I've done the Mosfet conversion a few times but was wondering if it's necessary if I fit LEDs instead of incandescemts.
Will I still get the heat generated by the 27ohm resistors?
Will I still get the heat generated by the 27ohm resistors?
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System 6 Mosfet Conversion
- Thread starter Andy_B
- Start date
Ragnar1057
Site Supporter
Hi,
I'm not familiar with what you mean by MOSFET conversion. Do you mean that you are replacing the old power transistor with MOSFETs?
If so, a (properly selected) MOSFET will be easily be able to drive an LED. Having said that, the original power transistors will be more than able to drive LEDs too. However, it's not as simple as that (it never is).
The main difference between a transistor and MOSFET is that transistors are current driven devices whereas MOSFETs are voltage driven. Another difference is that MOSFETs have some input capacitance. This means that they can draw a significant amount of current the instant the driver circuit attempts to turn them on. Another thing to consider is that you generally need to put a series resistor inline with the MOSFET gate connection or you can have problems switching them on and off - especially if you are switching them quickly (which a lamp matrix in a pinball machine does).
You asked about heat generated by resistors. Are you referring to current limiting resistors placed inline with the LED?
Resistors always generate heat. You just need to ensure that you have selected the correct resistor for the LED. Most LEDs nowadays take around 20-30mA each and have a forward voltage rating of around 2-3V. You can calculate the resistor value and power dissipation as follows.
Let's assume you have a 5V supply (Vs), the LED has a forward voltage (Vf) of 2V and a forward current (If) of 20mA (you can get these values from the data sheet for the LED you are using). The formula is R = (Vs-Vf) / If.
In the above example, this is (5-2)/0.02. This comes to 150 Ohms. Luckily, 150 Ohms is a 'preferred' value so they are easy to get hold of - but you might need to go up or down slightly depending on the LED you are using.
Now on to the power dissipation. We dropped 3V across the resistor in the above example (5V supply minus 2V LED forward voltage). In short, power = volts times amps. If we multiply 3V across the resistor by the LED forward current (20mA or 0.02A) we get 0.06W - or 60mW. In this case, a 1/4W resistor would be suitable for the job.
So, a 2V, 20mA LED connected to a 5V supply would need a 150 Ohm resistor rated at 1/4W (250mW).
That was a long answer to a short question so I hope I've not misunderstood what you were asking. If I have, please let me know and I'll do my best to answer correctly.
Before I go, I'd like to share a bit of information about LEDs. It is a popular misconception that you can just pop an LED into an incandescent bulb holder and it will work (and draw a lot less power than a bulb). Just like transistors and MOSFETs are different beasties, so are LEDs and bulbs.
A bulb is just a piece of wire (a resistance) that gets hot and glows when you pass current through it. An LED is an electronic device that needs to be current limited.
I don't know which machine you are referring to but I'll use a Williams WPC based machine as an example. These machines use an 18V supply across the bulb. This is because the bulb is 'strobed' as part of a lamp matrix. As a result, it is only driven ON by the power driver board for a maximum of 2ms in a 16ms period. This means it can only be driven ON 12.5% of the time - it will be OFF for 87.5% of the time. This means that it will only ever achieve a fraction of its expected brightness. To get around this, Williams used 6V bulbs but overdrove them at 18V to make them appear brighter - even though they are only driven for 12.5% of the time.
What does this mean for LEDs? I previously mentioned that LEDs need a current limiting resistor inline to ensure the LED doesn't draw too much current, overheat and die. If you put a 6V rated LED in there, it will be overdriven by an 18V supply - meaning it will take too much current. They have current limiting resistors built in to them but (IMHO), this isn't done correctly. Let's look at some measurements I took from bulbs and replacement LEDs.
A 'standard' pinball machine playfield bulb is rated at 6V. When connected to a 6V power supply, it takes 240mA. If I take a replacement LED and connect it to the same 6V supply, it draws 30mA. This is where people think that LEDs as a drop in replacement means less power consumption - this is not the case.
