Help in selecting transistor and heat sink

#12

Joined Nov 30, 2010
18,224
A Darlington trying to pass .1amp needs a microscopic amount if base current. Something less than .001 amps. It's probably impossible to buy an op-amp that can't do that.
 

studiot

Joined Nov 9, 2007
4,998
No one has taken an interest in my earlier question, so here is another one.

@jpborunda

With the proposed configuration you cannot connect one end of your load to ground.
Is this desirable?

If you require a ground connection you should revers the transitor polarity. ie use a pnp one.

O.1 amp pulses at 300 volts with american transistors?

MPSA 42 (NPN) or MPSA92 (PNP) will do the job nicely.
if you do not want to use power transistors.

MJE340 (NPN) MJE350 (PNP) would be a suitable choice of power transistor.

Also consider 2N6213 (NPN)
 

Thread Starter

jpborunda

Joined Apr 9, 2014
55
No one has taken an interest in my earlier question, so here is another one.

@jpborunda

With the proposed configuration you cannot connect one end of your load to ground.
Is this desirable?

If you require a ground connection you should revers the transitor polarity. ie use a pnp one.

O.1 amp pulses at 300 volts with american transistors?

MPSA 42 (NPN) or MPSA92 (PNP) will do the job nicely.
if you do not want to use power transistors.

MJE340 (NPN) MJE350 (PNP) would be a suitable choice of power transistor.

Also consider 2N6213 (NPN)
I apologize for not responding right away, I got carried away with the fast enlightenment provided by the other users. (Sorry)
What do you mean about the base emitter voltage? I think that the output voltage of the op amp would have to take into account the voltage drop, but since I need a relativeley small voltage, I think a 12 V supply would do.

And the floating load is actually a "requirement" here, but thank you for pointing it out. If I reverse the polarity of the bjt, the circuit would be a sourcing not sinking circuit right?

Thank you for the transistors options, I'll probably go with the MJE340.
 

studiot

Joined Nov 9, 2007
4,998
Vs goes from 0 to 3.3V in order to generate a current from 0 to 100mA(maximum) in R1
Vs is shown applied across R1 in series with the base emitter junction of the transistor by the follower amp.

So the voltage across R1 will not achieve your objective.

Note a darlington has two BE junctions and would make matters worse in this respect.
 
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AnalogKid

Joined Aug 1, 2013
10,225
Studiot, I answered your post #6 question in #19.

jp is correct, the output of the opamp automatically compensates for the turn-on voltage of whatever the pass transistor is, up to the point where the output stage clips. For a darlington, it would be 3.3V + (2 x Vbe), or approx. 4.6V at max Vs. Vcc = 12 V will be plenty of headroom; in fact, the opamp output current is so low that a 5V rail-to-rail part would work. For a MOSFET the max opamp output voltage would be 3.3V + Vgs (@Id = 0.1 A). Depending on the FET this could be less than 4.6 V, but maybe not.

Note to jp - if you change to a current source circuit with a PNP transistor, the interface between the transistor and the opamp gets much more complicated since that interface now has 200V across it.

I don't think I'd run a TO-92 transistor at 200 mW dissipation without knowing more about the available cooling environment.

ak
 

studiot

Joined Nov 9, 2007
4,998
3.3V + (2 x Vbe),
Is not 3.3 volts.

You obviously realise that both Vbe and Vr1 are within the follower loop, but Vs is shown as being applied to both so Vs will have to be greater by this amount, throughout its range.
The pulsed output with an average of 200mW is well within the capabilities of the stated small signal transistors. They were originally developed for driving discharge tube displays with just such pulses.

Obviously heatsinking would be required, and the OP has already asked about this. But 200mW is not so onerous.
 
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Thread Starter

jpborunda

Joined Apr 9, 2014
55
Studiot, I answered your post #6 question in #19.

jp is correct, the output of the opamp automatically compensates for the turn-on voltage of whatever the pass transistor is, up to the point where the output stage clips. For a darlington, it would be 3.3V + (2 x Vbe), or approx. 4.6V at max Vs. Vcc = 12 V will be plenty of headroom; in fact, the opamp output current is so low that a 5V rail-to-rail part would work. For a MOSFET the max opamp output voltage would be 3.3V + Vgs (@Id = 0.1 A). Depending on the FET this could be less than 4.6 V, but maybe not.

Note to jp - if you change to a current source circuit with a PNP transistor, the interface between the transistor and the opamp gets much more complicated since that interface now has 200V across it.

I don't think I'd run a TO-92 transistor at 200 mW dissipation without knowing more about the available cooling environment.

ak
Ok, well I wasn't really looking to change the configuration anyway. I was thinking of enclosing the circuit in a small case or something, with no forced air flow or anything like that, but only basic breathing holes/vents..

I see for example that the Darlington that #12 provided, is in a TO-220 and has 100 W power dissipation max (using a heat sink right?). How do I determine a safe power that a transistor can dissipate without a heatsink? Having such high power dissipation I would think that 1 W would represent no problem even without it, but then again, I dont really know how to aproach this.
 

studiot

Joined Nov 9, 2007
4,998
1Watt or even 0.1 watts in an enclosed space will soom warm things up.

