# DC-DC booster functionality

I am working on a DC-DC booster. I have already built it, boosting 1.7 V to almost 18 V. I am using an IRF540N as my transistor, an inductor, a diode, and two capacitors.

I was reading about how the booster raises the voltage. I was confused, specifically, on how the transistor works and the flow of current.

How does it raise the current?

How does the current really flow through the circuit and especially through the transistor? What is its scientific base?

• You cant raise both voltage and current at the same time. May 7 at 16:41
• You have a simulation. Look at the current and voltage through L1. When viewing this, understand the basic operation of an inductor. You will see the inductor charge up then dump. Rinse and repeat. May 7 at 16:50
• Useful search term : Lenz's Law. May 7 at 16:56
• Welcome! Have you tried to search for how a boost converter works? In your simulation, have you tried to probe the voltage and current in each node to correlate where current flows when and how that affects the voltage? Current will flow though the transistor when it’s turned on and no current will flow though it when it’s off. As for scientific base, it’s what @user_1818839 said, Lena’s law. May 7 at 16:58
• Cheap solar garden lights use an inductor in a transistor IC to boost the voltage from a 1.2V rechargeable battery cell to over 3.2V for a colored LED. May 7 at 18:46

The inductor alternately stores energy as magnetic flux, and releases energy as current.

• transistor on: ‘charging’ L1 (building flux)
• transistor off: ‘discharging’ L1 (collapsing flux)

During that second phase, as the flux collapses, the voltage across L1 flips, dumping current through the diode and into the load.

Here's a Falstad sim to try (try it here):

I modified the inductor to a smaller value (100uH) which is more appropriate for a 50kHz switching frequency. You can do the same for your sim, it should work then. 10mH is much too big.

The takeaway is, the same energy is loaded into the inductor during 'charge', then released during 'discharge' into the output. If, for example, you're stepping voltage up 10x, the output current will be about 1/10th the input current. But the power and energy are the same, less losses. So if you have 1A in, you get at 10x the voltage at a bit less than 100mA for your trouble.

With a buck it's the opposite: you're stepping down. If you're inputting 1A and stepping down 10:1, then the output is 10A but at 1/10th the output voltage.

In either case, the power in is the same as power out (less losses.)

What is ‘stepping ratio’? It’s the ratio of Vin to Vout. Stepping ratio roughly correlated with the PWM duty cycle. The relationship for a boost is expressed as:

• Duty cycle = 1 - Vin/Vout

So a 90% duty yields 10x the voltage and 1/10 the current.

Q1 turns on, connecting the end of L1 to GND. D1 is off, and so load current is (temporarily) supplied by C2. Current builds up in L1, and when Q1 switches off, L1 'wants' to keep the current flowing -- so it flows through D1 to top-off C2. L1's current will decay (quickly), so it doesn't deliver its current for a long time (that's why you need Q1 on for a high fraction of each cycle).

• I understand all of that. However, the IRF540N has a diode that allow the current to pass from the source to the drain, according to its symbol. Does that still mean the current will flow through the transistor from the drain to the source? May 8 at 16:49
• It does (nearly all FETs do) have that diode. However it doesn't ever conduct in this configuration. When the FET is 'on', current can flow from D-S (or S-D), when off current can't flow from D-S (but it could flow S-D via the diode). May 8 at 17:52