Remember I said that Williams overdrive the bulbs at 18V instead of 6V? Well, if I overdrive the LEDs using 18V, they take 207mA - nearly 7 times what we expect (and not far short of the amount of power a bulb would take).
The reason I mention this is that there may well be more than meets the eye when replacing playfield bulbs with LEDs.
I hope this helps with your original question. If not, please let me know and I'll do my best to help.
I'm not familiar with what you mean by MOSFET conversion. Do you mean that you are replacing the old power transistor with MOSFETs?
If so, a (properly selected) MOSFET will be easily be able to drive an LED. Having said that, the original power transistors will be more than able to drive LEDs too. However, it's not as simple as that (it never is).
The main difference between a transistor and MOSFET is that transistors are current driven devices whereas MOSFETs are voltage driven. Another difference is that MOSFETs have some input capacitance. This means that they can draw a significant amount of current the instant the driver circuit attempts to turn them on. Another thing to consider is that you generally need to put a series resistor inline with the MOSFET gate connection or you can have problems switching them on and off - especially if you are switching them quickly (which a lamp matrix in a pinball machine does).
You asked about heat generated by resistors. Are you referring to current limiting resistors placed inline with the LED?
Resistors always generate heat. You just need to ensure that you have selected the correct resistor for the LED. Most LEDs nowadays take around 20-30mA each and have a forward voltage rating of around 2-3V. You can calculate the resistor value and power dissipation as follows.
Let's assume you have a 5V supply (Vs), the LED has a forward voltage (Vf) of 2V and a forward current (If) of 20mA (you can get these values from the data sheet for the LED you are using). The formula is R = (Vs-Vf) / If.
In the above example, this is (5-2)/0.02. This comes to 150 Ohms. Luckily, 150 Ohms is a 'preferred' value so they are easy to get hold of - but you might need to go up or down slightly depending on the LED you are using.
Now on to the power dissipation. We dropped 3V across the resistor in the above example (5V supply minus 2V LED forward voltage). In short, power = volts times amps. If we multiply 3V across the resistor by the LED forward current (20mA or 0.02A) we get 0.06W - or 60mW. In this case, a 1/4W resistor would be suitable for the job.
So, a 2V, 20mA LED connected to a 5V supply would need a 150 Ohm resistor rated at 1/4W (250mW).
That was a long answer to a short question so I hope I've not misunderstood what you were asking. If I have, please let me know and I'll do my best to answer correctly.
Before I go, I'd like to share a bit of information about LEDs. It is a popular misconception that you can just pop an LED into an incandescent bulb holder and it will work (and draw a lot less power than a bulb). Just like transistors and MOSFETs are different beasties, so are LEDs and bulbs.
A bulb is just a piece of wire (a resistance) that gets hot and glows when you pass current through it. An LED is an electronic device that needs to be current limited.
I don't know which machine you are referring to but I'll use a Williams WPC based machine as an example. These machines use an 18V supply across the bulb. This is because the bulb is 'strobed' as part of a lamp matrix. As a result, it is only driven ON by the power driver board for a maximum of 2ms in a 16ms period. This means it can only be driven ON 12.5% of the time - it will be OFF for 87.5% of the time. This means that it will only ever achieve a fraction of its expected brightness. To get around this, Williams used 6V bulbs but overdrove them at 18V to make them appear brighter - even though they are only driven for 12.5% of the time.
What does this mean for LEDs? I previously mentioned that LEDs need a current limiting resistor inline to ensure the LED doesn't draw too much current, overheat and die. If you put a 6V rated LED in there, it will be overdriven by an 18V supply - meaning it will take too much current. They have current limiting resistors built in to them but (IMHO), this isn't done correctly. Let's look at some measurements I took from bulbs and replacement LEDs.
A 'standard' pinball machine playfield bulb is rated at 6V. When connected to a 6V power supply, it takes 240mA. If I take a replacement LED and connect it to the same 6V supply, it draws 30mA. This is where people think that LEDs as a drop in replacement means less power consumption - this is not the case.
