The object of this activity is to explore an inductor based circuit which can produce an output voltage which is higher than the supplied voltage. This class of circuits are referred to as DC to DC converters or boost regulators. In this activity the voltage from a 1.5 V supply ( battery ) will be boosted to a voltage high enough to drive two LEDs in series.
As in all the ALM labs we use the following terminology when referring to the connections to the M1000 connector and configuring the hardware. The green shaded rectangles indicate connections to the M1000 analog I/O connector. The analog I/O channel pins are referred to as CA and CB. When configured to force voltage / measure current –V is added as in CA-V or when configured to force current / measure voltage –I is added as in CA-I. When a channel is configured in the high impedance mode to only measure voltage –H is added as CA-H.
Scope traces are similarly referred to by channel and voltage / current. Such as CA-V , CB-V for the voltage waveforms and CA-I , CB-I for the current waveforms.
The circuits used in this Lab activity while generally low current can potentially produce voltages beyond the 0 to 5 V analog input range of the ALM1000. Input voltage divider techniques as discussed in the document on ALM1000 analog inputs would be required. Refer to the document and construct and use input voltage dividers as necessary when preforming these experiments with the ALM1000.
When the current flowing in an inductor is quickly interrupted a large voltage spike is observed across the inductor. This large voltage spike can in fact be useful in some cases. One example is the DC to DC boost converter, which is a circuit that can create a larger DC voltage from a smaller one with very high efficiency. The basic idea is to combine an inductive spike generator with a rectifier circuit, as shown in figure 2. Whenever the transistor is abruptly turned off the voltage at the drain spikes up, the diode D1 is forward biased and current will flow from the inductor to charge up the storage capacitors C3 and C4. When the drain voltage subsequently drops below the voltage on the capacitor, the diode is reverse biased and the output voltage remains constant. Just as in the chapter on AC power supplies, the output capacitor must be sized appropriately to minimize the ripple relating to the current flowing in the load. We will just use a small capacitor here and hence the circuit will not be able to source a large output current.
This exercise will expand on these concepts, deriving a converter that “boosts” a low voltage to a high voltage. Subsequent activities will close the loop around these circuits and examine loop stability and time-domain response.
* (This exercise will use the term “ideal” extensively. A more accurate term would be “almost ideal” - LTspice requires finite numbers in certain locations - switch on and off resistances can't be zero or infinity, so we're using values small enough and large enough to have negligible impact on the results.)
Open the OL_Boost_concept_ideal_sw.asc LTSpice file. Notice the differences between this circuit and the buck converter:
As with the buck basics lab, let's keep two things in mind at all times:
The figure below shows the “charge” state of the circuit’s operation, where S1 is closed and S2 is open.
Figure 2. Boost Converter Charge
When S1 closes, the left-hand side of the inductor is connected to the 5V supply, and the right-hand side is connected to ground. This means the voltage across the inductor is simply the 5V supply. This “charges” the inductor with a current that ramps up with a positive slope of:
Note: The polarity of the voltage across the inductor is arbitrary, we're using the convention that a positive voltage is one that causes an increase in energy stored in the inductor.
The next figure shows the other state, with S1 open and S2 closed.
Figure 3. Boost Converter Discharge
When S2 closes, the left-hand side of inductor L1 is still connected to Vin, while the right-hand side is now connected to Vout. The current through L1 is now flowing to the output, and decreasing with a negative slope of:
Similar to the buck converter basics activity 1 the “freq” and “duty” parameters set the frequency of the switching to 25kHz and the duty cycle of the voltages imposed on this switch node (sw_node) to 50%. That is, the righthand side of the inductor spends half of the time connected to ground (charging phase), and half of the time connected to the output (discharging phase). Run the simulation, and probe sw_node, Vout, and the current through inductor L1. Zoom in toward the end of the run after the startup transient damps out (after 8ms). (You can right-click, Auto range y-axis to line up the two waveforms.)
