This isn't homework, this is for a project I'm working on. The circuit is a darlington array, but it includes a base resistor already for protection. I need to calculate i1 (seen in the circuit diagram) to see what value of current results when I apply 5V to see if an additional resistor is needed. The only way I can think to do it, as this is what I do when calculating a base resistor for a single transistor, is to do a KVL from the base -> emitter. Can someone just check to make sure my equations are correct? I can solve the simultaneous equations no problem, I just want to make sure my logic is correct.

i1 loop: 5-(2.7k)(i1) - 7.2k(i1-i3) - 3k(i1 - i2) = 0
i2 loop: -0.7 - 7.2k(i3 - i1) = 0
i3 loop: -0.7 - 3k(i2 - i1) = 0

 

This isn't homework, this is for a project I'm working on. The circuit is a darlington array, but it includes a base resistor already for protection. I need to calculate i1 (seen in the circuit diagram) to see what value of current results when I apply 5V to see if an additional resistor is needed. The only way I can think to do it, as this is what I do when calculating a base resistor for a single transistor, is to do a KVL from the base -> emitter. Can someone just check to make sure my equations are correct? I can solve the simultaneous equations no problem, I just want to make sure my logic is correct.

i1 loop: 5-(2.7k)(i1) - 7.2k(i1-i3) - 3k(i1 - i2) = 0
i2 loop: -0.7 - 7.2k(i3 - i1) = 0
i3 loop: -0.7 - 3k(i2 - i1) = 0

In your first equation I believe it should be,

5-(2.7k)(i1) - 7.2k(i1-i2) - 3k(i1 - i3) = 0

You have similar mistakes in the other 2 loop equations.

 

Presumably your question relates to the input current you would see if a 5V input voltage were applied to the input.

A reasonable initial estimate would simply be I1=(5-1.4)/10.5k=343uA

The required input drive depends on the Darlington switch saturation current - you should perhaps check out the data sheet to find what value of input current to expect based on the anticipated switch current.

 

In your first equation I believe it should be,

5-(2.7k)(i1) - 7.2k(i1-i2) - 3k(i1 - i3) = 0

You have similar mistakes in the other 2 loop equations.

whoops, yes sorry that is what I meant! Do the loops translate ok? I haven't done transistor circuit analysis in a couple years, and basic dc circuit analysis for even longer.

Presumably your question relates to the input current you would see if a 5V input voltage were applied to the input.

A reasonable initial estimate would simply be I1=(5-1.4)/10.5k=343uA

The required input drive depends on the Darlington switch saturation current - you should perhaps check out the data sheet to find what value of input current to expect based on the anticipated switch current.

You can't add those resistors in series like that can you? Because of the short inbetween the diode drops. At least that was my thinking.

 

You can't add those resistors in series like that can you? Because of the short inbetween the diode drops. At least that was my thinking.

There is no short. There is simply a node at which the base shunting resistors are linked into the emitter-base interconnect.

There's nothing wrong with my approach. It's a standard method for Darlington analysis. If you think about it it's what your equivalent circuit effectively reduces to.

 

I'm somewhat puzzled by your concerns about limiting the input current to the device.

All of the (various) manufacturer's data sheets clearly state that these devices are compatible with all commonly used logic family levels. So a 5V (TTL?) input would be fine. A straight 5V input will be quite safe.

 

If you are interested I've attached a pdf datasheet for the device.

Refer to page 3

For the ULN2004A with 5V input the likely maximum input current is 0.5mA max. This would typically correspond to conditions for a load side switch on-state current of 125mA & an on-state VCE of 2V. Note that the Darlington doesn't saturate as does a single BJT switch [with VCE sat ~100mV]. So the on-state VCE might be of more interest to you with respect to your load side output requirements.

 

Sorry - forgot the attachment ...

 

I'm somewhat puzzled by your concerns about limiting the input current to the device.