Remember I said that Williams overdrive the bulbs at 18V instead of 6V? Well, if I overdrive the LEDs using 18V, they take 207mA - nearly 7 times what we expect (and not far short of the amount of power a bulb would take).
The reason I mention this is that there may well be more than meets the eye when replacing playfield bulbs with LEDs.
I hope this helps with your original question. If not, please let me know and I'll do my best to help.
OP
OP
Wow, what a fantastic reply!! I was only looking for a "yes" or "no" 
Yes, you are right, I am looking at swapping out the eight TIP42 lamp matrix transistors (Q63, Q65, Q67, Q69, Q71, Q73,Q75, Q77) with IRF9Z34N mosfets which means the large 27ohm power resistors at R149-R156 never get hot. Traditionally these power resistors get hot enough to char the circuit board they are mounted on and can de-solder themselves from the board.
However this techniques was written before LEDs were ever used in place of incandescents so I'm unsure if it's going to be safe to do if I go down the LED route.
I suppose what I am asking is if I swap to LEDs on the playfield will the heat issue with the 27 ohm resistors no longer occur? If it does still occur can I fit the MOSFETs?
Thanks again for the detailed reply. I will keep reading it and I'm sure, one day, it may make sense to me.
Regards
Andy
Yes, you are right, I am looking at swapping out the eight TIP42 lamp matrix transistors (Q63, Q65, Q67, Q69, Q71, Q73,Q75, Q77) with IRF9Z34N mosfets which means the large 27ohm power resistors at R149-R156 never get hot. Traditionally these power resistors get hot enough to char the circuit board they are mounted on and can de-solder themselves from the board.
However this techniques was written before LEDs were ever used in place of incandescents so I'm unsure if it's going to be safe to do if I go down the LED route.
I suppose what I am asking is if I swap to LEDs on the playfield will the heat issue with the 27 ohm resistors no longer occur? If it does still occur can I fit the MOSFETs?
Thanks again for the detailed reply. I will keep reading it and I'm sure, one day, it may make sense to me.
Regards
Andy
Ragnar1057
Site Supporter
Hi Andy,
Thanks for this. I'm at work at the moment but I hope to get onto this and type one of my (typically
) long replies in the morning.
Can you tell me what machine you are talking about so I can look at the schematics and offer a better answer to your question (a link to the schematics would be grand)?
Thanks for this. I'm at work at the moment but I hope to get onto this and type one of my (typically
) long replies in the morning.Can you tell me what machine you are talking about so I can look at the schematics and offer a better answer to your question (a link to the schematics would be grand)?
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Very comprehensive, and very interesting - is there a way to work out the forward voltage & current without the data sheets? The reason I ask is I am planning to omit a circuit & drive some LEDs directly under Arduino control (via a relay, not direct) & based on what you have said, I suspect it may be sensible to add a resistor in the circuit. There is a pic of the circuit that I will not be using in this thread https://www.pinballinfo.com/community/threads/building-a-topper.52422/ - can't make out what that resistor is, but I may just re-use it.
if you cant find the data sheet for your leds, connect your led to a 2-3.5v source (typical fwd voltage for most leds - if it doesn't light or is dim, increase voltage a bit, decrease if too bright) and measure it's current draw with your DMM. Once you know the current draw you can calculate the resistor value for whatever higher voltage you want to use. Typical Led is 20mA current but some bright ones are 150mA, so affects the resistor calc quite a bit.
OP
OP
Its a Firepower but I believe all Williams Sys 3 to 7 handle this circuitry in the same way.Hi Andy,
Thanks for this. I'm at work at the moment but I hope to get onto this and type one of my (typically
Can you tell me what machine you are talking about so I can look at the schematics and offer a better answer to your question (a link to the schematics would be grand)?
An excellent resource for schematics is www.firepowerpinball.com with many thanks to Richard Harvey and Phil Butcher.
Ragnar1057
Site Supporter
Hi mate. Drop me a PM and we can chat about this one. I suspect there are many ways of doing this - I can hopefully point you in the direction of an easy (and cheap) way.
Ragnar1057
Site Supporter
Thanks for pointing me in the right direction - what an excellent web site. This was just what I needed – clear schematics. Well done and all credit to the authors.
