thread Starter

ro169

Joined Oct 10, 2014
62
It's from my understanding that MOSFETs handle higher switching frequencies but less voltage. Since I'm using relatively low voltages and a low Fsw (50kHz) I figured that I could use either the TIP or the FET.

If you have no load and there's no feedback, then the output will go to maximum voltage.
Try adding some load resistance.

You do realize that a Darlington switch will have much lower efficiency than a MOSFET(?).
I thought of this already. Tried adding a load, didn't seem to change much. The reason I thought the setup would still buck with no load is because the TIP did switch with no connections on the Emmiter. But I suppose the storage elements possibly don't have a discharge path and could propose a problem.

Thanks for the advice:) I will consider using a MOSFET instead
 

Dyslexicbloke

Joined Sep 4, 2010
566
The transistor will work as a learning tool....
50Hz is very slow you will need a huge inductor...

Part from switching speed the primarry difference when using a fet is its ON impedance.
you darlington will always drop a fixed, nearly fixed, voltabe acros its junction wheras a FET will typicall have an RDS ON, that is Resistance between Drain and Source, of a few milliOhms.

I an assuming this is a project to learn about tbe concept given that the parts cost will significantly exceed even an expensive buck board.

Just as an asside industrial applications will oftetn use an IGBT, Insulated Gate Bipo,ar Transistor, rather than a FET where voltage is the factor limmiting FET use but at 12V it isnt going to make much difference.

If you try a FET remember to use a P Channel model or you will have to start playing with bootstrap driver circuits.
Be aware most, but not all PFET,s have higher RDS ON than their N Channel cousens.

You will need a discharge path, however small, try adding the devider even if the input isnt hooked up.

Hope this helps
Al
 

Thread Starter

ro169

Joined Oct 10, 2014
62
Hi, Thanks for your insight.

Possibly you misread? I'm switching at 50 000Hz, not 50.

I'm actually considering changing the switch to a FET. However I'm looking at N-channel FETs. I was under the impression that the higher Rd(on) in P-channel FETs reduces its efficiency and I don't mind having a driver circuit for the switch.
 

crutschow

Joined Mar 14, 2008
34,282
Hi, Thanks for your insight.

Possibly you misread? I'm switching at 50 000Hz, not 50.

I'm actually considering changing the switch to a FET. However I'm looking at N-channel FETs. I was under the impression that the higher Rd(on) in P-channel FETs reduces its efficiency and I don't mind having a driver circuit for the switch.
Note that an N-FET requires a boot-strap driver that can generate a gate voltage above the supply voltage to fully turn on the transistor.
 

ian field

Joined Oct 27, 2012
6,536
It's from my understanding that MOSFETs handle higher switching frequencies but less voltage. Since I'm using relatively low voltages and a low Fsw (50kHz) I figured that I could use either the TIP or the FET.
Some years back I salvaged 900V 9A MOSFETs from some scrap VGA monitors - I used one for a transistor assisted contactor ignition unit.

600V parts are common in flyback SMPSUs designed for 230V mains. The voltage rating can be as low as a mere 20V on PC motherboard MOSFETs.

The gate is voltage driven, not current driven like a transistor B/E. Some MOSFETs have a VGSmax of 20V, although many can only take 15V. Some specialised types even less.
Most common power MOSFETs need at least 6V on the gate to conduct fully, logic level types usually work down to a couple of volts.

As someone else mentioned; an N-channel MOSFET would need a bootstrapped gate supply. A P-channel MOSFET can be switched by pulling the gate toward GND, but N-channel gets precey if you want big specifications.
 

Thread Starter

ro169

Joined Oct 10, 2014
62
Yes I'm aware that the N-channel requires a driver, Thank You.

I suppose I will run some tests with a P-channel as well and compare their respective efficiencies. Could you recommend a P-Channel and/or a N-Channel MOSFET for my application.
I haven't worked with P-Channel MOSFETs before.
 
Last edited:

ian field

Joined Oct 27, 2012
6,536
Yes I'm aware that the N-channel requires a driver, Thank You.

I suppose I will run some tests with a P-channel as well and compare their respective efficiencies. Could you recommend a P-Channel and/or a N-Channel MOSFET for my application.
I haven't worked with P-Channel MOSFETs before.
Most manufacturers publish selection charts - the easier they make it to pick the right one, the more parts they sell.
 

ian field

Joined Oct 27, 2012
6,536
Yes! I've been browsing the International Rectifier website.
http://www.irf.com/product/_/N~1njcht

Settled on IRF540 N channel. Running some tests on it at the moment.
Looks promising :D
Your schematic shows a battery - if the real circuit runs off a mains transformer; you can rig a simple diode/capacitor charge pump on the secondary to provide the required gate supply. Its essentially a doubler that adds the peak voltage on top of the main rectified feed. You need to check the absolute maximum rating for the gate and clamp the bootstrap rail lower than that. Most gates can handle about 18 - 20V, but some published papers suggest much longer life if you don't exceed 15V.

The clamp can be as simple as a Zener and a current limiting resistor - the resistance can be fairly high unless you're going for fast switching.
 

Thread Starter

ro169

Joined Oct 10, 2014
62
Yes the source is a 12V battery.

I'm currently using the IRF540 as a switch.
And the TC4427 as a driver. (The source voltage for this is the 12V battery which drives the PWM signal to a square wave alternating between 0-12V).

The FET is switching quite cleanly with no output stage connected. However I still seem to be having a problem when the output stage is connected. AT 50% duty cycle the setup only bucks the output to ~8V and only reduces to 5V at a very low duty cycle (~10%). However at such a low duty cycle output is noisy and shows a ripple voltage ~5%.

I'm not quite sure as to why at 50% duty cycle the output is so high, and I don't want to reduce to duty cycle to such an extent where the ripple voltage is that high.
 

ian field

Joined Oct 27, 2012
6,536
Yes the source is a 12V battery.

I'm currently using the IRF540 as a switch.
And the TC4427 as a driver. (The source voltage for this is the 12V battery which drives the PWM signal to a square wave alternating between 0-12V).

.
Is that a high side MOSFET driver chip?

I'd probably use a blocking oscillator/inverter to generate the gate supply.
 

ian field

Joined Oct 27, 2012
6,536
The TC4427 is a non-inverting driver, so basically amplifies the signal voltage levels (microcontroller output of 0-5V) to that of its supply voltage (12V battery) to recreate the PWM signal alternating between 0-12V.

Here is the datasheet:
http://ww1.microchip.com/downloads/en/DeviceDoc/20001422G.pdf
That's for driving a complementary pair half bridge - no bootstrap for driving an N-channel high side MOSFET.

To switch on a regular power MOSFET; the gate must be at least 6V above the source - if you want the source to go all the way up to Vdd, the gate has to be supplied by more than 6V above Vdd.
 

Thread Starter

ro169

Joined Oct 10, 2014
62
I have the MOSFET connected in the following way:

PWM at the Gate (with a 10K pull down resistor)
12V at the Drain
Output from the Source
The FET switches fine without the output stage of the buck connected
 
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