Darlington array problem NPN PNP???

[steamer]

I have a circuit which uses negative wires to switch 8 light bulbs independently on and off, they all use a common positive. There is nothing i can do to change this part of the circuit, and it all works fine except...

The bulbs are dim because they take 400ma and the output from the circuit above is 20ma.

'What I need is a Darlington Array' I thought, but on looking at all the darlington arrays they seem to be PNP, do any NPN ones exist? Is there a workaround?

Here's one that is perfect appart from the fact that it has a common negative, which is the oposite to what i need.

https://www.chipcatalog.com/TI/ULN2803A.htm

Any help greatly apprieciated.

 

[steamer]

Sorry I meant PNP, I cant find PNP Darlington Transitors.

 

[steamer]

PNP Darlington Transitor Array that is

 

[TillEulenspiegel]

Sorry steamer, theres no schematic and you don't mention what the supply V and type is. More info would be helpful.

 

[Russlk]

Why not invert the drive signal and use the NPN array?

 

[steamer]

OK I will try to explain with some pictures.

This is how I wish the output from the black box were set up like

**broken link removed**


..if this was how it was, then to get 500ma for the bulbs from this 'black box' I would wire it up like (as far as i understand it, please let me know if I am completely wrong)...

**broken link removed**

But unfortunatly the power coming from the black box is not 'ideal' and is inverted, like this...

**broken link removed**

if I could find a darlington PNP array, then life for me and my limited electronics skills would be rosy. But I can find one.

PS: forgot to label righthand box, it would be any old standard darlington array like the ULN2803.

 

[steamer]

Why not invert the drive signal and use the NPN array?

How would I do that? is there an 'inverter' chip? a 'not gate array' would that be?

 

[Optikon]

Why not invert the drive signal and use the NPN array?

How would I do that? is there an 'inverter' chip? a 'not gate array' would that be?

Well, it looks like the problem is all the negative side points on the black box can only sink up to 20mA and he needs 500mA for the bulbs...

 

[Optikon]

OK I will try to explain with some pictures.

This is how I wish the output from the black box were set up like

**broken link removed**


..if this was how it was, then to get 500ma for the bulbs from this 'black box' I would wire it up like (as far as i understand it, please let me know if I am completely wrong)...

**broken link removed**

But unfortunatly the power coming from the black box is not 'ideal' and is inverted, like this...

**broken link removed**

if I could find a darlington PNP array, then life for me and my limited electronics skills would be rosy. But I can find one.

PS: forgot to label righthand box, it would be any old standard darlington array like the ULN2803.

Steamer, suppose you had a PNP array, are you sure you could even use that due to the power dissipation? A saturated darlington(when used as a switch) has about 1.2V VCEsat and if it carries your 500mA bulb current, that would mean each transistor in the array would dissipate 0.6W! so, if you had 10 devices in the array, and all ten bulbs were on, that would be 6W of power dissipation. At 6W of power dissipation, your array package options are limited to around none. You would need discrete components..

I'm not familiar with the array you meantioned earlier, can you get a package that can disspiate that much heat?

If you are willing to solve your problem with discrete components, then there are many ways to turn on your 500mA bulbs with that current returning directly to the "-" supply and not through your black box with the 20mA outputs.

 

[ljcox]

Firstly what you are calling negative is in fact the 0 Volt line. So the Black Box applies 0 Volt to light the lamps. Do you know if this device has open collector outputs? If so, you will need pull up resistors.

Someone suggested that you use an inverter. A suitable IC would be the CMOS CD4009 Hex Inverter. But you will need 2 of them since there are only 6 in the IC. You will need to connect the unused inputs to 0 Volt in order to prevent the possibility of unecessary current flow through them.

Then you can use a NPN Darlington, subject to the point that Optikon raised above. Does only one lamp come on at a time, or can there be several?

Len

 

[steamer]

Steamer, suppose you had a PNP array, are you sure you could even use that due to the power dissipation? A saturated darlington(when used as a switch) has about 1.2V VCEsat and if it carries your 500mA bulb current, that would mean each transistor in the array would dissipate 0.6W! so, if you had 10 devices in the array, and all ten bulbs were on, that would be 6W of power dissipation. At 6W of power dissipation, your array package options are limited to around none. You would need discrete components..

I'm not familiar with the array you meantioned earlier, can you get a package that can disspiate that much heat?

All bulbs 'could' come on at the same time, which would be 7 X 500ma(max, in reality about 400ma).
Lets take the L604C as an example, the datasheet is at https://www.mouser.com/index.cfm?ha...ductid=346050&e_categoryid=98&e_pcodeid=51113
Looking at the datasheet it would appear it can drive up to 8 500ma devices, if that is the case then why does heat dissipation become an issue? Had the manufacturer not thought of that and allowed that much heat dissipation into the design? Would a heat sink strapped to it not improve dissipation?

If you are willing to solve your problem with discrete components, then there are many ways to turn on your 500mA bulbs with that current returning directly to the "-" supply and not through your black box with the 20mA outputs.

In my second diagram, I thought that I had designed it so that the current returned directly to supply? bearing in mind that the bottom left pin of the darlington array is common ground? (please excuse my possible mis-use of terms).

The main reason I wish to use a darlington array is that the manufacturer that i'm thinking of using charges by the component to solder, having 8 darlingtons in a single chip reduces my costs drasticaly.

 

[steamer]

Here's an image which conveys my thought process a bit better.. **broken link removed**

 

[ljcox]

If the black box on the right hand side is a Darlington array, you do not show any resistors. Darlingtons will need resistors in series with the bases (if in common emitter) to limit the base current to a safe level.

