Transistor based AC amplifier - how to build with limited details.

[Ara Ghazaryan]

For example as far as I understand - Z-diodes should be changed.
Will for example **broken link removed** work?

I am asking too much probably, but among my problems is the research of components in stores (as they are German here, and I do not know the language). So once I know the exact part name - I can find it and order - any part, any price and quantity. But going online and researching those "Verstärker"s and "Kohleschicht-Widerstand"s is painful

So thanks again for all help!

 

[Ara Ghazaryan]

It is extremely easy to measure the resistance of the coil with a multimeter then use Ohm's Law to calculate its current when it has 24V. Current = 24V/Resistance. Its reactance (AC resistance) is higher at higher frequencies.

I managed to measure the resistance of my magnet coil. it is 100 Ohm (there is another one 130 Ohm). Operating on 25-30 V it will need then 300 mA of peak current output, right?

 

[Ara Ghazaryan]

Dublicate post

 

[MikeMl]

I managed to measure the resistance of my coil. Its 100 Ohm. so with the possible max voltage of 30 I will end up with 300 mA peak current, right?
Thanks

I=E/R = 30V/100Ω = 0.3A

To finalize the design of the amplifier, we need to know what the inductance of the coil is?
What is the coil used in or use for? Is it wound on an iron core? air core? few turns? many turns?

100Ω DC resistance bespeaks of hundreds to thousands of turns of very fine wire, ergo high inductance. Driving short duration pulses into a highly inductive load is not at all like an "audio amplifer".

 

[Ara Ghazaryan]

I dont have data on one of the coils - no marking. First one does not have a core - I have to design and make one myself - most likely a horse-shoe form something. The other one is this - again, no data on impedance as far as I could tell.

As for the purpose - those are magnet coils. And they will be pushing and pulling magnetic mechanical parts in a medium while other experiments will be contacted. No contact with these parts is permitted - hence the magnet to be used.

I=E/R = 30V/100Ω = 0.3A

To finalize the design of the amplifier, we need to know what the inductance of the coil is?
What is the coil used in or use for? Is it wound on an iron core? air core? few turns? many turns?

100Ω DC resistance bespeaks of hundreds to thousands of turns of very fine wire, ergo high inductance. Driving short duration pulses into a highly inductive load is not at all like an "audio amplifer".

 

[AnalogKid]

I would appreciate if you could suggest higher current level.

ak

 

[audioguru]

Yes, 30V/100 ohms= 300mA. Then the transistors need a base current of 30mA to turn on very well (saturate).
An electromagnet has inductance that will create a very high voltage when its driving transistor turns off, but then your other transistor will turn on and will short away the high voltage. But in case the power is turned off when a transistor is conducting, protection zener diodes should be used parallel with the electromagnet.

 

[MikeMl]

The solenoid you linked doesn't list the inductance, but I am guessing based on my experience that it is somewhere between 100mH and 1H. It shows that it requires 5.4W @24V under DC conditions, so I=P/E = 5.4W/24V = 0.255A, which means that its DC resistance is E/I = 24V/0.255A = 94Ω, which is consistent with your measurement.

Since this is a solenoid coil, I cannot see any reason to drive it with a bipolar current. The magnetic force it generates either attracts a metallic object or it doesn't; the polarity of the applied current makes no difference...

Here is the reality of switching a 1H 94Ω coil:



I model the solenoid with the values above and a "worst-case" inductance of ~1H. I use a 24Vdc supply. I use a simple low-side voltage-controlled switch (to represent an NPN or NFET). I add a 1nF (guess) capacitance C1 to represent stray winding, wiring and collector-to-emitter or drain-to-source capacitance...

Note what the simulation shows: it takes about 50ms for the current in the solenoid to go from zero to near its final value of 255mA. More interesting is what happens when the switch turns off at 60ms. Note that because of the energy stored in the inductance while turned on, the switch is subjected to a damped oscillating voltage of +- 8000V as the energy from the inductor is dissipated in its own internal 94Ω resistance. Where can you find a NPN or NFET that will tolerate being subject to +- 8kV?

Fortunately, there is way to solve this problem: I add a Silicon diode D1 (called a "snubber" or "catch" diode) to the running simulation. Note what happens to the voltage at node C V(c), and the current through the solenoid, I(L1). Note that if "switch" is replaced with either an NPN or NFET, the voltage at the collector or drain is now well behaved. The rated breakdown voltage would only have to be a bit higher than the supply voltage..., and it never goes negative...



Note the shape of the current through the solenoid. It takes about the same time for the current to build up (50ms) as it does for it to dissipate after the switch turns off (also 50ms). Note that if you expect to "pulse" the solenoid on/off, the shortest period would be about 100ms, meaning that you can only drive it at at 10Hz if you expect a full on, full off cycle. Driving the snubbed solenoid with a higher frequency might enable using Pulse-Width-Modulation to adjust the average current through the solenoid.

