Hello,

I have built a circuit like this to pulse 10 volts across the primary of a step up transformer, the pulses come from a micro controller, shown in the picture as just a square wave source.

The problem is that I measure about 2 volts across the primary and about 8 volts across the bottom portion of that circuit... across the transistor and diode.

What I would like is to have the full voltage across the primary... i do understand that there is a voltage drop across my diodes, but I don't understand why most of the voltage is across the transistor instead of the transformer.

Any help for me to understand this better or how to make a different circuit to do the same thing would be gratefully appreciated!

 

Do you have any biasing resistors on the transistor? It is possile it is not turning on fully. Also if you are pulsing at 40Hz and you are measuring with your meter on DC you may get some odd reading. Make sure your meter is on AC.

 

I do not have any biasing resistors, I am learning as I go on this project, so thanks for your help.

Here is a picture of the actual circuit, the frequency is variable in the audio range, say 1 Hz to 20,000 Hz.

Could you point me in the right direction so I can learn how to add biasing resistors to the circuit?

 

Unfortunately I cant save my schematics in a file format I can up load at the moment.

Try placing a 10K resistor between the base and +10V and a 220R resistor between the base and the micros O/P. I assume the 40Khz signal is either 3.3V or 5V.

 

Here is a PDF of how I think you should modify it.

 

Your microcontroller should not be subjected to high loads on it's I/O pins. 20mA is about the limit for most Microchip PIC uC's. If the supply to the uC is 5v, that means you need at least 220 Ohms between the I/O pin and the base of the transistor.

This also means that a typical transistor can only sink 10* the current through the base, or 200mA and remain saturated (Vce <= 0.2V).

The generic formula for operating a transistor as a saturated switch is:
Ic=Ib*10
Conversely, Ib=Ic/10

To calculate a base resistor, use this formula:
Rb = (Vin - Vbe) / (Ic/10)
where:
Vin = the voltage that will be applied to the end of the base resistor that is opposite the transistor's base; in your situation the uC's I/O pin.
Vbe = typically 0.7v to 0.8v. See the datasheet for the transistor in question for saturated Vbe values.
Ic = the desired collector current. Always rate transistors conservatively; if the datasheet says it's rated for 200mA, cut that in half; ie: the realistic limit is 100mA.

A logic level N-channel power MOSFET doesn't require constant current through it's gate, and can be driven directly from a uC's I/O pin at low frequencies (up to perhaps 1kHz, depending on gate charge). At higher frequencies, you will need a gate driver circuit or dedicated IC.

You will have a very hard time designing a transformer that can be successfully switched over that broad a range. At very low frequencies, the transformer will saturate, and your transistor will get fried. At the high end, the transformer won't work efficiently. You won't be able to use a laminated core transformer; you would need a toroidal transformer.

Your reverse-EMF diode across the primary will prevent the current from rapidly decaying, which is necessary for the secondary to produce an output. You will wind up with high current in a saturated primary, and very high power dissipation in your transistor or MOSFET.

What exactly is it that you are trying to do?

 

I just tried a 10k resistor between the 10 volt and the base. This still gives me about 7-8 volts across the transistor and ~2 volts across the primary.

I'll have to do some math here and look at the data sheet of the transistor, it is a ksh13009a out of a computer power supply.

I am trying to pulse a toroidal step up transformer that I spent a whole day winding, it has about 200:1000 turns, and this will be for something kind of like a small tesla coil, the frequency needs to be adjustable to tune into the resonant coils and capacitive load , but since i haven't build them yet i don't know where in that range the frequency is going to be, maybe 5 khz, maybe 10khz... my inductance/capacitance meter is in the mail!

I should have mentioned, there is an opto-coupler between the micro and the circuit, and the pin from the micro i am using has a built in resistor, so that can pulse the opto without me adding one. so the circuit looks like this...

I am extremely grateful for your assistance! Thank you.

Added a picture of the step up coil, it is on the yellow/white type powdered iron core.

 

Maybe this Transistor is not ideal? I have scavenged all the parts for this circuit except for the board, so I can try other transistors if you recommend.

The data sheet is here, http://semihow.com/product/data/KSH13009A_datasheet.pdf

It says..
Max Collector 12 amps (24 amps pulsed, I am pulsing)
Max Base is 6 amps

My power supply is a computer power supply, providing nominally 12 volts and 2 amps maxed, measured values is more like 10 volts, and have measured 2.3 amps out of it, but it does turn off automatically when I do something it doesn't like, which is nice to have the safety features.

I would be happy to get 2 amps through the primary.

Lets try this math,
Rb = (Vin - Vbe) / (Ic/10)
Vin, not sure how to calculate this, my guess is 10 volts minus the voltage drop across two diodes and the primary? minus the voltage drop across the 27 ohm resistor and minus the voltage drop across the opto, and minus 1.7 volts for the voltage drop across the green LED, so maybe there is not enough voltage there? The LED does light up and blink and pulse with the frequency coming out of the micro, and i can listen to the freq by touching a speaker from a microwave across the 27 ohm resistor, or i can hear the freq in the coils or the power supply...

Vbe is 1.2 volts at Ic 5 amps, Ib 1 amp, but I am not putting that much current through. so maybe it is closer to 0.7 like you mentioned?

Ic might be 2 amps

Lets try a few assumptions to feel out the values...

Rb = (Vin - 0.7) / (2/10)
Rb = (5 - 0.7) / (2/10) = 4.3/0.2 = 21 Ohms
Rb = (10 - 0.7) / (2/10) = 9.3/0.2 = 46.5 Ohms
Rb = (3 - 1.2) / (2/10) = 1.8/0.2 = 9 Ohms ...

So I may need some more hints, and help from someone with more experience, I am willing to learn and study and figure this out, Thanks for the help.

 

I added some contrast to your washed-out schematic and corrected its errors.
I calculate that the peak current in the primary will be only 50mA.
If you add another transistor to make a darlington then the peak current will be 500mA to 1A.