I can now see how the System 6 lamp control works. It’s heavy duty – phew. I can see why the 27 Ohm resistors get hot. I calculated that there is a significant amount of current on the base of those TIP42 transistors (remember I said that transistors are current driven). I think the only thing that stops those resistors jumping off the board and running out the door is the fact that they are strobed – not driven continuously. If my calculations are correct, I reckon the 27 Ohm resistors are significantly underrated (which is why they get hot). They should probably be 10W resistors, not 3W.
Now that I have an idea of how the circuit works, here’s my take on your proposal. First, we’ll go through the results if you change the transistors to MOSFETS, then I’ll talk about replacing lamps with LEDs.
Let’s start with the high side switching (the TIP42s you are proposing to replace with MOSFETS). Here’s a snippet of that section of the circuit.
This section selects each of the 8 columns in the 8x8 lamp matrix in turn. Each column will be strobed (switched on) briefly before moving on to the next. When a column is switched on, it applies 18V to one side of the 8 ‘row’ lamps that are connected to that column. The enabling/disabling of a column is achieved by the CPU turning on a 2N6427 Darlington transistor. This transistor turns on the TIP42 by allowing (a lot of) current to flow through the 27 Ohm resistor and the TIP42 base.
You could replace the TIP42 transistors by dropping in MOSFETs instead. The MOSFET Gate would then be driven by the voltage (not the current) from the Darlington transistor. Very little current would flow into the gate (meaning the 27 Ohm resistors would stay cool), the MOSFET should turn on and 18V would be placed on the selected lamp column output.
There are a couple of potential problems though.
- Although the current flowing through the 27 Ohm resistor and MOSFET gate would be very low, the voltage applied to that gate (with respect to ground) would be very close to the maximum rating for the MOSFET. It would be somewhere in the region of 17.1V. The absolute maximum for the MOSFET you are proposing is 20V – so you run the risk of stressing the device and reducing its lifetime.
- MOSFETs need careful consideration to their switching circuit. By that I mean, how the gate is driven. Firstly, they can draw a lot of instantaneous current (due to MOSFET input capacitance) when you switch them on. Although the existing circuit can already supply a lot of current, it may not be enough. The second issue is that the gate pin should be driven in a controlled way or the MOSFET may not switch on/off correctly – or at all. Nowadays, this is generally achieved using little MOSFET driver chips and a series resistor (Rg).
Now let’s look at the low-side switching. Here’s a snippet of that section of the circuit.
If we assume that the relevant column (1 of 8) has its ‘high side’ switched on, one side of the playfield lamp will be sitting at 18V. If we want to turn on a lamp that is connected to that column, we need to drive the other side of that lamp to ground (giving the current a path to flow through the lamp to ground - heating its filament as a result). This is done by turning on its low-side driver circuit.
The CPU will turn on the low-side Darlington transistor, which will turn on the low-side power transistor (2N6122 – or TIP41) by allowing sufficient current to flow through its base connection.
We now have a situation where the high-side of the playfield lamp is connected to 18V (via the high-side circuit), and the low-side of the playfield lamp has a path to ground (via the low-side circuit). Current will flow down this path, through the lamp filament, meaning it will light up.
If we look at this low-side circuit, it works in a similar (but not identical) way to the high-side circuit. The notable thing here is that it must handle the current flowing from the high-side circuit. In this case, I calculate that there is about 2.7W of power being dissipated by the 100 Ohm 3W resistor at position R95. This should be OK – but I suspect these resistors get hot too.
Let’s summarise where we are up to so far before I move on to LEDs.
- IMHO, the 27 Ohm 3W resistors in the high-side circuit are underrated and will run too hot. I think they should be rated at 10W.
- Replacing the high-side transistors with MOSFETS would address the overheating issue of the 27 Ohm resistors – but you may run into problems with the MOSFETs not turning on/off correctly if they don’t like being driven by the existing Darlington transistor setup.