Figure 4. Boost Converter Switch Node, inductor current, and Output
Observe the peak and valley of the waveform I(L1) (green waveform), noting the current ripple. Using the cursors from peak to peak, we can observe that the inductor is charging and discharging linearly (with the period of 1/25kHz, or 40us, with a duty cycle of 50% making ts1 and ts2 20us each).
The output voltage is almost exactly 10V - double the input voltage - with a small ripple imposed. Verify that the previously solved equations are true, using the cursors to measure the inductor current waveforms.
For the “charge” phase:
And for the “discharge” phase:
Revisiting the concept of zero DC across an inductor, how can we find the output voltage of a boost converter knowing the input voltage, frequency, and duty cycle? “Zero DC across an inductor” means that over a long period of time, the average volt-second product is zero. Thus:
Where tS1 is the time that S1 is closed, tS2 is the time that S2 is closed. Rearranging, we see that:
is the duty cycle of the switch node, we can rewrite the expression for VOUT as a function of duty cycle:
Since our duty cycle is based off ts1 and ts2, and the duty cycle is always between 0% and 100%, the above equation demonstrates that the average output voltage is always equal or larger than the input voltage, a basic property of a boost converter, and at a 50% duty cycle, the output voltage is double the input voltage.
Now change the duty cycle in the simulation and re-run. The following are screenshots show the output voltage at 20% duty cycle(expected output of 6.25V) and 80% duty cycle(expected output of 25V).
Figure 5. Boost converter output with a duty cycle of 20%
Figure 6. Boost converter output with a duty cycle of 80%
Can you boost to an arbitrarily high voltage? See Appendix: “Extreme Boosting” to find out.
So far we've operated the boost circuit unloaded. In this condition, the duty cycle to boost factor relationships held true, but what happens if you start to draw current from the output (as you would in a practical circuit - after all, a power supply exists to power stuff!) Furthermore, consider the boost converter's output switch (S2). If we look at the current waveform and the voltage waveform of the unloaded circuit, we see that for part of the cycle, the inductor current goes negative, and when this occurs, the output voltage is ramping DOWN! This seems counterproductive for a boost converter, doesn't it? This mode of operation has a name - “Forced Continuous Conduction Mode”. It is forced because the switches always impose a voltage across the inductor, so its current is always either ramping up or down.
Next, connect the 25 ohm load resistor to the output node (drawing an average current of 0.4 amps from the 10V output). Note that the impact on the output voltage is minimal, and the inductor current is still ramping up and down with the same peak-to-peak ripple, however the current is now always positive (flowing from input to output, according to our convention.)
Figure 7. Ideal Boost Converter with 25Ω load
But are we still FORCING this circuit to conduct continuously? We'll find out shortly…
ADALM1000 hardware module
Solder-less breadboard and jumper wire kit
1 - 2N3904 small signal NPN transistor
1 – ZVN3310 NMOS FET (ZVN2110A, 2N7000 or power FET device such as IRF510)
1 – 1 KΩ resistor
1 – 100 Ω resistor
1 – 47 Ω resistor
1 – 1 KΩ potentiometer
2 – 47 uF capacitors
2 – 0.1 uF capacitors
1 - HPH1-1400L (Coilcraft Hexapath inductor or other value from 1mH to 4.7mH)
2 – rectifier diodes (1N4001, 1N3064)
2 – LEDs ( one red one yellow )
Optional Additional Equipment:
1.5 V AA battery and holder
Small handheld DMM
First step is to build a 1.5V power supply ( to simulate a single cell battery ) as shown in figure 1. Build the circuit on one end of your solder-less breadboard being sure to leave space for the rest of this lab’s circuits. Note, if you have a 1.5 Volt battery (AA) and a battery holder with wires attached, you could substitute that as the 1.5 V supply.
