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I'm trying to make a buck converter with this mickey-mouse MC34063A chip that will step-down +5V to +3V3 up to 500 milliamps. I've gone through the calculations by hand and validated them with an online calculator as well. My PCB works great with small loads, but with loads of around 300+ milliamps the voltage starts to taper significantly.

Considering this, I'm guessing that the ESR and ripple characteristics of the output capacitor were chosen poorly by me. My question is about the output capacitor and inductor characteristics: what ESR values should I be looking for in the inductor and capacitor? People keep saying "keep it low", but what is actually "low"?

For reference here is the capacitor and inductor that I used:

Capacitor: https://www.digikey.com/product-detail/en/panasonic-electronic-components/EEE-FT1V680AP/PCE5015CT-ND/2652070

Inductor: https://www.digikey.com/product-detail/en/bourns-inc/SRP1250-6R0M/SRP1250-6R0MCT-ND/3767942

//------As requested, here is the schematic, PCB layout, and more context------//

// Frequency - 100KHz // Peak Current - 1A // Input - DC bench power supply (30V/10A max) // Load - Various loads drawing anywhere from 0 - 500mA.

My scope shows 200mV pk to pk ripple on the output, and the switching waveform on the inductor side looks like pork and beans. enter image description here enter image description here

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    \$\begingroup\$ And now, for the schematic...... \$\endgroup\$ – Andy aka Apr 22 at 15:49
  • \$\begingroup\$ What are you using as the input power source for your converter? How much power can it provide? \$\endgroup\$ – The Photon Apr 22 at 15:49
  • \$\begingroup\$ In a buck regulator, the inductor or power switch peak current should be set to the average output current you want (500 mA) plus half of the inductor ripple current (plus some margin of course). If the venerable MC34063 fails to deliver beyond 300 mA, what is the peak current setting with this circuit? The peak current limit is very crudely done in this IC as a transistor brutally lifts the timing capacitor to interrupt the current cycle. But it works and this chip has sold by millions and been replicated several times by other semis vendors. It's not a mickey-mouse chip! : ) \$\endgroup\$ – Verbal Kint Apr 22 at 15:53
  • \$\begingroup\$ The inductor is fine, looks like overkill for your application. The cap has enormous ESR, and will result in output ripple and a low frequency ESR zero. It may be that your loop is becoming unstable. Look at Vin when the output is dropping, and if you have a scope look at the switch node and Vout. And post the schematic. \$\endgroup\$ – John D Apr 22 at 15:53
  • \$\begingroup\$ Adding 10uF ceramic in parallel or switching to a tantalum capacitor would be much much better than the electrolytic. Also second the request for you to post your schematic, and add POST THE LAYOUT. Layout of a switching supply is CRITICAL. You can't guess your way into it - You have to do it with intent. \$\endgroup\$ – Kyle B Apr 22 at 16:25
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Your inductor is 6 uH ? The datasheet tells us that the max. frequency is 100 kHz and that the peak output current in the switch is 1.5 A.

How does the buck work? During the "on" time, energy is stored in the inductor, and during the subsequent "off" time, that energy is released into the load (powering it while the input is effectively disconnected).

Let us say you chose 50 kHz as your switching frequency. This means at 100% "duty cycle" the switch is on "20" μs. 100% is not realistic -- let's choose a very generous 90% duty cycle. Then the switch powers the inductor, every second, for 50,000 bursts of 18μs each (where 18 = 90% of 20).

The energy stored in an inductor is 0.5 L I². Let's suppose that you're running at the "maxed out" level and the switch fully is delivering 1.5A (datasheet maximum) when it turns off. Then, the energy stored in the inductor is: $$ E = \frac{1}{2} L I^2 = \frac{1}{2}(6 \mu H) (1.5A^2) = 6.75 \mu J $$ This happens 50,000 times per second. Thus, the total power available from the inductor is $$ P = (6.75 \mu J) \times (50,000 /sec) = 0.3375 W $$ Now think about this: Your load is 3V3 at 500mA = 1.65W total. Now, some of this comes directly from the source (when the transistor is ON), but you can see from this quick calculation that your maximum output is very, very affected by the choices of operating frequency, inductor size, and the 1.5A maximum switch current.