All of the (various) manufacturer's data sheets clearly state that these devices are compatible with all commonly used logic family levels. So a 5V (TTL?) input would be fine. A straight 5V input will be quite safe.

Well my first concern was to see what input current would result if I just applied 5V into the base. I didn't want to pull too much current through the load, so I was planning on placing a resistor in series into the base.

Looking at the datasheet, looks as if I'm running into the opposite problem I thought I would run into. I actually tested it earlier with 5V input to the base, and it resulted in roughly 150mA through the collector, but I don't think the collector was fully saturated since the load was designed to drop 0.2Vce, and not the 1.2Vce that the datasheet calls for. What could I do to increase the current into the base if I'm taking the 5V from a microcontroller?

 

Not sure what the exact nature of your problem is. Do you mean you need to obtain a lower Vce than you are getting?

Pushing more current into the input probably won't reduce Vce (on state) any further. In a Darlington connection, the output transistor collector cannot go into "conventional" saturation. The high current side transistor's collector on-state voltage will always be just higher than its base voltage.

If the base voltage is 1V then the collector will be somewhat higher - say 1.1V. The low side transistor will be saturated, with its Vce typically ~ 100mV.

As I say - I'm unsure if that is the gist of your problem or I have misunderstood your intention in the design ...

 

Not sure what the exact nature of your problem is. Do you mean you need to obtain a lower Vce than you are getting?

Pushing more current into the input probably won't reduce Vce (on state) any further. In a Darlington connection, the output transistor collector cannot go into "conventional" saturation. The high current side transistor's collector on-state voltage will always be just higher than its base voltage.

If the base voltage is 1V then the collector will be somewhat higher - say 1.1V. The low side transistor will be saturated, with its Vce typically ~ 100mV.

As I say - I'm unsure if that is the gist of your problem or I have misunderstood your intention in the design ...

It may be that my understanding of transistors isn't clear, but I'll try and elaborate. I have the collector biased so that Vce ends up being 0.5V or so. I couldn't figure out why even if I changed the input current why the output current would not change. But if you think about the graph of Ic vs Vce, at low values of Vce (in the triode or saturation region), all Ic is virtually the same for any value of Ib. I was assuming full saturation for a single transistor was Vce=0.2V as convention, but you can't assume that with the darlington as you've stated. It doesn't fully saturate until Vce=1.2V, which means I need to bias the collector so enough voltage drops at the collector since my emitter is grounded.

Do you see what I'm talking about? Or am I going about this the wrong way? I'm basically deriving what I need from the device graphs in the datasheet, then finding the values of resistance from those response values.

 

Do you see what I'm talking about?

Only partially - basically because you haven't clearly stated what you are trying to achieve.

Are you simply experimenting to understand what are the device properties or do you have a specific design goal in mind?

The latter can be quite simply stated in the case of a switching operation.

For a switching situation you have a maximum switch current you envisage and possibly a maximum switch on-state voltage you can tolerate. With respect to the appropriate base drive current for a transistor in switch mode, a rule of thumb is take the switching β as one fifth (or less) the typical linear mode β and calculate the required base drive off that assumption. In the case of the devices you are looking at, the inclusion of the input base resistance by the manufacturer is presumably intended to remove most of the guesswork from the exercise - provided the user adheres to the device specifications / limitations outlined in the data sheet.

If you are designing a specific circuit then post your schematic and the discussion might progress more effectively. State what drive source you intend to use - is it a micro-controller with nominal 5V TTL output levels? What is the device switching as a load?

 

Only partially - basically because you haven't clearly stated what you are trying to achieve.

Are you simply experimenting to understand what are the device properties or do you have a specific design goal in mind?

Its a little of both. When I took my electronics course in my undergraduate, I never got a firm understanding of transistors. I can do the analysis on paper, but when I've tried using them in real life it seems all the little things I didn't quite understand manifest. Thats why I wanted to do this project. The project is simply to drive a high power led. The led needs 350mA, so I need some amplification as I cannot even get close to driving this from a microcontroller. I chose the darlington array because it is able to sink the required levels of current quite easily (or supposed to lol).

