John's TADD-2-mini [1] uses the Wenzel sine-to-square converter. It performs very well but requires +10 V.
I'm looking for a solution that works at 5 V (e.g., USB powered) and also uses fewer parts. Wenzel also mentions using a differential line receiver [2]. That would be an ideal single-chip 5 V solution for me but the two parts he mentions, MC1489 [3] and SN55182 [4], don't appear fast enough for a 10 MHz input.
Can any of you circuit experts suggest some line receivers that would work? Maybe DS9637 [5]? This isn't for cesium work so it doesn't have to be quite as good as the TADD-2.
Thanks,
/tvb
[1] http://www.tapr.org/~n8ur/T2_Mini_Manual.pdf
[2] http://www.wenzel.com/documents/waveform.html
[3] https://www.onsemi.com/pub/Collateral/MC1489-D.PDF
[4] http://www.ti.com/lit/ds/slls092d/slls092d.pdf
[5] http://www.ti.com/lit/ds/symlink/ds9637a.pdf
On 1/19/18 11:31 AM, Tom Van Baak wrote:
John's TADD-2-mini [1] uses the Wenzel sine-to-square converter. It performs very well but requires +10 V.
I'm looking for a solution that works at 5 V (e.g., USB powered) and also uses fewer parts. Wenzel also mentions using a differential line receiver [2]. That would be an ideal single-chip 5 V solution for me but the two parts he mentions, MC1489 [3] and SN55182 [4], don't appear fast enough for a 10 MHz input.
Can any of you circuit experts suggest some line receivers that would work? Maybe DS9637 [5]? This isn't for cesium work so it doesn't have to be quite as good as the TADD-2.
What about the plethora of LVDS receivers - they're basically a
differential input thresholder, with deliberate hysteresis, looking for
a 300 mV shift across a 100 ohm resistor.
Plenty fast (GHz)
Thanks,
/tvb
[1] http://www.tapr.org/~n8ur/T2_Mini_Manual.pdf
[2] http://www.wenzel.com/documents/waveform.html
[3] https://www.onsemi.com/pub/Collateral/MC1489-D.PDF
[4] http://www.ti.com/lit/ds/slls092d/slls092d.pdf
[5] http://www.ti.com/lit/ds/symlink/ds9637a.pdf
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On 1/19/18 11:31 AM, Tom Van Baak wrote:
John's TADD-2-mini [1] uses the Wenzel sine-to-square converter. It performs very well but requires +10 V.
I'm looking for a solution that works at 5 V (e.g., USB powered) and also uses fewer parts. Wenzel also mentions using a differential line receiver [2]. That would be an ideal single-chip 5 V solution for me but the two parts he mentions, MC1489 [3] and SN55182 [4], don't appear fast enough for a 10 MHz input.
There are tons of LVDS that run on 3.3 or 5V, Maxim, TI both have parts.
The first Maxim part google turned up with "LVDS receiver" was a quad,
esd protected unit good to 400 MHz, 2ns prop delay, 50mV hysteresis.
You do want to watch the common mode voltages - some of the parts are
not good about having the signals swing near the rails (or beyond).
Tom,
I recently built such a device using a couple of stages of a 74HC04,
with the RF input to the 1st stage AC coupled and the input biased
by a high-value (a couple of megohms) resistor feeding back from
the output of the first inverter stage to the input of that stage. It's so
dirt simple, it's hard to beat.
I was not particularly concerned about jitter, so if you're not already
familiar enough with the circuit to know whether or not it's good
enough, it might be worth a try. BTW, I am running at 10 MHz from
a PRS-10 Rb, and just wanted something to drive a count-down
chain. And there's always the 74AC04 if you wanted faster edges.
Dana Whitlow
On Fri, Jan 19, 2018 at 1:31 PM, Tom Van Baak tvb@leapsecond.com wrote:
John's TADD-2-mini [1] uses the Wenzel sine-to-square converter. It
performs very well but requires +10 V.
I'm looking for a solution that works at 5 V (e.g., USB powered) and also
uses fewer parts. Wenzel also mentions using a differential line receiver
[2]. That would be an ideal single-chip 5 V solution for me but the two
parts he mentions, MC1489 [3] and SN55182 [4], don't appear fast enough for
a 10 MHz input.