Let's say every cycle, the inductor totally "empties" i.e. its stored energy goes to zero (just before the switch turns on to begin replenishing it). When replenishment begins, one end of the inductor is at the load, at 3.3V. The other end is hard-switched to the +5 rail thru the switch. If your input is only 5V, life will be hard because the switch in 34063 is not one but two diode drops (at least) below the +5V supply. At 0.7V/diode-drop, that means the energy must be supplied to the inductor with only (5-2*0.7-3.3)V across it, which is only 0.3V !

The switch is on for 18μs and the current will rise from zero to its final value at a rate $$ i(t) = \frac{1}{L} \int v(t) dt = \frac{1}{6 \mu H} \int_0^{18\mu s} (0.3V) dt = 0.9A $$ which is quite a bit lower than the 1.5A maximum above. And the energy stored is proportional to the square of this value too! This means that the inductor is doing very little work during the "off" time.

So life is difficult. I would say, based on this, that the 34063 is not a good choice for this particular application due to its low operating frequency and its rather lossy switch transistor. That 0.3V differential across L is a real difficulty, and is likely responsible for the trouble you're seeing at the high end.

But I hope working thru these equations is useful to see how designing SMPS's can be quite an undertaking...

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  • \$\begingroup\$ +1, this is a very good first answer. (Welcome Atomique!) \$\endgroup\$ – Adam Lawrence Apr 23 at 13:41
  • \$\begingroup\$ Thank you for the detailed answer. That makes sense with that darlington setup I see in the data sheet. I am getting a new appreciation for voltage and current regulation networks. \$\endgroup\$ – jm567 Apr 25 at 0:42
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If you are stepping down from 5.0 V to 3.3 V, the converter is operating with a duty cycle of 66%. At 100 kHz, your on-time is 6.6 us. With 6 uH inductor and 500 mA load current, you can expect a peak inductor current of 2.12 A, which is much higher than your load current. I assume here that you are operating in continuous conduction mode. I think for such low load currents you should be using a larger inductor or you should use a different converter with higher switching frequency. You are probably running the converter in current limit and the output voltage is not regulated anymore. You can verify if this is the case by measuring the inductor current. In case you can't do this, you can probe the switching node and check how the duty cycle changes with the load. There should be a slight change with load. If the duty cycle is being reduced agressively with the load current is because you are hitting the current limit.

PS: The inductor value defines the current ripple and therefore the peak current through the switches. The ESR of the inductor generates additional losses. If you don't care much about efficiency, this is not important. Just use one with a ESR of the same order of magnitude of the on resistance of the switches. The current ripple goes through the output capacitor. You should determine its capacitance based on a ripple requirement. The larger the output capacitor the lower is the ripple. Have in mind that the ripple current also goes through the ESR of the output cap. If the ESR is large it might be the main contributor to the ripple magnitude. If you look at the ripple waveform and it looks triangular, then you know the ESR is too large (the triangular shape is the inductor current flowing through the ESR of the output cap). This will also affect the load transient performance of the converter, but I assume it is not a major concern for you.

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  • \$\begingroup\$ +1, this is a very good first answer. (Welcome Ricardo!) \$\endgroup\$ – Adam Lawrence Apr 23 at 13:40
  • \$\begingroup\$ Hi, Ricardo and thank you for your reply. I will consider using a larger inductor and tantalum cap for improving my design. As for the recommendation in higher switching frequency - is this because the faster the control signal switches from on to off, the smaller the peak current and ripple on the inductor? I want to be sure I understand the why. Thanks! \$\endgroup\$ – jm567 Apr 25 at 0:49
  • \$\begingroup\$ The voltage across the inductor sets the slope of the change of current in the inductor. The slope is V/L. When the SW node is pulled to Vin, the voltage across the inductor is Vin-Vout. When the SW is pulled to ground, then it is just Vout. The average inductor current is equal to the load current. And then you have the ripple current on top. The peak current is the load current plus half the ripple current. The ripple current is (Vin-Vout)/L×Ton = (Vin-Vout)/L×Vout/Vin×T = (Vin-Vout)/L×Vout/(Vin×f). \$\endgroup\$ – Ricardo Nunes Apr 25 at 6:20
  • \$\begingroup\$ You can see in the expression that the lower the switching frequency and the lower the inductor, the larger is the inductor current ripple. You don't want to be in the extremes. Too large peak current can trigger the current limit. Also, most likely you are using a current mode converter. If you are, this converter senses the inductor current and uses it to regulate the output voltage. A small current ripple is also not good because you have a smaller sense signal and noise in the converter will create jitter. I think people usually set the ripple current to be one third of load current. \$\endgroup\$ – Ricardo Nunes Apr 25 at 6:28

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