Darlington Pair using BC 547

Thread Starter

pal114525

Joined Jul 3, 2015
46
If you add a 220nF in series with the coil it will be resonant at near 16kHz.
Driving that with a 0-12V, 16kHz square-wave showed a peak inductor current of about 350mA in simulation.
To get more current you could drive the resonant circuit with a push-pull audio bridge circuit such as the TDA2005, which would near double the supply voltage across the load (about 20Vpp with a 12V supply).
That should give a maximum peak current in the resonant circuit (220nF in series with the coil) of over 600mA.

Edit: If you can tolerate some frequency deviation then you could add a series capacitor to form a resonant circuit at about 16kHz and use feedback to form a high current resonant oscillator at this resonant frequency. You can do that with a fairly simple bridge driver circuit, such as below:
The LTspice simulation shows a peak inductor current of a little over 700mA.
Note: R4 needs to be rated for at least 2W.

View attachment 88160
Hi,

Thanks for your valuable feedback.
If I don't use 0.5ohm resistor (R4), then also, I am getting the same result.
Would you please explain me the advantage of using 0.5 ohm resistor (R4) in the schematic?

Thanks & Regards.
 

crutschow

Joined Mar 14, 2008
34,285
........................
If I don't use 0.5ohm resistor (R4), then also, I am getting the same result.
Would you please explain me the advantage of using 0.5 ohm resistor (R4) in the schematic?
R4 is to limit the shoot-through current as the bridge transistors switch state (when both are momentarily ON at the same time).

If you increase its value to about 7 ohms, the peak inductor current will be reduced to 500mA.
 

Thread Starter

pal114525

Joined Jul 3, 2015
46
R4 is to limit the shoot-through current as the bridge transistors switch state (when both are momentarily ON at the same time).

If you increase its value to about 7 ohms, the peak inductor current will be reduced to 500mA.
Hi,

Thanks for your valuable feedback.
There are two CMOS inverters in the design (M1, M2) & (M3 & M4).
The inputs to these inverters are pulses and the output from these inverters are inverted pulses which are applied to the RLC circuit from both the sides.
Would please tell me why two inverters are required in this circuit?

Thanks & Regards.

Thanks & Regards.
 

Thread Starter

pal114525

Joined Jul 3, 2015
46
Hi,

Thanks for your valuable feedback.
There are two CMOS inverters in the design (M1, M2) & (M3 & M4).
The inputs to these inverters are pulses and the output from these inverters are inverted pulses which are applied to the RLC circuit from both the sides.
Would please tell me why two inverters are required in this circuit?

Thanks & Regards.

Thanks & Regards.
Hi,

Thanks for your valuable feedback.
Please see the attached schematic. Here, I have used single CMOS Inverter and getting the result.
Would you please explain me why there are two CMOS Inverter in the previous schematic?

Thanks & Regards.
 

Attachments

Thread Starter

pal114525

Joined Jul 3, 2015
46
Hi,

Thanks for your valuable feedback.
Please see the attached schematic. Here, I have used single CMOS Inverter and getting the result.
Would you please explain me why there are two CMOS Inverter in the previous schematic?

Thanks & Regards.
R4 is to limit the shoot-through current as the bridge transistors switch state (when both are momentarily ON at the same time).

If you increase its value to about 7 ohms, the peak inductor current will be reduced to 500mA.
Hi,

Thanks for your valuable feedback.
Would you be kind to explain me the following point?
1. The two MOSFETs ( FDC638P and IRF7831) are available as SMD which cannot be used in breadboard.
Are there any substitutes for them?

Thanks & Regards.
 

Thread Starter

pal114525

Joined Jul 3, 2015
46
Hi,

Would you please tell me the substitutes for FDC638P and IRF7831. As these are SMD, hence cannot be tested on breadboard.
I need to use them on breadboard in my circuit.

Thanks & Regards.
 

crutschow

Joined Mar 14, 2008
34,285
.......................
Please see the attached schematic. Here, I have used single CMOS Inverter and getting the result.
Would you please explain me why there are two CMOS Inverter in the previous schematic?
I used two Inverters (bridge configuration) to get double the AC voltage across the coil and the high coil current you originally mentioned.
If you don't need current to be that high then a single Inverter is fine.
Would you please tell me the substitutes for FDC638P and IRF7831. As these are SMD, hence cannot be tested on breadboard.
I need to use them on breadboard in my circuit.
You can use any MOSFETs with similar voltage, gate-charge, and current ratings.
A search of Digi-Key or any electronics vendor you prefer should give you some choices with through-hole connections.
 

Thread Starter

pal114525

Joined Jul 3, 2015
46
I used two Inverters (bridge configuration) to get double the AC voltage across the coil and the high coil current you originally mentioned.
If you don't need current to be that high then a single Inverter is fine.
You can use any MOSFETs with similar voltage, gate-charge, and current ratings.
A search of Digi-Key or any electronics vendor you prefer should give you some choices with through-hole connections.
Hi,

Thanks for your valuable feedback. I believe two CMOS inverters form H-Bridge and which allows current to flow through the coil in either direction. As a consequence, the magnetic field will collapse each other and there would be no net magnetic field.

But, I want a magnetic field.

Would you be kind to help me in this regard.

Thanks & Regards.
 

Thread Starter

pal114525

Joined Jul 3, 2015
46
Hi,

Thanks for your valuable feedback. I believe two CMOS inverters form H-Bridge and which allows current to flow through the coil in either direction. As a consequence, the magnetic field will collapse each other and there would be no net magnetic field.

But, I want a magnetic field.

Would you be kind to help me in this regard.

Thanks & Regards.
The schematic is attached here.
 

Attachments

Thread Starter

pal114525

Joined Jul 3, 2015
46
Hi,

Thanks for your valuable feedback.
Would you be kind to explain me the following point?

1. If the inductor value (inductance) is changed from 458uH to some other (say 1.1 mH) value, then what are the components would require a change? What are the equations involved in that?

Thanks & Regards.
Hi,

Thanks for your valuable feedback. The current through the inductor is sinusoidal in nature. But the voltage across the inductor is no longer a sinusoidal signal.
How is that possible?

Thanks & Regards.
 
Top