- The gate connection of the MOSFET is voltage driven. A MOSFET in the TIP42 position may become stressed (have a reduced life) by being driven by a voltage close to its maximum rated voltage. You could try looking for an alternate MOSFET with a higher gate-to-source (Vgs) value.
). I’ll post a separate reply about LEDs.
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Ragnar1057
Site Supporter
Now let’s talk about replacing incandescent lamps with LEDs in a Williams/Bally pinball machine. The following post will make statements that some people may find controversial. If so, please feel free to reply and explain why you disagree. I’m happy to be proved wrong, but let’s keep things friendly and constructive. Here goes: –
Incandescent lamp driver circuits in older Williams/Bally pinball machines are not designed to drive LEDs. Although LEDs may appear to work, they are being driven incorrectly, will run hotter than they are designed to, will have a reduced lifespan, and will not offer the current consumption reductions that many people mention as a selling point.
Legacy playfield lamp matrix driver electronics work in a way that I like to refer to as ‘rubber banding’. Imagine that a rubber band is connected between the high-side and low-side driver circuits.
When the high-side (column) driver is activated, it pulls one side of the lamp up to 18V; and when the low-side (row) driver is activated, it drags the other side of the lamp down to ground (0V). A potential difference exists across the lamp, current will flow, the lamp filament will get hot and emit light. If only one (or neither) driver is turned on, that connection will ‘float’ – meaning no potential difference exists, no current will flow, and the lamp will stay off.
Why did I even bother to mention this? Because it is relevant when we consider retrofitting LEDs into a circuit that was designed to drive incandescent lamps. We need to further understand how the driver circuits were designed.
A lamp is just a piece of resistance wire that heats up and emits light when current flows through it. The amount of light it emits depends on the current that flows through it. Lamps have a voltage rating (6V in the case of a pinball playfield lamp) and a resistance (of the filament wire – generally a couple of Ohms). Using Ohm’s law, the amount of current flowing through a lamp is equal to the voltage across it divided by the resistance of the filament. If we assume a 6V / 2 Ohm lamp, we can expect 3 Amps to flow (I know that the resistance changes as the filament heats up). This is relevant as you will have noticed that the high-side driver electronics will pull one side of the lamp to 18V – but the bulb is rated at 6V. Using the same calculation, we can see that a 6V / 2 Ohm playfield lamp driven by an 18V supply will draw 9 Amps (18V / 2 Ohms).
If we connected a 6V playfield bulb to an 18V battery, it would burn out in seconds. Why doesn’t this happen in a pinball machine?
This is where the driver designers were clever. They wanted to reduce the number of wires under a playfield (you need 2 wires per lamp – and we’re dealing with 64 lamps in this example). So, they created a matrix of 8 columns and 8 rows per column. At the intersection of each column/row, they fitted a lamp. This means that they only need 16 wires – 8 for the columns, 8 for the rows.
They then wrote the software so it would only address one column’s lamps at a time – column 1, then column 2, on to column 8 then back to 1. When a column was turned on, they could then turn the lamps connected to that column on or off by turning the driver circuit for each of the 8 rows on or off. If a column AND row driver were on, the lamp would light, if either the column OR row driver were off, the lamp would stay off. Hopefully, you can see where my rubber band analogy comes in. The only unanswered question is why they drove 6V lamps with 18V.
To use WPC pinball machines as an example, each column driver turns on for 2ms (2 milliseconds, or 2 one-thousandths of a second). This means that any given lamp on the playfield can only have current flowing though it for 2ms out of every 16ms (or 12.5% of the time). Because the filament in the lamp takes time to heat up and emit light, you’d hardly see anything at all. So, the designers hit the lamps with 18V instead of 6. The filaments would heat up a lot quicker and therefore glow brighter even though they were only on for a maximum of 12.5% of the time.
Finally, we can bring LEDs into the equation. As I’ve mentioned before, LEDs are nothing like lamps. The only thing they really have in common is that they give out light. An LED emits light when current flows through it – meaning the current flow through them needs to be limited. If you look at an LED data sheet, there are a couple of important parameters that we need to know – its forward voltage (Vf) and its forward current (If).
