Can I cascade two different shift register

arduinodigital-logicnixieshift-register

I would like to drive multiple nixie tubes using shift register. I wonder that can I cascade two different shift registers to control them using arduino or another MCU? My plan is to connect DATA OUT of HV5522 to SER IN of TPIC6B595.

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The other answer already covers daisy chaining shift registers, with considerations about potentially needing to add a buffer on the clock and latch line according to the length of the line.

Given the use of HV5522 and TPIC6B595 also need to consider the different supply voltages and digital signal voltage levels.

From the HV5522 datasheet Recommended Operating Conditions:

Parameter	Sym.	Min.	Max.	Unit
Logic Supply Voltage	V_DD	10.8	13.2	V
High-level Input Voltage	V_IH	V_DD–2	V_DD	V
Low-level Input Voltage	V_IL	0	2	V

And DC Electrical Specifications:

Parameter	Sym.	Min.	Max.	Unit	Conditions
High-level Output Data Out	V_OH	V_DD–1V	-	V	I_DOUT = –100 μA
Low-level Output Voltage - Data Out	V_OL	-	1	V	I_DOUT = 100 μA

Which means the HV5522 logic supply voltage and input/output voltages are not 5 V TTL or CMOS standard compatible, due to operating in the range around 12 V.

Whereas the TPIC6B595 datasheet Recommended Operating Conditions look 5 V CMOS compatible:

Parameter	Min.	Max.	Unit
VCC Logic supply voltage	4.5	5.5	V
VIH High-level input voltage	0.85	VCC	V
VIL Low-level input voltage	0.15	VCC	V

As a result, it will not be possible to directly daisy chain the HV5522 and TPIC6B595 devices, since logic level translators will also be necessary. If a 5 V MCU is used then depending upon which order the devices are daisy chained will affect the number of level translators needed. E.g.:

5 V MCU -> level translator -> 12 V HV5522 -> level translator -> 5 V TPIC6B595
5 V MCU -> 5 V TPIC6B595 -> level translator -> 12 V HV5522

If a 3.3 V MCU is used, it's output levels might be directly compatible with the input thresholds of the TPIC6B595.

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You should be fine driving the ULN2803 directly from a 74HC595; the latter can source the current that the Darlington driver needs.

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This is only a rough answer because there are areas I'm not sure of, so I'd appreciate it if anyone could point out mistakes. I assume one of your issues is that the delay function only allows for a resolution of milliseconds and the delayMicroseconds function only has a minimum delay time of 2μs. However, the ATmega32U4 on the Arduino Micro (and others as well) has a clock speed of 16MHz, we can therefore calculate the time between cycles:

$$\Delta t = \frac{1}{16\text{ MHz}}=0.0625\: \mu\text{s}$$

Therefore, we can calculate how many clock cycles need to elapse before changing the output from HIGH to LOW and back again. We want a delay of 1μs and therefore we calculate:

$$n=\frac{1}{0.0625}=16$$

Where n is the number of clock cycles. We therefore want the CPU to wait for 16 cycles, we can achieve this with some inline assembly:

__asm__("nop\n\t"/*14 more of these*/"nop\n\t");

However, I haven't taken into account the number of clock cycles digitalWrite requires before changing the output pin voltage, so you would need to look into that too.

As I said, this is just a sketch answer, so I'd appreciate validity checks and suggestions/criticisms; I'm quite new to this!

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