Can any of you circuit experts suggest some line receivers that would
work? Maybe DS9637 [5]? This isn't for cesium work so it doesn't have to be
quite as good as the TADD-2.
Thanks,
/tvb
[1] http://www.tapr.org/~n8ur/T2_Mini_Manual.pdf
[2] http://www.wenzel.com/documents/waveform.html
[3] https://www.onsemi.com/pub/Collateral/MC1489-D.PDF
[4] http://www.ti.com/lit/ds/slls092d/slls092d.pdf
[5] http://www.ti.com/lit/ds/symlink/ds9637a.pdf
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In message 898171c2-0e9a-6a2a-dcfc-b7d893f8942f@earthlink.net, jimlux writes:
What about the plethora of LVDS receivers - they're basically a
differential input thresholder, with deliberate hysteresis, looking for
a 300 mV shift across a 100 ohm resistor.
I played with that, I used a small transformer to balance the signal
and then into LVDS receiver through a voltage divider. Worked well,
but I didn't measure the jitter, it was just for a micro-controller.
--
Poul-Henning Kamp | UNIX since Zilog Zeus 3.20
phk@FreeBSD.ORG | TCP/IP since RFC 956
FreeBSD committer | BSD since 4.3-tahoe
Never attribute to malice what can adequately be explained by incompetence.
In message 63ae173b-93f4-ffe4-ddf1-655761665df6@earthlink.net, jimlux writes:
On 1/19/18 11:31 AM, Tom Van Baak wrote:
You do want to watch the common mode voltages - some of the parts are
not good about having the signals swing near the rails (or beyond).
Also be aware that specs are for balanced input signals, if you tie
one of the inputs to a threshold voltage, published specs may not apply,
in particular speed.
--
Poul-Henning Kamp | UNIX since Zilog Zeus 3.20
phk@FreeBSD.ORG | TCP/IP since RFC 956
FreeBSD committer | BSD since 4.3-tahoe
Never attribute to malice what can adequately be explained by incompetence.
On 1/19/18 12:54 PM, Poul-Henning Kamp wrote:
In message 898171c2-0e9a-6a2a-dcfc-b7d893f8942f@earthlink.net, jimlux writes:
What about the plethora of LVDS receivers - they're basically a
differential input thresholder, with deliberate hysteresis, looking for
a 300 mV shift across a 100 ohm resistor.
I played with that, I used a small transformer to balance the signal
and then into LVDS receiver through a voltage divider. Worked well,
but I didn't measure the jitter, it was just for a micro-controller.
You can also do it with capacitive dc block to one side, and some
resistors - the ap notes describe it. The receivers are a fairly high Z
input, so you pick the voltage divider resistors to make the termination
resistance right for the incoming signal.
On 2018-01-19 14:31, Tom Van Baak wrote:
John's TADD-2-mini [1] uses the Wenzel sine-to-square converter. It
performs very well but requires +10 V.
I am using Wenzel approach without modification to drive it from +5V. I
didn't see any issues for my applications (usually it perform
sine-to-square conversion from OCXO to feed MCU clock).
--
WBW,
V.P.
On 2018-01-19 14:31, Tom Van Baak wrote:
John's TADD-2-mini [1] uses the Wenzel sine-to-square converter. It
performs very well but requires +10 V.
I am using Wenzel approach without modification to drive it from +5V. I
didn't see any issues for my applications (usually it perform
sine-to-square conversion from OCXO to feed MCU clock).
--
WBW,
V.P.
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Tom
What's the input signal amplitude?
What's the desired output signal (eg 5V CMOS, 3.3V CMOS etc)?
Bruce
On 20 January 2018 at 08:31 Tom Van Baak tvb@LeapSecond.com wrote:
John's TADD-2-mini [1] uses the Wenzel sine-to-square converter. It performs very well but requires +10 V.
I'm looking for a solution that works at 5 V (e.g., USB powered) and also uses fewer parts. Wenzel also mentions using a differential line receiver [2]. That would be an ideal single-chip 5 V solution for me but the two parts he mentions, MC1489 [3] and SN55182 [4], don't appear fast enough for a 10 MHz input.
Can any of you circuit experts suggest some line receivers that would work? Maybe DS9637 [5]? This isn't for cesium work so it doesn't have to be quite as good as the TADD-2.
