Darlington pair

[malc9141]

Just to keep things clear here-
30revs/sec*360°/rev=10,800°/sec. That's 10.8°/millisec. Not sure how you came up with 12°.

Too right. Sorry. Sloppy.

 

[malc9141]

Hello again, This is important: in your diagram, what would the voltage be across R3 if the pot U2 was 100Ω, ie 0.1 of the 1kΩ in the diagram and the source voltage V1, instead of 36 V, was only 18 V?
Because that voltage>current is the minimum to open the bleed.

I estimate that this would be 16.7 V but am not confident.

 

[Roff]

Hello again, This is important: in your diagram, what would the voltage be across R3 if the pot U2 was 100Ω, ie 0.1 of the 1kΩ in the diagram and the source voltage V1, instead of 36 V, was only 18 V?
Because that voltage>current is the minimum to open the bleed.

Which diagram?

 

[malc9141]

Which diagram?

It was RONV's diagram Post 13, above (I was just returning to edit that in)

 

[Roff]

EDIT: This is just information, NOT a response to your last question. Maybe you already know this.

The general equation for the current through an inductor is
[latex]I=I_f+(I_i-I_f)e^\frac{-tR}{L}[/latex]
where [latex]I_i[/latex] is the initial current, and [latex]I_f[/latex] is the final current, that is, the current at t=∞.
[latex]I_f=V_L/R_L[/latex], where [latex]V_L[/latex] is the voltage across the inductor, and [latex]R_L[/latex] is the resistance in series with the inductor, including the resistance of the coil.

So, [latex]I=\frac{V_L}{R_L}+(I_i-\frac{V_L}{R_L})e^\frac{-tR}{L}[/latex].
If [latex]I_i[/latex] is zero, then [latex]I=\frac{V_L}{R_L}(1-e^\frac{-tR}{L})[/latex].
Solving for [latex]V_L[/latex],

[latex]V_L=\frac{IR_L}{1-e^\frac{-tR}{L}}[/latex]

When I=5A, L=2mH, R=4.5Ω, and t=0.5mS,

[latex]V_L=33.3V[/latex]
This is the required voltage across the inductor. Add the voltage drop across the switch (e.g., MOSFET, Darlington) to get the required supply voltage.

 

[malc9141]

Pleased with that, tho' I do know it. A long time ago in school but I've revisited more recently.

What I'm not sure about (inter alia) is how the PD drops across a transistor when it's not fully on. I suppose it depends on the exact transistor, to be precise, but if (with the circuit on post 13) "battery" voltage is 18 V, and the Pot P is 100 Ω, what would be the voltage across the Coil?

In this exact case, I don't know t, and I depends on this V question

Malc

I see from your foto that you are a squash player!

 

[kinarfi]

I ran several simulations using the diagram in 13, pn transistors, and again with a FET and both gave approximately the same square wave with the same current rise time, but the quickest current decay was with 3 1000v, 6a diodes in series with the inductor, 160 μs, with 2, 165 μs, with 1, 172μs, worst was with a shunt diode, decay was as slow as build up. A snubber appeared to be good, reduced the oscillating.
Time starts at .1ms, Current flat lined at 4.43A after 4.211ms, at .2ms current was 897ms, at .3ms - 1.61A, .4ms - 2.19A, .5ms - 2.64a, .6ms - 3a, .7ms - 3.3a, .8ms - 3.5a, .9ms - 3.7a, 1ms - 3.85a
I'm tired, think I'll go to bed!!
g'nite

 

[Roff]

I ran several simulations using the diagram in 13, pn transistors, and again with a FET and both gave approximately the same square wave with the same current rise time, but the quickest current decay was with 3 1000v, 6a diodes in series with the inductor, 160 μs, with 2, 165 μs, with 1, 172μs, worst was with a shunt diode, decay was as slow as build up. A snubber appeared to be good, reduced the oscillating.
Time starts at .1ms, Current flat lined at 4.43A after 4.211ms, at .2ms current was 897ms, at .3ms - 1.61A, .4ms - 2.19A, .5ms - 2.64a, .6ms - 3a, .7ms - 3.3a, .8ms - 3.5a, .9ms - 3.7a, 1ms - 3.85a
I'm tired, think I'll go to bed!!
g'nite

Please post the .asc file so we can see what you did.

 

[kinarfi]

I used the attached and the one from post 15 except I used the same switch set up on both to get the square wave. There are notes on the attached file.

 

[ronv]

I think that circuit (13) will always be slow to turn off as the inductive kick keeps the transistor on. Kind of just like the diode across the coil.
Here is a kludge circuit with a power zener that might do what you want. It needs work but you can get some ideas.

 

[malc9141]

Post 45. Actually, it does help! (I can play around with t. I hadn't thought of that). Schlafen gut

 

[malc9141]

I ran several simulations using the diagram in 13, pn transistors, and again with a FET and both gave approximately the same square wave with the same current rise time, but the quickest current decay was with 3 1000v, 6a diodes in series with the inductor, 160 μs, with 2, 165 μs, with 1, 172μs, worst was with a shunt diode, decay was as slow as build up. A snubber appeared to be good, reduced the oscillating.
Time starts at .1ms, Current flat lined at 4.43A after 4.211ms, at .2ms current was 897ms, at .3ms - 1.61A, .4ms - 2.19A, .5ms - 2.64a, .6ms - 3a, .7ms - 3.3a, .8ms - 3.5a, .9ms - 3.7a, 1ms - 3.85a
I'm tired, think I'll go to bed!!
g'nite

Oh, that's what those diodes were all about. I couldn't see why they were there but I accept the idea even if I don't understand it - so far, anyway.

My original questions remains: if you have a Darlington pair, and a line voltage of X, how do you estimate the output voltage.

You gave clear answers to line PD 36 V in, but with 25 V, what would the PD beyond the final transistor?

 

[kinarfi]

It would be helpful to know the frequency of the injector and the dwell, then I can give you more exact data. The transistors you are using are not in the spice data, so the data could be off a tad.
This is ONE over built, very robust circuit!!! Should last forever, IMHO.