I opened the link you posted and when to the mouser site, but when I tried to download the data sheet, it did not arrive.

It would be helpful if you showed the transistors (ie. the ones inside the Darlington package) instead of just a black box.

Len

 

[steamer]

From the PDF...Eight darlingtons per package
output current 400ma per driver (500ma peak)
output voltage 90v
integral suppression diods for inductive loads
outputs can be parraleled for higher current
TTL/CMOS inputs

Description:
The L603 and 604 are high voltage, high current darlington arrays each containing eight open collector darlington pairs with common emmitters. Each channel is rated at 400ma and can with stand peak currents of 500ma. "

Improvement to diagram...

**broken link removed**

 

[steamer]

If the black box on the right hand side is a Darlington array, you do not show any resistors. Darlingtons will need resistors in series with the bases (if in common emitter) to limit the base current to a safe level.

I thought the 24ohm resistor would limit the current to 500ma for all the Darlington outputs? is this not the case then?

 

[Optikon]

From the PDF...Eight darlingtons per package
output current 400ma per driver (500ma peak)
output voltage 90v
integral suppression diods for inductive loads
outputs can be parraleled for higher current
TTL/CMOS inputs

Description:
The L603 and 604 are high voltage, high current darlington arrays each containing eight open collector darlington pairs with common emmitters. Each channel is rated at 400ma and can with stand peak currents of 500ma. "

Improvement to diagram...

**broken link removed**

A couple of things to note.. the 18pin dip pack says max watts is 1.8 which is pretty decent but that assumes a 25C temp environment. If where you install this thing it is normally warmer than room temp, you'll have to de-rate this power rating if you want your design to be reliable and long-lived.

I personally, would make sure the dissipation doesn't exceed 1.5W on that package and I would also heat sink it with a small heat sink exposed to air.

I think you are on the right track. Use those 20mA outputs to drive your power switches (possibly using an inversion in there) and then intentionally limit the lamp current to 400mA, When eight of these are running, your power dissipation will be ok.

 

[Russlk]

The L604 is rated at 1.8 watts total and each channel can disapate 0.8 watts, so a maximum of two lamps can be on at one time. It may be possible, by using wide traces (because most of the heat goes out thru the leads) and glueing a heat sink to the top, to double the rating. But, you are still limited to 4 lamps on at one time.

If assembly cost is a big factor, look into surface mount with pick and place equipment. A PC board with few holes is much cheaper, the parts are not more expensive, and labor is less.

10 SOT-89 darlingtons may not be that costly, all things considered.

 

[ljcox]

If the black box on the right hand side is a Darlington array, you do not show any resistors. Darlingtons will need resistors in series with the bases (if in common emitter) to limit the base current to a safe level.

I thought the 24ohm resistor would limit the current to 500ma for all the Darlington outputs? is this not the case then?

You don't need (or want) a resistor in series with the lamps. I assume they are 12 Volt lamps. If not, and say they were 6 Volt, then you would need a resistor for each lamp. If you only had a single resistor as per your diagram, the lamp brightness would decrease as more and more lamps are turned on.

The attachment shows the circuit I had in mind. I have drawn the Darlington as a single transistor for simplicity.

You did not answer my question posted a day or two ago about whether the "Black Box" has open collector (or drain) outputs. If it does, then you will need pull up resistors.

Edit.
After writing the above, I studied your latest diagram again. 2 points.

1. the diode common in the Darlington package should be left open. Otherwise all lamps will be on permanently.

2. the data you posted states that the Darlington array can handle CMOS or TTL inputs which implies that the base resistors (10k in my diagram) are internal.

As for your question on the diagram about removing the connection to the "-" pin on the "Black Box", I don't know what is in the Black Box so I can't comment other than to ask:- Is this pin meant to be the return connection for the lamps? If so, then you can probably leave it open.

Len

 

[steamer]

Sorry for the late response, have been busy over the weekend so could not get to the internet. Thanks very much BTW for your time, it is greatly appreciated.

You did not answer my question posted a day or two ago about whether the "Black Box" has open collector (or drain) outputs. If it does, then you will need pull up resistors.

Sorry missed that Q. Yes it does have open collectors. So if it does need resistors, thats going to bump up manufacturing costs by a fair old margin, which is a shame. You know, I'm sure I spotted a driver once which was basicaly a darlington array but with resistors, i shall have a look.

Edit.
After writing the above, I studied your latest diagram again. 2 points.

1. the diode common in the Darlington package should be left open. Otherwise all lamps will be on permanently.

I thought that looked a bit odd, have now left open as you suggest.

2. the data you posted states that the Darlington array can handle CMOS or TTL inputs which implies that the base resistors (10k in my diagram) are internal.

So if i read you correctly, there are resistors built into the darlington array, meaning that no 'extra' resistors will be necessary between "black box" and darlington Array... This is great news!

The diagram now stands as this...

**broken link removed**

...would it be fair to say this (IF the polarity of the 'black box' was correct) is now a 'fully working circuit'?

 

[ljcox]

Your circuit looks good, but you have not shown the pull up resistors (you said the black box has open collectors).

STOP PRESS. A thought has just occurred to me. Your circuit shows a + at each output so I assume that output transistors are PNP with open collectors (or drains if they are FETs).

This means that pull up resistors won't be necessary. Also, if you check what is inside the Darlington package, you may find that it has internal pull down resistors. If so, you don't need external pull downs.

If you do need external resistors, note that you can buy resistor arrays in a SIL or DIL package which would be ideal for your application.

Len