Thanks,
/tvb
[1] http://www.tapr.org/~n8ur/T2_Mini_Manual.pdf
[2] http://www.wenzel.com/documents/waveform.html
[3] https://www.onsemi.com/pub/Collateral/MC1489-D.PDF
[4] http://www.ti.com/lit/ds/slls092d/slls092d.pdf
[5] http://www.ti.com/lit/ds/symlink/ds9637a.pdf
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Am 19.01.2018 um 20:31 schrieb Tom Van Baak:
John's TADD-2-mini [1] uses the Wenzel sine-to-square converter. It performs very well but requires +10 V.
I'm looking for a solution that works at 5 V (e.g., USB powered) and also uses fewer parts. Wenzel also mentions using a differential line receiver [2]. That would be an ideal single-chip 5 V solution for me but the two parts he mentions, MC1489 [3] and SN55182 [4], don't appear fast enough for a 10 MHz input.
Can any of you circuit experts suggest some line receivers that would work? Maybe DS9637 [5]? This isn't for cesium work so it doesn't have to be quite as good as the TADD-2.
< http://cds.linear.com/docs/en/datasheet/6957fb.pdf >
< http://cds.linear.com/docs/en/design-note/dn514f.pdf >
I have used it, found no problems. It is somewhat small :-)
regards, Gerhard
Tom
What's the input signal amplitude?
What's the desired output signal (eg 5V CMOS, 3.3V CMOS etc)?
Bruce
It's for a typical 5 or 10 MHz OCXO / Rb / Cs with sinewave output; say, 1 Vpp. The output should be 3.3 or 5 V depending on what the MCU needs. It doesn't have to have stunning performance: think breadboard, PIC, Arduino sort of stuff. I was looking for something in a PDIP-8 package; the same as all the picDIV or picPET chips I use. That's why older parts like µA9637 / DS9637 came to mind.
/tvb
Now, if one added an optional TTL threshold 5V CMOS single gate inverter/buffer to produce 5V output when required and added an SMT 5V->3.3V regulator and mounted it all on a small PCB with pins to make it DIP compatible that would meet the brief and then some.
Otherwise an SMT (not many DIP ones left) comparator mounted on a DIP/SMT converter board would also work albeit with a PN/jitter performance penalty.
Bruce
On 20 January 2018 at 11:37 Gerhard Hoffmann dk4xp@arcor.de wrote:
Am 19.01.2018 um 20:31 schrieb Tom Van Baak:
John's TADD-2-mini [1] uses the Wenzel sine-to-square converter. It performs very well but requires +10 V.
I'm looking for a solution that works at 5 V (e.g., USB powered) and also uses fewer parts. Wenzel also mentions using a differential line receiver [2]. That would be an ideal single-chip 5 V solution for me but the two parts he mentions, MC1489 [3] and SN55182 [4], don't appear fast enough for a 10 MHz input.
Can any of you circuit experts suggest some line receivers that would work? Maybe DS9637 [5]? This isn't for cesium work so it doesn't have to be quite as good as the TADD-2.
< http://cds.linear.com/docs/en/datasheet/6957fb.pdf >
< http://cds.linear.com/docs/en/design-note/dn514f.pdf >
I have used it, found no problems. It is somewhat small :-)
regards, Gerhard
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A fast DIP comparator such as an LT1016 should work but it won't perform well without an effective ground plane.
If a CMOS gate is used then a low Q LC impedance step up network or equivalent will be needed to increase the signal swing at the gate input. Add a couple of schottky diode clamps for overvoltage clamping (internal CMOS gate clamp devices usually degrade performance if they conduct).
Bruce
On 20 January 2018 at 11:40 Tom Van Baak <tvb@LeapSecond.com> wrote:
Tom
What's the input signal amplitude?
What's the desired output signal (eg 5V CMOS, 3.3V CMOS etc)?
Bruce
It's for a typical 5 or 10 MHz OCXO / Rb / Cs with sinewave output; say, 1 Vpp. The output should be 3.3 or 5 V depending on what the MCU needs. It doesn't have to have stunning performance: think breadboard, PIC, Arduino sort of stuff. I was looking for something in a PDIP-8 package; the same as all the picDIV or picPET chips I use. That's why older parts like µA9637 / DS9637 came to mind.
/tvb
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Bob
With a 1V p-p sort of output, a simple matching network will get you into the 4 to 6V p-p range.
