# Designing oscillating circuit for DIY DC -DC converter

Am building a power supply that uses a transformer which outputs 24v rms. The problem is it uses various sensitive components such as an LCD. I think I have thrown together a buck converter to solve this issue. However am stuck on how to drive the MOSFET since I can't use a 555 timer because I don't have a 5v rail.

I have a feeling am trying to run before I can walk but hey. Ofcourse I can just buy a ready built dc-dc converter but I thought I'll have a go at building a crude DIY one. Not too bothered about efficiency.

Here's what I've done but am not getting 5v. As previously mentioned the output voltage vin * duty cycle which is 5.1v. Am going to stick this into a 3.3v reg.

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What do you want with your conveter? A 5V voltage? Be clearer. –  Kristoffon Aug 8 '12 at 21:30
Let's make it work... for fun... If you use two equal bias resistors on the gate to 32V & ground . it will turn on this MOSFET and not exceed the +/-20V limit on Vgs. Use open drain or open collector with correct resistor to bias Q1 for 0V Vgs shutoff condition (anything <2V=Vgs) Now use any relaxation Schmitt oscillator biased to give 70% d.f. and choose Cap to ground on input for 31KHz. In a SMPS chip the clock sets a flip flop and then resets when voltage rises to regulate a buck convertor. (simplified)using 1.25V bandgap reference diode and comparator. –  user11355 Aug 9 '12 at 15:17

I agree with Russell and David: toss the N-channel FET and use a P-channel instead. The IRF9530 looks OK if we keep an eye on the gate voltage, that should be less than 20 V, and since you have 32 V in that means you can't pull the gate all the way to ground.

Drive an NPN transistor or N-channel FET with the rectangle wave, and connect collector/drain via a resistor divider to the +32 V. 10 kΩ on the +32 V side and 10 kΩ on the low side will give you 16 V gate voltage when on, and a few mV when off. 16 V is good for more than 10 A. For Q1 you can use a BC546, for example.

You'll have to adjust the duty cycle of the rectangle wave to 16 %. Keep in mind that this is a DC/DC converter, but not a regulator: the output voltage will vary with varying input voltage. A next step may be to add feedback from the output voltage so that the duty cycle can vary to keep the output voltage constant.

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I changed a few components' values after Russell's comment. Total gate capacitance of the IRF9530 is about 3.6 nF, and with the original 10 kΩ resistors this gave a charge/discharge time constant of 18 µs, which is much too slow for a 100 kHz clock. With the current 100 Ω resistors this is 180 ns.

C1 gives a short base current peak to start charging the gate faster. This wasn't required with the 10 kΩ resistors, as Q1 would be in saturation quickly, but with the lower load extra base current is welcome.

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Stevens's circuit is a starting point. That MAY be a bit slow switching at 100 kHz. Adding a say 1 NF capacitor across R3 will make it switch faster. Alter cap value for best switched waveform. || Even that does not provide much gate drive current to the MOSFET. If you get serious about this circuit ask and I'll supply a higher gate current drive version. –  Russell McMahon Aug 9 '12 at 13:13
@Russell - You're right about the gate current drive; R1 and R2 should be much lower resistance. The capacitor is a good idea, I'll add it. Thanks for the feedback. –  stevenvh Aug 9 '12 at 13:24
Thanks for the reply steven. But I think you have misunderstood what am asking for. I need a 5v rail from my 32v input. The circuit you gave me gives me 16v what must I do to change this? Also am sorry if I wasn't clear "by design a oscillator circuit" I meant a circuit that generates a 5v square wave @ the required duty cycle. The function gen was a quick and easy way for me to play around. –  Ageis Aug 9 '12 at 23:14
@Ageis - No, the 16 V is only the gate voltage. I had to use the divider to get the 16 V because the gate-source voltage shouldn't exceed 20 V. That's true for most MOSFETs (I couldn't find others). But the 16 V would turn it on just like when you wouldn't have the voltage divider, and pull the gate all the way to ground. Do you mean you need ideas for the actual rectangle wave generator? The 555 will do, but you'll have to use a voltage regulator for it, since it can't work at 32 V. Use a LM317L or a 12 V zener diode to get 12 V. Let us know if you need more details on this. –  stevenvh Aug 10 '12 at 6:44
thanks for your help too!! –  Ageis Aug 15 '12 at 15:55
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You have shown an N Channel MOSFET, whose gate must be driven positive relative to its source to turn it on.
The MOSFET gate must be driven ABOVE source by typically 5V or more. This value varies with device.

To use an N Channel MOSFET in that circuit you need to reference the 5V gate drive to +32V so it swings 32/37V.
Vout here ~+ Dutycycle x Vin = 70% x 32V ~= 22V.

As David Kessner says, a P Channel MOSFET may be an easier-to-use choice here.
Connect source to Vcc = +32V, drain to L1 and drive gate negative going relative to source and to +32V to turn the FET on.

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This is true for an N-channel MOSFET, which is what is used in most switching power supplies. A P-channel MOSFET gate need to be driven about 5v lower, not higher, and might be easier in this case. The disadvantage is less efficiency and more heat. –  user3624 Aug 9 '12 at 0:20
@DavidKessner - Yes. He showed an N Channel FET. I was rushing out so he got a 3 line answer. Now expanded. –  Russell McMahon Aug 9 '12 at 4:56
I know what it's like. +1 anyway! :) –  user3624 Aug 9 '12 at 5:05

The biggest hurdle I see with your buck is that the 32V input is above the input range of the majority of control ICs out there. Most power supply PWM controllers operate at 18V or lower. You will most likely need a lower-power rail if you intend to use a cheap off-the-shelf controller.

Having a P-channel MOSFET is a good idea if you're trying a do-it-yourself approach. A buck controller for N-channel MOSFETs will have a bootstrap supply driver included, so you won't need to worry about the higher-than-the-rail supply yourself.

As others have pointed out, the duty cycle for the switch will be the output voltage divided by input voltage if the buck is running in continuous mode, and will be less that that if it's discontinuous.

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thanks for that. but that's not what am getting. See updated schematic –  Ageis Aug 10 '12 at 12:57
Your buck is at zero load. The duty cycle is a function of conversion ratio and output load in discontinuous mode. Either reduce the duty cycle or add enough output load to make the converter go into continuous mode. –  Madmanguruman Aug 10 '12 at 19:21
Am I correct in assuming the converter will be more efficient if it's in continuous mode? Is there a way I can estimate this load or is it trial and error? –  Ageis Aug 14 '12 at 22:07
CCM is less efficient than DCM. In DCM the inductor current is zero when the switch turns on, there's no turn-on loss (just gate drive loss). The DCM/CCM threshold is a function of the inductance and switching frequency - a higher inductor will go into CCM at lighter load. –  Madmanguruman Aug 15 '12 at 3:03
+1 for Madmanguruman. Thank you very much for your help. –  Ageis Aug 15 '12 at 15:55
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