Coil driver simulation and questions

I think your numbers are wrong

75nH is very low and at a on time of 3us at 15V will give you a peak current of 666A

Please check your simulation and of possible share a screenshot, lot easier to help then

Cheers

Klaus

I don't think the 2N3055 is specified to be fast enough for that, though most recently manufactured ones are much faster than spec. I'm sure Phil A will chime in if I'm wrong.

Could you please be a bit more specific ? I would be happy to send a screenshot of the emiiter output current and/or voltage, but I would then need a email address or something, as this news group does not allow attachments.

The base of each 2N3055 is driven by a relaxation oscillator at 150 Kilohertz. What is so fast about that ?

If it's LTspice, post the .asc file.

Recently-made 2N3055s aren't as slow as the originals, but they're still slow. I don't know what speed your model runs at.

30 mA in 3 us from 15V corresponds to

3 us * 15V / 30 mA = 1500 uH

so it isn't inductive.

I suspect that you're dragging down your driver stage. Try using a voltage source to drive one of the bases and see.

When you fix it, watch out for the magic pixel smoke. ;)

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC / Hobbs ElectroOptics 
Optics, Electro-optics, Photonics, Analog Electronics 
Briarcliff Manor NY 10510 

http://electrooptical.net 
http://hobbs-eo.com

If those transistors go into saturation, the rise time when they turn off is very poor, by switchmode standards. A bit of capacitance across the 10 ohm resistors would help. So would a different transistor, with lower Rbb.

A schematic would always help. What are you trying to do with the coils? At 30 mA why do you need such a beefy transistor?

George H.

Thanks, but I already tried out your suggestions before I made my initial post. For example, using a 0.01uF capacitor across the 10 Ohm and BC548(instead of 2N3055) boosts the current to 125 mA and the emitter voltage to 500 mV. The emitter output voltage consists of nice square wave and the collector current is also a nuce square wave.

snipped-for-privacy@gmail.com wrote

The 2N3055 was not even up to audio frequencies with a beta of 20 to 70.. Ft of < 2.5 MHz The BC548 is low power with a beta of 200 (-B and -C version even higher), and an Ft of 100 MHz.

For the current in an inductor i = t / L where t is the time in seconds the transistor is 100% on (as a switch). A 3055 will likely not be 100% on any time soon in less than 1 / 150.000 seconds ;-)

A current limiting resistor in the emitter will create a current source, You do not want that if you use the transistor as a switch, unless it is a very low value and perhaps used to detect the current to dynamically trigger a cutoff. A small current transformer in the collector for feedback is better.

Example of current sense feedback to a micro to dynamically current limit Ic every cycle: + | L | _____|_ feedback | }|| [ ] }|( 10:1 current transformer ___|__ }|| | c -- b e | ///

What are you trying to do, circuit diagram?

I have used IRLZ34N MOSFETS as switch at that frequency. There are many suitable MOSFETS.

For original 2N3055 the fT was just 800 kHz at small current, so the current gain for a 150 kHz sine wave would be only 5. The edges of a non-sine waveform would be severely rounded, hence requiring considerable base current. Modern 2N3055 are perhaps 3 times faster, but still very slow.

I agree that an emitter current limiting resistor acts as a current source, but connecting the emitter directly to ground makes me queasy. So, I have a small resistance, 10 Ohm with a bypass capacitor. This is a side project, and I am trying to understand the workings of an air-core transformer.

snipped-for-privacy@gmail.com wrote

The way I go about transformers is like this:

First the no-load situation, get the primary inductance.

If a switcher look at the maximum on-time of the switch. _______ | | |+ )L U ) |- | | / switch | |

--------

i ^ / | / | /

--------->t

The current will then linear rise as U.t/L, where t is the time the switch his on. so in a 1 Henry coil after 1 second the current will be 1 Ampere.

L should be so big as to not create a huge no-load current (maybe 10% or less of full load).

So from L you can now calculate the number of primary turns,

Air-coils do not go into saturation, but if you keep the switch on too long the current will not rise linear as it will be limited by the R of the coil.

And that is really, all there is too it, for a switcher that is.

The fun starts when the switch opens..... and a large flyback voltage appears across L if there is no load.