If we connect an LED to a battery without limiting the current that can flow through it, it will burn out in an instant. How do we limit current? With a resistor connected in series with the LED. In the following examples, I will assume the LED has a forward voltage of 2V and a forward current of 30mA (0.03A). I won’t explain the equations as I’ve done that in an earlier post in this thread.
First, here’s the LED connected to a 6V supply. It needs a 130 Ohm resistor (actually, 133.33 Ohms) to ensure that the correct current flows through the LED. The resistor will dissipate 120mW of power by limiting this current flow – the overall power consumption is 180mW.
Now let’s change the supply voltage (for the same LED) to 18V and see what happens.
We now need a 510 Ohm resistor (actually, 533.33 Ohms) to ensure that the correct current flows through the LED. The resistor will dissipate 480mW (was 120mW) of power by limiting this current flow – the overall power consumption is has increased from 180mW to 540mW, and this is for a single LED.
Let’s summarise where we are up to with playfield lamps: -
The LED is the load. When Q2 turns on (because a very small amount of current flows through R1), the LED will light. Note that there is no current limiting resistor like we had before – so what prevents the LED from burning out?
The clever part is Q1 and Rset. The value of Rset is chosen to only allow 30mA to flow through it. When more than 30mA flows, Q1 turns on – which diverts current away from the base of Q2 - which starts to turn it off and therefore limits the current flow. For those who are struggling with this, it is a simple means of limiting the current that flows through a playfield LED to ensure it is driven as designed and does not get hot and burn out. This circuit doesn’t care about the supply voltage. I could use a 2V LED with a supply voltage of 6V or 18V without changes.
Hopefully, you can now understand why I’m not a fan of putting LEDs into a driver circuit that was designed for lamps. But this doesn’t help with the original question – “If I swap to LEDs on the playfield, will the heat issue with the 27 Ohm resistors no longer occur? If it does still occur, can I fit the MOSFETs?”
The answer to the first part is – you will probably still have a resistor heating problem, but perhaps not to the same degree. Remember that an aftermarket LED sold as a playfield lamp replacement is rated at 6V. It will contain a series current limiting resistor whose value was probably based on a 6V supply. LEDs switch on an off instantly. This means that when they are driven by an 18V supply, they will almost certainly have much more current flowing through than you expect (perhaps 7 times greater than they would when used with a 6V supply) and will therefore consume more power. As such, I still feel the 27 Ohm resistors will get hot – but probably not as hot as they would with incandescent lamps.
The answer to the second part is hopefully covered in my previous post.
As always, please let me know if you have any questions or if you disagree with something.
Incandescent lamp driver circuits in older Williams/Bally pinball machines are not designed to drive LEDs. Although LEDs may appear to work, they are being driven incorrectly, will run hotter than they are designed to, will have a reduced lifespan, and will not offer the current consumption reductions that many people mention as a selling point.
Legacy playfield lamp matrix driver electronics work in a way that I like to refer to as ‘rubber banding’. Imagine that a rubber band is connected between the high-side and low-side driver circuits.
- If we pull the high-side of the rubber band up and the low-side of the rubber band down far enough, the band is fully stretched between those two points. In electrical terms, it means we have a difference in potential when we compare one side of the playfield lamp with the other.
- If we only pull one end of the rubber band (or neither), then one (or both) ends float around - and there is no distance between each end. Electrically speaking, we don’t have that potential difference across the lamp.
When the high-side (column) driver is activated, it pulls one side of the lamp up to 18V; and when the low-side (row) driver is activated, it drags the other side of the lamp down to ground (0V). A potential difference exists across the lamp, current will flow, the lamp filament will get hot and emit light. If only one (or neither) driver is turned on, that connection will ‘float’ – meaning no potential difference exists, no current will flow, and the lamp will stay off.
Why did I even bother to mention this? Because it is relevant when we consider retrofitting LEDs into a circuit that was designed to drive incandescent lamps. We need to further understand how the driver circuits were designed.
