# Impedance matching open collector output

I am designing a circuit that will receive a 20 MHz square wave (LVDS), convert the LVDS to single ended, and drive a 75 Ohm line. I've found some 75 Ohm drivers from TI, but the applications engineers are telling me they are not sure if they can handle those speeds at that load. I've decided to use a Darlington Array type of driver with open collector outputs based on some feedback from the applications engineers.

Some transmission line calcs (not my specialty, correct if needed):

λ = 300 / fMHz = 300 / 20 = 15 meters

Transmission line if:

≥ λ / 8 = 1.875 meters ~= 6 feet

So, if I use greater than a 6' BNC cable, I need to worry about impedance matching, correct? Assuming the BNC cable is 75 Ohm, the receiver is 75 Ohms, I need to make the source (open collector output) 75 Ohms as well? I get confused about the ability to drive a 75 Ohm load (i.e. current drive capability) with the need for impedance matching. If this is the case, how do you impedance match an open collector output?

So, if I use greater than a 6' BNC cable, I need to worry about impedance matching, correct?

Yes. For really good results, you might use a $\lambda/10$ or $\lambda/20$ rule instead of $\lambda/8$.

Assuming the BNC cable is 75 Ohm, the receiver is 75 Ohms, I need to make the source (open collector output) 75 Ohms as well?

Matching the source is the best way to minimize ringing.

But it's also possible to design systems that terminate only the source or only the receiver. Then you are allowing the signal to reflect once and counting on the termination at the other end to eliminate the reflection when it reaches there.

I get confused about the ability to drive a 75 Ohm load (i.e. current drive capability) with the need for impedance matching.

Yes, it takes a reasonably high-power driver to drive a 75 or 50 ohm line.

If this is the case, how do you impedance match an open collector output?

With open collector, you can do this with a 75-ohm pull-up. The transistor itself has fairly high output impedance. This is essentially how CML logic outputs work.

Note this means the transistor itself is driving a load equivalent to 37.5 ohms, and needs to provide current accordingly.

• Thank you. With a 75 Ohm pull-up, this will match when the transistor is off. When the transistor is on (signal grounded), do I still need to match the impedance? Or can I assume, since the signal is grounded, there is nothing to reflect back in the first place? 75 Ohm pull-up was my first idea, but got confused about the grounded state. Feb 14 '17 at 17:14
• @jareddbh Don't expect to drive any 3-terminal open-collector amplifier to saturation (as a switch). Their high-frequency ability requires linear operation. Observe their S22 spec - you'll see that output Z is quite low - lower than you'd expect from a transistor open-collector. Feb 14 '17 at 17:21
• If the pullup is 75 Ohms, the transmission line is 75 Ohms, and the load is 75 Ohms, don't I end up with voltage dividers? Feb 14 '17 at 17:35
• @jareddbh, yes, the final signal amplitude will be half of the transmitter's power supply voltage. This is the price for high-quality signaling. I would recommend to use LTspice to model your transmission line with various component values. Feb 14 '17 at 17:43
• @jareddbh, the characteristic impedance of the transmission line is 75 ohms. That doesn't mean you can model it as a 75-ohm resistor. Feb 14 '17 at 17:49

A wavelength of an EM wave in free space might be 15 metres but, in a cable it might be 70% of this i.e. 11 metres. This is because electricity doesn't travel at the speed of light in a cable.

The general rule is to ensure the cable length is no more than one tenth the wavelength of the signal frequency. Given that your signal is a square wave, another rule of thumb suggests that you should consider the fifth harmonic as the highest relevant frequency so now you need to be thinking of applying terminations should the cable exceed 300 mm.

Should you decide to implement a termination, a good method is to have 75 ohms in series with the transmit end. This ensures that the voltage at the receive end is the same as the send end. You can get away with this providing the input impedance of your receiver is significantly higher than the characteristic impedance i.e. 1 kohm or higher (see [b] below): - • I plan on terminating to ensure performance. The rule of thumbs you provided are very helpful. The load is a 75 Ohm load, but that creates a voltage divider with the 75 Ohm series resistor, giving me half of the transmit voltage, correct? Feb 14 '17 at 22:33
• Yes, however, all you need for a simple send-receive circuit is either a transmit terminator or a receive terminator and, apart from cable looses (small), there is no degradation of signal. Look-up transmission line termination methods on google. Feb 14 '17 at 22:58
• I'm assuming that the device I am driving is terminated like (b) above. I need to verify this as the data sheet only tells me that the impedance is 75 Ohm (bit synchronizer). If that is the case, you're saying that the built-in termination in the bit synchronizer should be sufficient? Feb 15 '17 at 16:30
• No, I'm saying that you don't need a terminator if you have one at the sending end. I know this may sound odd but if you do the math, providing you have a terminator at the send end you can get full Vp_p at the receive end with an open circuit. Feb 15 '17 at 17:13
• I talked to a buddy and he had a simple solution. What if I made the pull-up rail 10V instead of 5V? Then my divide by two still gives me 5V at the receiver. I can add a 10V rail if that solves the problem. Feb 15 '17 at 17:59