# SMPS experiment problem

Discussion in 'Homework Help' started by bug13, Aug 8, 2012.

1. ### bug13 Thread Starter Senior Member

Feb 13, 2012
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Hi guys

Circuit is as seen in the picture, when I wired the circuit up today in the lab, I can only adjust the output voltage from around 10.5v to 11v, with 50% duty cycle to 90% duty cycle.

I expect the output should be half of the input @ 50% duty cycle and around 10V @90% duty cycle, am I right?

or what have I done wrong?

Thanks

Zhuhua

2. ### Jony130 AAC Fanatic!

Feb 17, 2009
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The circuit that you post will not work properly because the BJT don't work as a switch.
In you diagram BJT work in emitter follower configuration.
So the emitter voltage never will be greater then 2V - Vbe.

All so you should be aware that this equation
Vout = D * Vin hold only if the coil "work" in steady state. In steady state
ΔIon = ΔIoff , Von * ton = Voff * toff and the average voltages across the coil must be equal 0V. We say that circuit is work in Continious Current Mode (CCM). Current through inductor flows continuously.

For example if we modify the circuit that he can work properly

The circuit will give Vout = 6V for D = 50% only when we change load resistor and we force the circuit to work in CCM.

When the switch is ON, the current that is flow through the coil at the end of a ton time (at time equal to 500us) reach the value equal to:
ΔIon = (Von * ton)/L = (12V - 6V) * 500μs)/30mH = 0.1A
So the minimum average load current must be equal to Iload_min = ΔIon/2 = 50mA.
This give as RL_min = 6V/50mA = 120Ω

In conclusion, this circuit will give as 6V at D=50% only when working in CCM. The coil entry CCM only if load current is greater then 50mA.
And this means that we must change RL < 120Ω.

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3. ### bug13 Thread Starter Senior Member

Feb 13, 2012
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Hi Jony130,

Thanks for you details explanation, it really solve my problem, just a little question, can you explain why the Von =12v-6v please, as I don't quite get that part.

Thanks

Zhuhua

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Just in case you need an answer sooner than later ...

When the BJT turns on it applies ~12V via the emitter to the left hand terminal of the inductor. The other terminal of the inductor connects to the +ve side of capacitor and load in parallel. If it assumed the converter is producing +6V output at 50% duty cycle [CCM], then the capacitor and load are notionally at a mean value of 6V. So the drop across the inductor at the BJT switch on time is notionally (12V-6V) or 6V - which is the Von term Jony130 is referring to.

In practice it wont be that simple, since the assumption about the capacitor voltage being 6V at the instant of BJT turn on will not be exact - particularly where the load ripple voltage is high. If you consider your example situation with the load changed to 120Ω to give almost CCM operation then my simulation shows the instantaneous load voltage to be ~3V at the switch-on point. The simulation also shows a peak-to-peak ripple of ~7V on a mean value of 6V.

As mentioned above a 120Ω load doesn't quite give CCM operation at 50% duty cycle. This only appears to happen in my simulation when the load is reduced to around 90Ω.

Last edited: Aug 9, 2012
5. ### bug13 Thread Starter Senior Member

Feb 13, 2012
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so say I want 4v @50% is:
ΔIon=(Vs-Vl)*Ton/H=(12-4)*500us/30mH=133mA?
or
4V @40% is:
ΔIon=(Vs-Vl)*Ton/H=(12-4)*400us/30mH=107mA?

Just need to confirm my understand is correct.

Thanks

Zhuhua

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Again if the objective is CCM operation then the required duty cycle for 4V output would be 4/12=1/3 or 33.3% rather than 50%.

In addition the lower duty cycle 33.3% would mean setting the load resistance even lower [around 75Ω] to achieve CCM with the other components the same at 1kHz operating frequency.

Again from my simulation the actual load voltage at BJT turn-on with a 75Ω load at 33% duty cycle is ~950mV which is a long way from 4V. For the simple model proposed one would have to get the ripple down by an order of magnitude at least to get some agreement in practice and theory.

I note also from my simulation that 33% isn't sufficient duty cycle to obtain 4V output at 75Ω load. I had to increase the duty to 39%. This is probably due in part to conduction losses in the BJT and limitations of the simple [CCM] model.

Last edited: Aug 9, 2012
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Mar 6, 2009
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8. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Today after the work I build this circuit on the breadboard.

And for F = 10KHz and D = 50% the output voltage was equal 4.45V and I have to change the duty to 56% to obtain 5V output at 39Ω load.
I also measure voltage across the inductor and use Rs resistor to measure inductor current.

And if we reduce the load resistance the inductor current decreases.
And circuit enters discontinuous current mode DCM, inductor current drop to 0A before the end of a switching cycle.

And the circuit start to behave very strange.

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9. ### bug13 Thread Starter Senior Member

Feb 13, 2012
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I tried your circuit at school today, and I got those strange waveform as well, my tutor said it might be the resonant of the inductor and capacitor.

10. ### bug13 Thread Starter Senior Member

Feb 13, 2012
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what is the reason for a 1nF cap on the R2 in Jony's circuit anyway?

11. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Probably to speed up Q2 switching.

12. ### bug13 Thread Starter Senior Member

Feb 13, 2012
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how's that work? by producing a pulse of larger initial current through the cap?

I actually see this kind of connection a lot, are they for the same reason?

Last edited: Aug 11, 2012
13. ### Jony130 AAC Fanatic!

Feb 17, 2009
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I do some measurement to show you how this speed-up capacitor helps.

I use a signal gen F = 100KHz for Vcc = 5V i Vin = 5V; Rc = 1K
I use Rb = 100K as a base current limiter resistor.
As you can see the delay time is equal ts = 4µs

To reduce this delay time I decrease RB resistor value.

Rb = 10K, ts = 2.5µs

Rb = 1k, ts = 1.5µs

And finally with speed-up capacitor and RB = 100KΩ

As you can see the additional capacitor really helps.
This additional capacitor provides additional base current. In the capacitor current is proportional the rate of voltage change across it (proportional to how quickly the voltage across capacitor is changing). The faster the voltage change the large the current flow through capacitor. I = C * dV/dt
And in our circuit the input voltage change very quickly from 0V to 5V.
So this fast change in input voltage produces a flow of a capacitor current. And this additional current flow into the base and speed-up switching time from OFF to saturation.

Last edited: Aug 11, 2012
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