A lamp is just a piece of resistance wire that heats up and emits light when current flows through it. The amount of light it emits depends on the current that flows through it. Lamps have a voltage rating (6V in the case of a pinball playfield lamp) and a resistance (of the filament wire – generally a couple of Ohms). Using Ohm’s law, the amount of current flowing through a lamp is equal to the voltage across it divided by the resistance of the filament. If we assume a 6V / 2 Ohm lamp, we can expect 3 Amps to flow (I know that the resistance changes as the filament heats up). This is relevant as you will have noticed that the high-side driver electronics will pull one side of the lamp to 18V – but the bulb is rated at 6V. Using the same calculation, we can see that a 6V / 2 Ohm playfield lamp driven by an 18V supply will draw 9 Amps (18V / 2 Ohms).
If we connected a 6V playfield bulb to an 18V battery, it would burn out in seconds. Why doesn’t this happen in a pinball machine?
This is where the driver designers were clever. They wanted to reduce the number of wires under a playfield (you need 2 wires per lamp – and we’re dealing with 64 lamps in this example). So, they created a matrix of 8 columns and 8 rows per column. At the intersection of each column/row, they fitted a lamp. This means that they only need 16 wires – 8 for the columns, 8 for the rows.
They then wrote the software so it would only address one column’s lamps at a time – column 1, then column 2, on to column 8 then back to 1. When a column was turned on, they could then turn the lamps connected to that column on or off by turning the driver circuit for each of the 8 rows on or off. If a column AND row driver were on, the lamp would light, if either the column OR row driver were off, the lamp would stay off. Hopefully, you can see where my rubber band analogy comes in. The only unanswered question is why they drove 6V lamps with 18V.
To use WPC pinball machines as an example, each column driver turns on for 2ms (2 milliseconds, or 2 one-thousandths of a second). This means that any given lamp on the playfield can only have current flowing though it for 2ms out of every 16ms (or 12.5% of the time). Because the filament in the lamp takes time to heat up and emit light, you’d hardly see anything at all. So, the designers hit the lamps with 18V instead of 6. The filaments would heat up a lot quicker and therefore glow brighter even though they were only on for a maximum of 12.5% of the time.
Finally, we can bring LEDs into the equation. As I’ve mentioned before, LEDs are nothing like lamps. The only thing they really have in common is that they give out light. An LED emits light when current flows through it – meaning the current flow through them needs to be limited. If you look at an LED data sheet, there are a couple of important parameters that we need to know – its forward voltage (Vf) and its forward current (If).
If we connect an LED to a battery without limiting the current that can flow through it, it will burn out in an instant. How do we limit current? With a resistor connected in series with the LED. In the following examples, I will assume the LED has a forward voltage of 2V and a forward current of 30mA (0.03A). I won’t explain the equations as I’ve done that in an earlier post in this thread.
First, here’s the LED connected to a 6V supply. It needs a 130 Ohm resistor (actually, 133.33 Ohms) to ensure that the correct current flows through the LED. The resistor will dissipate 120mW of power by limiting this current flow – the overall power consumption is 180mW.
Now let’s change the supply voltage (for the same LED) to 18V and see what happens.
We now need a 510 Ohm resistor (actually, 533.33 Ohms) to ensure that the correct current flows through the LED. The resistor will dissipate 480mW (was 120mW) of power by limiting this current flow – the overall power consumption is has increased from 180mW to 540mW, and this is for a single LED.
Let’s summarise where we are up to with playfield lamps: -
- The machine drives lamps at 18V rather than their rated 6V – but they are only driven ON for a maximum of 12.5% of the time.
- The lamps are only ON for 2ms out of every 16ms, but they take time to heat up and cool down. Because of this (and something called ‘persistence of vision’) the human brain doesn’t notice that the lamps are flashing.
- During the time that a lamp is driven ON, it draws a lot more current than it should (almost certainly reducing its lifespan) – but they don’t burn out straight away as they are only driven ON for 2ms out of every 16ms.
- The driver circuits do not limit the current that can flow through the lamp in any way.
- LEDs are driven at very low voltages.
