# Why can't you use a coil to transform 5 volts DC to 12 volts DC?

Why can't you just use a coil transformer, like when you want to get 240 volts out of a car battery instead of a boost converter? I found a thread that sparked my interest - How to transform 5V into 12V?. I want to do a similar thing so I can charge things like a laptop from a portable 5V phone battery pack.

• Transformers don't have a fixed direction. You could, if you apply a low voltage high current AC signal on the secondary, get a high voltage low current signal on the primary. But it depends on load, efficiency is low, bulky. Like i.stack.imgur.com/IsFNt.gif, it's a transformer used "backwards" to ring a telephone's ringer from a 9V source. – Passerby Mar 8 '16 at 0:14
• The 5V battery for charging a phone outputs a DC voltage. You would need to generate an AC voltage into a coil to get a higher or lower AC voltage. – tokamak Mar 8 '16 at 0:22
• I designed this: falstad.com/circuit/… in a circut simulator would this work? – J.Clarke Mar 8 '16 at 0:38
• @J.Clarke no. no it won't. Not from a DC battery. Also, 18 Amps in at 12V for 100mA at 5V? wow thats bad. – Passerby Mar 8 '16 at 0:40
• @J.Clarke There are some concepts that need to be understood before stepping up to transformers, AC, DC, etc.. I feel that if I try to explain you why DC voltages can't be raised or decreased by transformers I would create more doubts than knowledge. So, my suggestion is that you try to study simpler components than going directly to transformers. It may not appear that complicated, but transformers are indeed a complex subject. Check resistors, capacitors and mainly inductors, which is the basis of transfomers. – PDuarte Mar 8 '16 at 1:00

Just to be clear: we are talking about converting 5V DC into 12V DC.
The 2013 question is about DC-DC. Plus, there is a .

For a transformer to function, a varying magnetic field must be created inside it. The varying current in the first winding (the so-called "primary" winding) creates a varying magnetic field, which in turn creates an electromotive force in the secondary winding. This does not happen when the primary is fed with pure DC.

Nevertheless, it is possible to make a transformer-based DC-DC converter. DC is chopped so that the transformer is fed with varying current. This is normally done with semiconductor switches. The output of the transformer secondary is rectified, which produces a DC output voltage.

For a small increase in voltage (factor of 6x or less), an inductor-based boost converter is usually more practical than transformer-based (in general, although there are important exceptions).

• I didn't realize you could embed tags. Thanks! :) – bitsmack Mar 8 '16 at 0:41
• So if i put an oscillator between the 5v dc supply and the coil it could work? – J.Clarke Mar 8 '16 at 0:43
• If you feed the output of the oscillator into the transformer primary, you will see AC on the secondary. So, it could work. – Nick Alexeev Mar 8 '16 at 0:47

You're probably wondering why your Falstad circuit seems to work. The problem is not with the simulator. CircuitLab shows the same behavior.

simulate this circuit – Schematic created using CircuitLab

This happens under the following conditions:

1. The DC source is ideal (no resistance).
2. The transformer is ideal (no loss or primary-side resistance).

When you apply a voltage to an inductor, the current increases over time. If there's any resistance, the current will eventually level out (DC), at which point the inductor does nothing. But with no resistance, the current can rise forever.

In an ideal transformer, this endlessly rising current produces a DC voltage and current on the secondary side. So ideal transformers do work at DC. But there's no such thing as an ideal transformer. Add a small resistance like 0.1 ohms between your voltage source and the transformer, and you'll see the secondary-side output decay pretty quickly.

I'm guessing this has been asked before, but I didn't find it in a cursory search, so I'll give a short answer here.

Coils only effectively transfer energy while the current through the coil is changing. In a dc circuit, the current through the coil(s) only changes noticeably immediately after the circuit is first closed, then again right after the circuit is re-opened. The current builds up until the resistive losses in the coil, or the source impedance, waste all of the available power, with no meaningful energy transfer from one coil to another.

• Could this: falstad.com/circuit/… Work? – J.Clarke Mar 8 '16 at 0:41
• I desinged it in a circuit simulator the light bulb is just a dummy load and the dc voltage supply would be a 5v phone battery pack, could i use this to charge low wattage power 12v appliances (Very low low wattage less than 10 w) – J.Clarke Mar 8 '16 at 0:41
• @J.Clarke Nope, transformers don't work for DC. You can't just blindly throw things into a simulator, you need to design a circuit and then use the simulator to verify that your design behaves as you expect. – uint128_t Mar 8 '16 at 1:11
• +1 @uint128_t I agree. Simulators may well catch your minor errors, but often aren't programmed to "catch" major stuff (like feeding DC to a transformer) that the designers simply didn't expect anyone to try. In this case, it looks like that's exactly what happened to your simulation. – Robherc KV5ROB Mar 8 '16 at 1:21
• @J.Clarke This is also an appropriate time to point out that that simulator is rubbish, as it clearly doesn't accurately model transformers. There are many other simulators which are considerably more capable, LTSpice is free and relatively easy to use. However, my point above still applies: a simulator is better used to check for errors in a design, not design a circuit from scratch. – uint128_t Mar 8 '16 at 1:22

I think i solved my question just use a boost converter (stupid me :p) because doing it any other way is not very optimal. However could you do:
Dc power source -> Oscillator -> Transformer -> Rectifier -> Output

• If you use a flyback-topology dc-dc converter, then this is pretty much what it does internally (although a full description of flyback topology function is a bit more complex). - FYI, most portable (battery-powered) cell phone 5V chargers are already using a dc-dc converter to boost the 1.2-4.2 battery output (depending on battery(s) used) to the 5V output. You might find it more efficient to buy a battery & 12V boost converter to put together on your own, as opposed to cascading dc-dc converters by "re-boosting" the cell-charger's output. – Robherc KV5ROB Mar 8 '16 at 16:00
• @RobhercKV5ROB Thanks that is exactly the sort of answer i'm looking for. So a boost converter simply builds up an electric charge (Electromagnetic field) in the inductor coil then the mosfet cuts the power to it causing it to crash and the charge surges back through the wire at a higher voltage/ampage getting smoothed out by a capacitor along the way? is that right? – J.Clarke Mar 8 '16 at 22:47
• That's how a single-inductor dc-dc converter (whether boost, buck, or buck-boost topology) works (well, technically it's a magnetic field, rather than an electric charge). In those circuits, the inductor gets "charged" while the switching device (usually MOSFET) is 'on', then discharges through the load & ad diode, when the switching device is "off." - This wikipedia article probably explains the theory better than I can in comments here: en.wikipedia.org/wiki/DC-to-DC_converter#Electronic_conversion – Robherc KV5ROB Mar 8 '16 at 23:12
• On second read, another technical detail is that in a boost converter, the mosfet usually connects the inductor to GND when conducting, and the source is always feeding into the inductor (while the diode keeps the capacitor from discharging through the MOSFET. In most/all "normal" single-inductor converters, the current through the inductor never changes directions, though in buck-boost single-inductor topologies, the inductor current's polarity appears reversed to the load, as opposed to its appearance to the power supply. – Robherc KV5ROB Mar 8 '16 at 23:17