Figure 1, 1.5 Volt power supply
Once you have the 1.5 V supply constructed, you will need to adjust the potentiometer, R1, such that the output is set to 1.5 V. Use one or the other of the ALM1000 inputs in Hi-Z mode to measure the voltage. ( display the AVG voltage for the channel you choose ). An optional DMM could also be used to measure the DC voltage. Note the dashed green box in figure 1 surrounding the ground connections. Later you will be measuring the current in ground for different sections of the circuit. The ground connections in figure 1 will always be connected directly to the ground of the ALM1000. The other sections of ground will, at various times, be either connected to the main ground or the CH-B connector pin on the ALM1000. So as you construct the circuit keep these “ground” connections separate, i.e. don’t use one of the common power bus strips for all the “grounds”.
Temporarily connect one of your LEDs from the 1.5 V output to ground. Be careful to note the polarity of the diode so it will be forward biased. Does it light up? Probably not since 1.5 V is generally not enough to turn on an LED. We need a way to boost the 1.5 V to a higher voltage to light a single LED let alone two LEDs connected in series. Disconnect the 2.5 V supply and remove the LED before moving to the next construction step.
Next, on your solderless breadboard construct the DC-DC boost circuit section as shown in figure 2.
The 6 winding HPH1-1400L inductor can be configured for 6 different inductance values depending on how many windings are connected in series. Connecting all 6 windings in series will give 36 times ( N2 ) the inductance of a single winding ( 0.2 mH ) or about 7 mH. 5 windings = 25 X 0.2 or about 5 mH, 4 windings = 16 X 0.2 or about 3.6mH. Any or all of these configurations should work for L1.
You can use a 1N4001 or a 1N3064 for the rectifier diode D1 and the snubber diode D2. Be sure to connect the left end of L1 to the 1.5V supply from the section in figure 1. The gate of the NMOS switch transistor M1 is connected to the CH-A output of the ALM1000.
The ground connections in figure 2 shown in the dashed green boxes will sometimes be connected directly to the ground of the ALM1000. At other times, they will either be connected to the main ground or the CH-B connector pin on the ALM1000 depending which branch current is being measured. So as you construct the circuit keep the “ground #2” connections separate from the “ground #3” connections (and the “ground #1” connections from figure 1), i.e. don’t use one of the common power bus strips for all the “grounds”.
Figure 2, DC to DC boost converter section
Start with the three sections of ground ( #1 and #2 and #3 ) all connected to the main ALM1000 ground. With CH-B in Hi-Z mode you will be using it to observe various voltage waveforms around the circuit.
Start with a switching frequency of 2 kHz which is supplied by AWG A, CA-V. Set the Min value to 0 and the Max value to 3.5 ( enough to turn on M1 ). Set the mode to SVMI and the shape to square.
Using CH-B in Hi-Z mode observe and save the voltage waveforms seen at the following circuit nodes. First, the 1.5 V input source. Second, the drain of M1. Third the “boosted” output, VOUTat the top of LED1. You should also measure the current in the LEDs by taking the voltage at the junction of R3 and LED2 divided by the value of R3, 100 Ω.
What is the average DC voltage of the “boosted” output? What is the p-p ripple seen on the waveform? What is the average DC current in the LEDs and what is the p-p ripple seen in the current? Record the values in your lab report.
You may want to save snap-shots of the voltage traces or save the VBuffB array to another temporary data array to be plotted later when you are making the current measurements.
Now that we have measured all the interesting voltage waveforms associated with the circuit we need to measure the relevant current waveforms as well.
If we set the channel B voltage to a DC value we can think of it as an AC ( and DC ) ammeter with one end connected to the set DC voltage. The DC voltage we set could be a supply voltage or it could be 0V and we would then be measuring the current flowing into or out of ground.
The first current we will measure is the current in the inductor L1. Be sure the frequency of the clock CH-A is still set to 2 KHz. Set AWG CH-B to SVMI mode and Shape to DC with a Max value of 1.5 V, to match the 1.5 V from the 1.5 V supply from figure 1 ( or AA battery if you are using that ). Because the shape is set to DC the Min and frequency values are ignored. Disconnect the left end of L1 from the 1.5V supply and connect L1 to CH-B. Under the curves menu be sure to select the CB-I trace to be displayed.