- LEDs are easily damaged or destroyed if too much current flows through them – they need to be current limited somehow.
- Resistors can be used to limiting the amount of current flowing through an LED – but the amount of power consumed increases as the power supply voltage increases.
- LEDs do not emit light by a filament heating up and glowing. They emit light the instant that current flows through them and turn off the instant the current flow stops – meaning they are more prone to appearing to flash on and off.
The LED is the load. When Q2 turns on (because a very small amount of current flows through R1), the LED will light. Note that there is no current limiting resistor like we had before – so what prevents the LED from burning out?
The clever part is Q1 and Rset. The value of Rset is chosen to only allow 30mA to flow through it. When more than 30mA flows, Q1 turns on – which diverts current away from the base of Q2 - which starts to turn it off and therefore limits the current flow. For those who are struggling with this, it is a simple means of limiting the current that flows through a playfield LED to ensure it is driven as designed and does not get hot and burn out. This circuit doesn’t care about the supply voltage. I could use a 2V LED with a supply voltage of 6V or 18V without changes.
Hopefully, you can now understand why I’m not a fan of putting LEDs into a driver circuit that was designed for lamps. But this doesn’t help with the original question – “If I swap to LEDs on the playfield, will the heat issue with the 27 Ohm resistors no longer occur? If it does still occur, can I fit the MOSFETs?”
The answer to the first part is – you will probably still have a resistor heating problem, but perhaps not to the same degree. Remember that an aftermarket LED sold as a playfield lamp replacement is rated at 6V. It will contain a series current limiting resistor whose value was probably based on a 6V supply. LEDs switch on an off instantly. This means that when they are driven by an 18V supply, they will almost certainly have much more current flowing through than you expect (perhaps 7 times greater than they would when used with a 6V supply) and will therefore consume more power. As such, I still feel the 27 Ohm resistors will get hot – but probably not as hot as they would with incandescent lamps.
The answer to the second part is hopefully covered in my previous post.
As always, please let me know if you have any questions or if you disagree with something.
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Hi Alan - I've made the measurements (across a set of 3 LEDs), but I'm not sure how to do the calculation. The measurements are
7.5V - very dim
9V - 10.6mA
10.5V - 27.6mA
12V - 46.4mA
Using V = IR doesn't give a consistent value for R across these three readings. However, by playing with the numbers [ie (V-x) = IR] using the first two gives a value for x roughly equal to 8, and substituting in the third keeps it pretty much the same. Unfortunately I don't know if this is a useful figure or just an artefact of the numbers.
How do I use these figures to decide on the size of resistor I need (or if I need one at all) - does it make a difference that I'll have 4 sets of the 3 LEDs in parallel
I wonder if the LEDOCD or Afterglow boards would work with sys6 ?
when you say “3 leds” i guess you mean 3 in series with each other? If so voltages you used seem to make sense.
Cheers Alan - yes, 3 in series - still not sure what to enter for the forward voltage/current from the measurements I have taken.
If 3 in series were dim at 7.5v then their fwd voltage is greater than 2.5v each, you didn't say which of the other voltages gave the best brightness? If there was no real difference in brightness between 9v and 10.5v and 12v, then use the lower (as the higher will be overdriving the LEDS, resulting in a shorter working life), but also its possible to sometimes drive leds really bright - but this is also overdriving them and will result in a shorter life too.
Looking at the current draw, it looks like around 9.5v is what you should be looking at. Lets say with 9.5v, your 3 leds draw 18mA current. You decide you are powering with a 12v source. So:
Supply voltage = 12v
Forward voltage (desired voltage) = 9.5v
current 18mA = 0.018A
Desired voltage drop = 12 - 9.5 = 2.5v
Resistance = V/I = 2.5 / 0.018 = 139 Ohms.
Thanks Alan, so roughly speaking, the forward voltage is the voltage the leds need to work, and the forward current is the current they carry when run at that voltage, & I'm calculating the value of the resistor that will limit the current to that value when running at higher voltages?
Asking to increase understanding (mine & anyone else reading the thread)
Stew
that it you got it