Drive that into a 5V compatible CMOS gate and move on …. If you have a super hot output, put
a 3 db pad on it.
Bob
On Jan 19, 2018, at 5:40 PM, Tom Van Baak tvb@LeapSecond.com wrote:
Tom
What's the input signal amplitude?
What's the desired output signal (eg 5V CMOS, 3.3V CMOS etc)?
Bruce
It's for a typical 5 or 10 MHz OCXO / Rb / Cs with sinewave output; say, 1 Vpp. The output should be 3.3 or 5 V depending on what the MCU needs. It doesn't have to have stunning performance: think breadboard, PIC, Arduino sort of stuff. I was looking for something in a PDIP-8 package; the same as all the picDIV or picPET chips I use. That's why older parts like µA9637 / DS9637 came to mind.
/tvb
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Something like the attached circuit is suitable for driving the MCU clock input directly.
The diodes should be schottky signal diodes like the 1N5711 series. The series resistors limit the diode peak current and the CLK input protection network current. It should work with inputs from 1V pp to 8Vpp. If SMT components were used it should all fit on a DIP compatible daughter board.
Bruce
On 20 January 2018 at 12:37 Bob kb8tq kb8tq@n1k.org wrote:
Bob
With a 1V p-p sort of output, a simple matching network will get you into the 4 to 6V p-p range.
Drive that into a 5V compatible CMOS gate and move on …. If you have a super hot output, put
a 3 db pad on it.
Bob
On Jan 19, 2018, at 5:40 PM, Tom Van Baak tvb@LeapSecond.com wrote:
Tom
What's the input signal amplitude?
What's the desired output signal (eg 5V CMOS, 3.3V CMOS etc)?
Bruce
It's for a typical 5 or 10 MHz OCXO / Rb / Cs with sinewave output; say, 1 Vpp. The output should be 3.3 or 5 V depending on what the MCU needs. It doesn't have to have stunning performance: think breadboard, PIC, Arduino sort of stuff. I was looking for something in a PDIP-8 package; the same as all the picDIV or picPET chips I use. That's why older parts like µA9637 / DS9637 came to mind.
/tvb
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Hi
Unless you really beat on the thing for days on end, you can do without the 330 ohm and 100 ohm
resistors (along with the two diodes). Most modern gates have pretty robust protection diodes. The
source impedance is high enough after the transform that the available current is pretty low. On a
NC7SZ125 the negative diode is rated for 50 ma max and the positive diode is rated for 20 ma
Some math:
If the two 1K’s properly terminate the circuit, you have a 250 ohm source. (500 ohm load and 500 ohm
transformed from the sine input). A 1V overdrive (1/2 V + and 1/2 V -) will put 2 ma into the diodes on the
peaks. The more likely case is that the negative is hit a bit harder. The bias is most likely a bit below
1/2 Vcc for best symmetry.
None of this is to say you should hit the diodes. No matter what sort they are, the performance will
degrade a bit when you do. How much is of course a “that depends”. Most of us are not driving the
gate with a -180 dbc/Hz source and expecting -177 out of the gate.
Bob
On Jan 19, 2018, at 8:14 PM, Bruce Griffiths bruce.griffiths@xtra.co.nz wrote:
Something like the attached circuit is suitable for driving the MCU clock input directly.
The diodes should be schottky signal diodes like the 1N5711 series. The series resistors limit the diode peak current and the CLK input protection network current. It should work with inputs from 1V pp to 8Vpp. If SMT components were used it should all fit on a DIP compatible daughter board.
Bruce
On 20 January 2018 at 12:37 Bob kb8tq kb8tq@n1k.org wrote:
Bob
With a 1V p-p sort of output, a simple matching network will get you into the 4 to 6V p-p range.
Drive that into a 5V compatible CMOS gate and move on …. If you have a super hot output, put
a 3 db pad on it.
Bob
On Jan 19, 2018, at 5:40 PM, Tom Van Baak tvb@LeapSecond.com wrote:
Tom
What's the input signal amplitude?
What's the desired output signal (eg 5V CMOS, 3.3V CMOS etc)?
