minimalist sine to square
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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<PIC_CLK_Network.gif>