The heat needs to be transferred to the external world somehow.
That is why cases have cooling fins and other devices.
I'm quite sure force cooling (fans) would be way over the top.

The thermal calculation is fairly easy.
I would say that mounting your circuit in the inside of the lid of a small metal (diecast?) case with the transistor in good thermal contact would do you just fine.

http://www1.electusdistribution.com.au/images_uploaded/heatsink.pdf
 

AnalogKid

Joined Aug 1, 2013
10,225
Is not 3.3 volts.

You obviously realise that both Vbe and Vr1 are within the follower loop, but Vs is shown as being applied to both so Vs will have to be greater by this amount, throughout its range.
Don't think so. It is a *follower* loop, so the output follows the input (or tries to) no matter what is in between. The opamp output does whatever it takes to make V+in equal to V-in, no matter what the output is connected to. Even though the overall circuit is a current sink, the part is a voltage feedback opamp and that is how it thinks. The opamp compensates for the Vbe drops exactly the same way it does in an ideal rectifier circuit, where diode Vf drops are in the feedback loop.

That's the nice thing about this circuit. If R1 is accurate and stable, the Vbe's and transistor gain can drift with temperature, vary from part to part, whatever, and the output performance is the same. It actually is the same as a standard non-inverting gain circuit, with the series element replaced by a constant-voltage element.

The pulsed output with an average of 200mW is well within the capabilities of the stated small signal transistors.
Forgot my own post about average power. oops.

ak
 

studiot

Joined Nov 9, 2007
4,998
Don't think so. It is a *follower* loop, so the output follows the input (or tries to) no matter what is in between. The opamp output does whatever it takes to make V+in equal to V-in, no matter what the output is connected to.
You are right, I didn't read the initial circuit properly, I thought the op amp was ground referenced.

However the circuit is inherently non-linear since it depends upon the gain of the pass transistor, which in turn depends upon the collector current.

The correct expression is


\({I_c} = \alpha \left[ {\frac{{{V_s} \pm {V_{inputoffset}}}}{{R1}} + {I_b}} \right]\)
 

AnalogKid

Joined Aug 1, 2013
10,225
Again, don't think so. While a transistor is inherently non-linear, the *circuit* is just fine. In your equation, alpha varies with age and temperature, but Ib is controlled by the opamp's differential action and the overall feedback loop to compensate for all circuit nonlinearities except input offset voltage. Since the overall circuit gain is unity, the offset voltage is not multiplied by the transistor alpha. Or, more correctly, it is reduced by the opamp functional gain of 1/alpha before it is multiplied by alpha. So alpha barely matters (see below), Ib is trivial (see previous posts), and the offset is less than 0.31% of full scale.

To produce a relatively "square" output pulse. the circuit open loop gain should be at least +20 dB at the highest frequency of interest. A pulse width of 500 us is an equivalent squarewave freq of 2 KHz. 10 x this is 20 KHz and captures the first four harmonics, enough for some squareness. A problem is that post #1 states a minimum pulse width of zero, so the spec needs some work. But as a starting point, +20 dB at 20 KHz.

A 741 (ooooooohhhhh) would have a problem with this, except that the opamp isn't doing any of the work. This is a compound amplifier, with a fixed output stage gain of approx +40 dB plus the opamp open loop gain at 20 KHz which is less. But the circuit requirement is a gain of 1, so there's lotsa open loop gain to play with at the higher bandwidths required for narrower pulse widths.

And for all of you concrete thinkers out there, no, I'm not suggesting using an LM741 in this application. I'm using Our Father to make the point that the output transistor is doing most of the work, and just about anything with a diff amp input will work for the opamp.

ak
 

studiot

Joined Nov 9, 2007
4,998
In your equation, alpha varies with age and temperature,
I don't see either age or temperature in my equation.

I did say that alpha varies with collector current, and bearing in mind that the purpose of the circuit is to control collector current, this will have a bearing.

I did not say it would be large and may not be significant, but the effect is there. The effect will be worst at low currents using a single power transistor, which is not designed to run such levels.
 

THE_RB

Joined Feb 11, 2008
5,438
...
However the circuit is inherently non-linear since it depends upon the gain of the pass transistor, which in turn depends upon the collector current.
...
I hate to step into someone else's argument but if the opamp has a lowish output impedance the gain of the transistor is not the main source of non-linearity and instability, it is the Vbe drop. (Or ratio of Vbe : Vrs)

That was why i suggested a bipolar transistor which has less Vbe drop and more stability than a darlington, and much less voltage drop than the FET Vgs.
:)
 

AnalogKid

Joined Aug 1, 2013
10,225
Agree, but... Ib could amount to a 1% error depending on the transistor selected. Circuits like this usually are not in high-precision applications; still...

And an extra 0.6 Vbe still is less than a FET Vgs (again depending on the transistor selected).

ak
 

studiot

Joined Nov 9, 2007
4,998
In order to proceed with the OP here is the datasheet for the MJE340.

The sheet does not give an ft but other sources state 10MHz.
 

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