Bruce
It's for a typical 5 or 10 MHz OCXO / Rb / Cs with sinewave output; say, 1 Vpp. The output should be 3.3 or 5 V depending on what the MCU needs. It doesn't have to have stunning performance: think breadboard, PIC, Arduino sort of stuff. I was looking for something in a PDIP-8 package; the same as all the picDIV or picPET chips I use. That's why older parts like µA9637 / DS9637 came to mind.
/tvb
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<PIC_CLK_Network.gif>
That network was for a 10MHz input.
For 5MHz double the L and C values.
For 3.3V one could use a lower step up say from 50 to 400 ohms rather than from 50 to 800 ohms.
1uH and 150pF and change 1k6 resistors to 820R.
Swap the 5V supply for a 3.3V supply.
L and C values aren't critical 5% or somewhat looser tolerance should be suffice.
Bruce
On 20 January 2018 at 14:14 Bruce Griffiths <bruce.griffiths@xtra.co.nz> wrote:
Something like the attached circuit is suitable for driving the MCU clock input directly.
The diodes should be schottky signal diodes like the 1N5711 series. The series resistors limit the diode peak current and the CLK input protection network current. It should work with inputs from 1V pp to 8Vpp. If SMT components were used it should all fit on a DIP compatible daughter board.
Bruce
On 20 January 2018 at 12:37 Bob kb8tq <kb8tq@n1k.org> wrote:
Bob
With a 1V p-p sort of output, a simple matching network will get you into the 4 to 6V p-p range.
Drive that into a 5V compatible CMOS gate and move on …. If you have a super hot output, put
a 3 db pad on it.
Bob
On Jan 19, 2018, at 5:40 PM, Tom Van Baak <tvb@LeapSecond.com> wrote:
Tom
What's the input signal amplitude?
What's the desired output signal (eg 5V CMOS, 3.3V CMOS etc)?
Bruce
It's for a typical 5 or 10 MHz OCXO / Rb / Cs with sinewave output; say, 1 Vpp. The output should be 3.3 or 5 V depending on what the MCU needs. It doesn't have to have stunning performance: think breadboard, PIC, Arduino sort of stuff. I was looking for something in a PDIP-8 package; the same as all the picDIV or picPET chips I use. That's why older parts like µA9637 / DS9637 came to mind.
/tvb
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Even the modern PICs spec 50mA max input currents.
Simulation indicates 20mA peak diode currents without the 330 ohm resistors for a 2V pp input, even more for higher input signal levels. If one can guarantee that input is around 1V pp then the extra diodes and resistors aren't required. If its possible that an input of 16dBm or more may be used then the extra diodes and resistors are required. I simulated the circuit for inputs up to +22dBm.
Current flowing in the IC protection diodes can degrade the timing jitter substantially (tens of picosec for HCMOS).
Bruce
On 20 January 2018 at 14:34 Bob kb8tq kb8tq@n1k.org wrote:
Hi
Unless you really beat on the thing for days on end, you can do without the 330 ohm and 100 ohm
resistors (along with the two diodes). Most modern gates have pretty robust protection diodes. The
source impedance is high enough after the transform that the available current is pretty low. On a
NC7SZ125 the negative diode is rated for 50 ma max and the positive diode is rated for 20 ma
Some math:
If the two 1K’s properly terminate the circuit, you have a 250 ohm source. (500 ohm load and 500 ohm
transformed from the sine input). A 1V overdrive (1/2 V + and 1/2 V -) will put 2 ma into the diodes on the
peaks. The more likely case is that the negative is hit a bit harder. The bias is most likely a bit below
1/2 Vcc for best symmetry.
None of this is to say you should hit the diodes. No matter what sort they are, the performance will
degrade a bit when you do. How much is of course a “that depends”. Most of us are not driving the
gate with a -180 dbc/Hz source and expecting -177 out of the gate.
Bob
On Jan 19, 2018, at 8:14 PM, Bruce Griffiths bruce.griffiths@xtra.co.nz wrote:
Something like the attached circuit is suitable for driving the MCU clock input directly.
The diodes should be schottky signal diodes like the 1N5711 series. The series resistors limit the diode peak current and the CLK input protection network current. It should work with inputs from 1V pp to 8Vpp. If SMT components were used it should all fit on a DIP compatible daughter board.
Bruce
On 20 January 2018 at 12:37 Bob kb8tq kb8tq@n1k.org wrote:
Bob
With a 1V p-p sort of output, a simple matching network will get you into the 4 to 6V p-p range.