In light of what I show here, what are you trying to do? Why do you think you need to drive the Solenoid current in both directions when a unipolar driver with a snubber diode seems to do everything?

 

[Ara Ghazaryan]

thanks for detailed explanation. i will go through it once I have my PC (I will need to research a bit to understand all statements of yours). But meanwhile I will try to answere some of your key questions - Why bipolar driving?

it will be applied on a magnet particle - with a constant polarisation. so it will be pushing and pulling. Also i will need to use a horseshoe extention as the solenoid itself is too big to accomodate it where i beed it. with an extention i will be able to squeze it in the setup I have now. And the materials I have currently application of same polarity field rezults in magnetisation of latter and I end up working on the apper (not full) part of magnetization histeresis curve.
hence the desire of AC driving.

 

[audioguru]

He said he is driving the electromagnet with an AC signal of only 4Hz so it might vibrate a permanent magnet moving core back and forth. The frequency is so low that the inductance will not reduce the current, but narrow pulses might not have full amplitude since the inductance will slow current rise times.

I see that Mike said the same things as me but his snubber diode assumes that the power supply is NEVER disconnected when the coil is energized. The high voltage will have nowhere to dissipate except to zap a transistor.

 

[Ara Ghazaryan]

He said he is driving the electromagnet with an AC signal of only 4Hz so it might vibrate a permanent magnet moving core back and forth. The frequency is so low that the inductance will not reduce the current, but narrow pulses might not have full amplitude since the inductance will slow current rise times.

I see that Mike said the same things as me but his snubber diode assumes that the power supply is NEVER disconnected when the coil is energized. The high voltage will have nowhere to dissipate except to zap a transistor.

yes, I guess thats what happened with one of the transistors when I tried to built and operate one of the similar AB series pushpull curcuits downloaded from internet - snapped transistor. And thats when i decided to apply for an experienced help

 

[MikeMl]

I have designed and built a magnetic deflection system that was used in a linear particle accelerator used to implant ions in semiconductor wafers. It had coils with an inductance of several Henries. The correct way of thinking about driving such deflection coils is that the driver is a bi-polar constant-current source (with a compliance of hundreds of volts); not a switch connected to a fixed voltage power supply like 24V. A much different approach...

 

[Ara Ghazaryan]

I have designed and built a magnetic deflection system that was used in a linear particle accelerator used to implant ions in semiconductor wafers. It had coils with an inductance of several Henries. The correct way of thinking about driving such deflection coils is that the driver is a bi-polar constant-current source (with a compliance of hundreds of volts); not a switch connected to a fixed voltage power supply like 24V. A much different approach...

You must be right - the approach i am working on is far from optimal. but thats partially due to limitation of resources (power supplies, coil, function generator, not the cicuit components which can be ordered no problem). and partially because i will need to control the ramping "style" (sharp, trianle, sin, etc) and frequency. as well as power.

 

[MikeMl]

...
I see that Mike said the same things as me but his snubber diode assumes that the power supply is NEVER disconnected when the coil is energized. The high voltage will have nowhere to dissipate except to zap a transistor.

Huh? I dont understand what audioguru is trying to say? Am I not "disconnecting" the supply by opening the switch at the 60ms point in the simulation. The snubber diode (if present) provides a current path regardless if the switch is open or closed... Only in the first case (without the snubber diode) does the transistor get zapped.

 

[MikeMl]

Ara Ghazaryan, I would just buy a LM675 from Digikey for US$5.50, or the others shown on this page.. Set it up with a gain of ~5 operating on split supplies of +25 and -25V to +30V and -30V (see Fig 2 on the data sheet). Put it on a 3W heatsink. Done...

 

[Colin]

Will this work:

 

[AnalogKid]

Yes.

 

[Colin]

You have a "slop of 3v." Do you think you will get much out of the circuit with 5v input?

 

[AnalogKid]

You didn't understand this circuit the last time I posted it, even after I wrote a detailed analysis. I infer from your statement that you still don't understand it.

The peak to peak output voltage swing = Vcc-(2 x Vcesat)

In this country, slop is not an engineering term. What do you mean by it in the context of this circuit, and why is it in quotes?

ak

 

[Colin]

To start with, the two 13v zeners will have a "dead spot" of 2v because they add up to 26v for the 24v supply.
On top of this the base-emitter junctions at the top and bottom will add another 1v.
This means the input voltage will not have any effect when it is rising and falling by 3v because this is the "gap" between turning on the bottom zener then the voltage falling and turning on the top zener.
Normally "biasing diodes" are just at the point of turning ON, so the rising or falling signal will pass through the diode, even when it is a few millivolts.
Any voltage below 3v will not be transferred and a 5v signal will have some attenuation in the 1k as well as the 2u2. Most 5v signals from a digital source are less than 5v as the output of most chips is 4.5v max and 0.5v min. This gives a 4v p-p signal and the other two losses makes this circuit very unreliable.
On top of this, zeners have a 5% tolerance and another 1v may disappear.