History has fixed in the minds of membrane-switch engineers the belief that 100 ohms is the maximum circuit resistance they should build into their products. The ability to achieve this low level of circuit resistance was made possible through the availability of highly conductive silver inks with volume resistivities of 0.03 ohms/square/mil or less. But these inks are expensive and add to the cost of a completed switch.
Where did the 100-ohm barrier come from? Should we be putting this limitation on each membrane switch we manufacturer? The answer is a profound NO!
Circuit Resistance as defined by ASTM F1680 is “electrical resistance as measured between two terminations whose internal contacts, when held closed, complete a circuit.” Since today’s microcontroller input current requirements are much lower than they were ten years ago, acceptable resistance levels are consequently higher than the 100-ohm limit we’ve traditionally followed.
I recently asked an electronics designer about this. He explained it in terms of a project he recently completed, which involved a small numerical keypad switch. He had not specified the switch resistance to the membrane-switch manufacturer, but said the switches he received worked fine and had an average circuit resistance of about 20 ohms. The interesting fact is that after quickly looking over the circuit design, he told me the actual circuit resistance could be as high as 2500 ohms without affecting switch performance. That is more than 100 times higher than the actual resistance of the switch he is now purchasing. More importantly, it means the switch could have been produced more economically with a less-expensive conductive ink that has a higher volume resistivity.
What factors did the engineer consider before establishing this high circuit resistance value. In order to understand, lets look at a typical interface circuit, such as the 3 x 4 matrix keypad shown in Figure 1. Note that in this example, all circuit traces in rows are labeled with “driver,” while the traces in columns are labeled “receiver.” The interface circuitry constantly scans these row and column lines for switch closure. The receiver lines are set at a high logic potential, while the rows have a low potential.
Figure 2 represents one switch line of the matrix. The microprocessor is constantly scanning, one at a time, all possible membrane switch lines. This means that the circuit driver is applying a low-logic current of approximately 0.1 volts (DC) to each row in sequence and repeating the cycle continuously.
When the switch is open, no current passes through it, so switch current (Is) equals zero. Note, however, that a small amount of current will leak into the receiver when the switch is open. Consequently, the voltage at the receiver, Vr, is dependent on the value of the logic level system voltage (Vcc), the amount of current leaking to the receiver (Ir), and the resistance from the pull-out resistor (Rp). For a correctly functioning switch, the equation Vr = Vcc – (Ir x Rp) must be valid when the switch is open. In other words, the designer must keep the resistance value of Rp at an appropriate level to maintain a receiver voltage above the circuit’s “turn-on” value, which is determined by the logic chip specified for the switch. If the receiver voltage drops below the circuit’s turn-on value, the receiver will read switch closure even though the switch is still open.
Each of the row (driver) lines are being pulled to a low logic level sequentially as mentioned previously. Scanning of each driver line occurs very quickly, so even momentary switch closure is enough to pull the appropriate row to a low logic level. Let’s take a look at the instant in time when the switch is closed and the correct row is pulled to a logic level low.
Upon closure, current moves through the membrane switch. In fact, the current through the pull-up resistor (Ip) is equal to the current passing through the membrane switch (Is) plus the leakage current to the receiver (Ir). In other words, Ip = Is + Ir.
Now it’s time to consider the membrane-switch resistance determined by the circuit resistivity of the ink and the dimensions of circuit traces. The switch resistance (Rs) must be adequate to ensure that during switch closure, voltage at the receiver (Vr) is lower than the circuit’s turn-on value. Otherwise, the switch will still appear open to the receiver.
With modern CMOS chip technology, we can provide higher receiver turn-on voltages than with older TTL technology. This allows a larger voltage drop across the Membrane Switch, which, in turn, allows higher switch resistance, often up to 50,000 ohms or more if noise is neglected. Still higher switch resistance could be achieved by using receivers with even better input specifications, which are now available as separate chips or integrated into some microprocessors.
Determining ink resistivity
Now imagine you’ve been asked to quote the cost for particular switch to a potential customer. You know the final selling price will be greatly affected by the cost of the conductive ink you use. To come up with the most economical solution, you’ll need to determine the maximum resistivity of ink you can use to meet customer requirements. Remember, the higher the resistivity of the ink, the less it will cost.
Use the following procedure to determine maximum ink resistivity:
Ask the customer to provide a true maximum switch resistance requirement. Don’t accept an answer of 100 ohms. Your best bet is to check with the person designing the interface. Call this value Rs.
Determine which switch is the farthest away from the tail connector. Measure the distance between the tail connector and the switch, then multiply this distance by two (because the switch consists of both receiver and driver lines combined). Call the resulting value L.
Determine the average printed and cured thickness of the conductive ink in mils. Call this value t.
Determine how wide the printed trace will be? Call this value W.
Calculate the maximum volume resistivity (Rv) of the ink required for the job, using the formula Rv = (Rs x t x W) ÷ L.
So if the customer specifies that R = 10,000 ohms, and you determine that L = 35 in., t = 0.5 mils, and W = 0.04 in., then Rv = (10,000 x 0.5 x 0.04) ÷ 35 = 5.71 ohms/square/mil.
Competition among membrane-switch manufacturers continues to escalate, so now is an ideal time to look at alternative methods and materials for switch production. Why not start with a component that represents a substantial portion of the switch’s cost–the conductive ink. A wide range of silver- and carbon-based conductive inks are available with volume resistivities of 0.008 ohms/square/mil and up. The price per gram drops dramatically as the resistivity increases, because the higher the volume resistivity, the fewer silver particles the ink contains. Be smart, be competitive, and don’t forget to question the 100-ohm switch resistance limit.
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Post time: Jun-04-2016