Hit the run button and you should see the square wave ( switch drive ) on the CA-V trace, the constant 1.5V DC voltage on the CB-V trace and the inductor current on the CB-I trace. Adjust the CB-I vertical scale as needed to fit the peak amplitude of the current on the screen.
Note the time when switch M1 is on and when M1 is off and the current in L1. Take a snap-shot of the CB-I current trace for future reference ( or save the IBuffB array into another array so it can be plotted later).
Disconnect L1 from CH-B and reconnect it to the 1.5V supply.
Now we want to measure the currents in the two sections of “ground” you separated out when you built the breadboard. First, disconnect the ground #2 section from the main ground bus and connect it to CH-B. Set CH-B Max value to 0 ( the same voltage as ground ). The current in ground #2 is just the current in M1, the switch.
Hit the run button and you should see the square wave ( switch drive ) on the CA-V trace, the constant 0V DC voltage on the CB-V trace and the M1 source current on the CB-I trace. Adjust the CB-I vertical scale as needed to fit the peak amplitude of the current on the screen. Note the polarity ( direction ) of the current. Display the saved L1 current trace and compare the two. What part of the L1 current flows in M1 and where in time does it flow? Save the CB-I trace of ground section #2. (save the IBuffB array into another array ).
Now we will measure the current in the LED load, including the by-pass filter capacitors C3 and C4. Disconnect the ground #2 section from CH-B and reconnect it to the main ground bus. Disconnect the ground #3 section from the main ground bus and connect it to CH-B. The current in ground #3 combines the LED current and the current in the capacitors C3 and C4 which is also the current in diode D1.
Hit the run button and you should see the square wave ( switch drive ) on the CA-V trace, the constant 0V DC voltage on the CB-V trace and the LED load current on the CB-I trace. Adjust the CB-I vertical scale as needed to fit the peak amplitude of the current on the screen. Note the polarity ( direction ) of the current. Display the saved L1 current trace and compare the two. What part of the L1 current flows in load and where in time does it flow? Save the CB-I trace of ground section #3. (save the IBuffB array into another array ).
Conservation of current says that the sum of the current in M1 and D1 should be the same as the current in L1. Use the Math plotting function to plot the sum of the saved ground #2 and ground #3 data buffers. Compare this to the measured L1 current waveform.
As additional investigation.
Now increase the frequency to 4, 6 and 8 KHz. Measure and record the boosted output voltage waveform and the current waveforms. Explain what has changed and why?
With a 2 KHz clock frequency, reduce the duty-cycle of the CH-A square wave switch drive waveform from 50% to 45%, 35% and 25%. How does the duty cycle change the voltage and current waveforms?
The output impedance ( load regulation ) of the simple emitter follower in figure 1 can be improved through the addition of feedback. Shown in figure A1 is a follower circuit where the single NPN ( Q1 is a 2N3904) transistor is replaced with a NPN/PNP compound transistor ( Q2 is a 2N3906 ). The load regulation of a single transistor follower is a function of how much the VBE changes as the emitter current changes. In the case of the compound transistor configuration a small increase in the current of Q1 is amplified and causes a much larger current to flow in PNP transistor Q2.
There is also the added benefit that the base current of Q1 is now much smaller and its effect on the voltage divider of R1 and R2 is much smaller.
Figure A1, Improved follower
Even better load regulation can be achieved by using an op-amp to provide high gain. As shown in figure A2, the output of the op-amp drives the base of transistor Q1 to whatever bias voltage is need to maintain the emitter voltage ( negative input of the op-amp ) equal to the voltage set at the positive input of the op-amp by the voltage divider. The circuit can essentially source current up to the maximum current limit of the transistor or the +5V power supply with very little change in the +1.5V output.
Figure A2, Precision follower
For Further Reading:
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