I might not be communicating what I'm talking about, but I think I have figured out what exactly to do, and I will test this tomorrow morning. I'll be sure to share my results.

For a switching situation you have a maximum switch current you envisage and possibly a maximum switch on-state voltage you can tolerate. With respect to the appropriate base drive current for a transistor in switch mode, a rule of thumb is take the switching β as one fifth (or less) the typical linear mode β and calculate the required base drive off that assumption.

Can you elaborate on this? Why is this assumption made? Just to make sure, I'm not to confuse the linear region with the saturation region correct? The different regions are opposite for bipolar transistors and mosfets.

 

Driving the high current LED should not require any special consideration of the achievable Vce(sat).

At 350mA one might expect an LED forward drop of (say) 3.5-4.0V. So depending upon the supply voltage from which you intend to switch the LED you may or may not require some series limiting resistance.

With a 5V DC source to supply the 350mA current, then the LED voltage drop plus the Darlington on-state drop is going to be close to 5V anyway. So in that case you may not need any series limiting resistance. It would be a different story if you were using a 15V DC source to power the LED. The data sheet suggests you might need as much as 500uA input current to drive the Darlington Vce(sat) to typically 1.4V at 350mA on-state current. A 5V drive to the Darlington gate may therefore not be sufficient to get the switch + LED conditions you require. Or you just live with what you get with a 5V input. The Darlington thermal dissipation conditions should be OK.

Working with a switching βsw = βlin/5 is just a means of ensuring a BJT switches on reliably. There is no hard and fast rule. Some people even suggest making the base drive one tenth of the switch on-state current - irrespective of the working β. As the BJT moves into the saturation region a point is reached where further increasing the base current has little or no effect. In the end, the one thing you want to avoid with a high load switching current case is the likelihood of the device coming out of saturation and departing from the recommended safe operating area.

 

Driving the high current LED should not require any special consideration of the achievable Vce(sat).

At 350mA one might expect an LED forward drop of (say) 3.5-4.0V. So depending upon the supply voltage from which you intend to switch the LED you may or may not require some series limiting resistance.

With a 5V DC source to supply the 350mA current, then the LED voltage drop plus the Darlington on-state drop is going to be close to 5V anyway. So in that case you may not need any series limiting resistance. It would be a different story if you were using a 15V DC source to power the LED. The data sheet suggests you might need as much as 500uA input current to drive the Darlington Vce(sat) to typically 1.4V at 350mA on-state current. A 5V drive to the Darlington gate may therefore not be sufficient to get the switch + LED conditions you require. Or you just live with what you get with a 5V input. The Darlington thermal dissipation conditions should be OK.

Working with a switching βsw = βlin/5 is just a means of ensuring a BJT switches on reliably. There is no hard and fast rule. Some people even suggest making the base drive one tenth of the switch on-state current - irrespective of the working β. As the BJT moves into the saturation region a point is reached where further increasing the base current has little or no effect. In the end, the one thing you want to avoid with a high load switching current case is the likelihood of the device coming out of saturation and departing from the recommended safe operating area.

For some reason, I couldn't wrap my head around the fact that some voltage has to be dropped onto the collector for Vce. I'm just reading this now, but I realized what you said yesterday when I was playing with the circuit physically. With no resistor, the darlington sinks about 375mA, which is too high. I had some power resistors laying around, and with a 2Ω resistor in series it got my Vce at about 1.2V, placing it perfectly in saturation and limiting the current to roughly 330mA.

Moral of the story, I feel like I finally understand transistors decently well. You don't happen to know of any good analog design books where they reference design techniques like what you have stated, do you? I really want to learn a lot of the tricks and techniques, but none of my undergrad electronics books really get into the design aspect. Its more of just analysis.

By the way, thanks for the help! You've helped me reassure some things I felt a little shaky about.