Drive that into a 5V compatible CMOS gate and move on …. If you have a super hot output, put
a 3 db pad on it.
Bob
On Jan 19, 2018, at 5:40 PM, Tom Van Baak tvb@LeapSecond.com wrote:
Tom
What's the input signal amplitude?
What's the desired output signal (eg 5V CMOS, 3.3V CMOS etc)?
Bruce
It's for a typical 5 or 10 MHz OCXO / Rb / Cs with sinewave output; say, 1 Vpp. The output should be 3.3 or 5 V depending on what the MCU needs. It doesn't have to have stunning performance: think breadboard, PIC, Arduino sort of stuff. I was looking for something in a PDIP-8 package; the same as all the picDIV or picPET chips I use. That's why older parts like µA9637 / DS9637 came to mind.
/tvb
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<PIC_CLK_Network.gif>
Hi
My main point is that a +22 dbm (or even 16 dbm) OCXO is a very rare item. If your
signal generator is set to +22 dbm … shame on you. If the part can do well over +7 to
+13 dbm, that will cover the vast majority of the 10 MHz oscillators / signal sources out there.
Bob
On Jan 19, 2018, at 8:51 PM, Bruce Griffiths bruce.griffiths@xtra.co.nz wrote:
Even the modern PICs spec 50mA max input currents.
Simulation indicates 20mA peak diode currents without the 330 ohm resistors for a 2V pp input, even more for higher input signal levels. If one can guarantee that input is around 1V pp then the extra diodes and resistors aren't required. If its possible that an input of 16dBm or more may be used then the extra diodes and resistors are required. I simulated the circuit for inputs up to +22dBm.
Current flowing in the IC protection diodes can degrade the timing jitter substantially (tens of picosec for HCMOS).
Bruce
On 20 January 2018 at 14:34 Bob kb8tq kb8tq@n1k.org wrote:
Hi
Unless you really beat on the thing for days on end, you can do without the 330 ohm and 100 ohm
resistors (along with the two diodes). Most modern gates have pretty robust protection diodes. The
source impedance is high enough after the transform that the available current is pretty low. On a
NC7SZ125 the negative diode is rated for 50 ma max and the positive diode is rated for 20 ma
Some math:
If the two 1K’s properly terminate the circuit, you have a 250 ohm source. (500 ohm load and 500 ohm
transformed from the sine input). A 1V overdrive (1/2 V + and 1/2 V -) will put 2 ma into the diodes on the
peaks. The more likely case is that the negative is hit a bit harder. The bias is most likely a bit below
1/2 Vcc for best symmetry.
None of this is to say you should hit the diodes. No matter what sort they are, the performance will
degrade a bit when you do. How much is of course a “that depends”. Most of us are not driving the
gate with a -180 dbc/Hz source and expecting -177 out of the gate.
Bob
On Jan 19, 2018, at 8:14 PM, Bruce Griffiths bruce.griffiths@xtra.co.nz wrote:
Something like the attached circuit is suitable for driving the MCU clock input directly.
The diodes should be schottky signal diodes like the 1N5711 series. The series resistors limit the diode peak current and the CLK input protection network current. It should work with inputs from 1V pp to 8Vpp. If SMT components were used it should all fit on a DIP compatible daughter board.
Bruce
On 20 January 2018 at 12:37 Bob kb8tq kb8tq@n1k.org wrote:
Bob
With a 1V p-p sort of output, a simple matching network will get you into the 4 to 6V p-p range.
Drive that into a 5V compatible CMOS gate and move on …. If you have a super hot output, put
a 3 db pad on it.
Bob
On Jan 19, 2018, at 5:40 PM, Tom Van Baak tvb@LeapSecond.com wrote:
Tom
What's the input signal amplitude?
What's the desired output signal (eg 5V CMOS, 3.3V CMOS etc)?
Bruce
It's for a typical 5 or 10 MHz OCXO / Rb / Cs with sinewave output; say, 1 Vpp. The output should be 3.3 or 5 V depending on what the MCU needs. It doesn't have to have stunning performance: think breadboard, PIC, Arduino sort of stuff. I was looking for something in a PDIP-8 package; the same as all the picDIV or picPET chips I use. That's why older parts like µA9637 / DS9637 came to mind.
/tvb
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