# Divide by 2 and divide by 10 on one 74hc390?

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#### dr pepper

##### Well-Known Member
I've used the '390 in a few circuits and it works well, however this time I need seperate divide by 2 and a divide by 10 dividers, with seperate clock inputs and divisor outputs.
I see the '390 has 2 counters internally, a divide by 2, ok I guess this done one of my tasks, but is it possible to use the other as a seperate divide by 10?, the datasheet is a bit vague when it comes to this.

#### ericgibbs

##### Well-Known Member
hi,
The LS390 is a divide by 2 and divide by 5 sections.
E

#### MikeMl

##### Well-Known Member

Since there are two sets of counters per package, you can independently use one set as a div by 2, and the other set as a div by 10

#### AnalogKid

##### Well-Known Member
Since there are two sets of counters per package, you can independently use one set as a div by 2, and the other set as a div by 10
Not quite. If the /5 part is driven by the output of the /2 part, then you can get /2 and /10 outputs *from a single clock signal.*
I need seperate divide by 2 and a divide by 10 dividers, with seperate clock inputs and divisor outputs.
If you want the /2 and /10 outputs to come from separate clock signals, then a single 390 will not do what you want.

Edit: Never mind.

ak

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#### MikeMl

##### Well-Known Member
I take this to mean that there are two divide-by-2 and two divide-by-5 stages per package. Why can't you create separately clocked divide-by-2 and divide-by-10 counters out of them?

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#### AnalogKid

##### Well-Known Member
I missed the parenthetical pin numbers in post #3, and thought the 390 was simply an updated version of the 90, not a dual component. oops.

ak

#### dr pepper

##### Well-Known Member
Thats Ok I did the same thing, thats why I couldnt work out how the heck you get /100 from one of these.

I need 50Kc and 10Kc from 10Mc, so /100 with one '390, then /2 with one quarter of a '390, and /5 with t'other quarter, that'll get me what I need, not sure if the 10kc will be a square wave or not.
My original thought was /100 then /2 to get 50Kc, and /100 then /10 to get 10Kc, but seeing as these things are not synchronous /2 and /5 might be better phase wise.

#### MikeMl

##### Well-Known Member
output from divide by 5 is not a square wave; divide by 2 is.

#### Les Jones

##### Well-Known Member
Why not use the first 390 to divide the 10 Mhz down to 100 Khz then one of the second 390's divide by two to 50 Khz square wave then use the other half of the second 390 to divide the 100 Khz by 5 then 2 to get the 10 Khz square wave ?
Edit. I have just realised that this was suggested in post #7 Another suggestion would be to configure the first 390 so 500 Khz and 100 Khz outputs were available and then divide each of these by 10 using each half of the second 390 (Both halfs configured /5 /2)

Les

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#### dr pepper

##### Well-Known Member

Yep get it now, can do that, the 10kc doesnt need to be symmetrical, I'll lash it up on breadboard and see if it'll work.
There may be a way to get both 10kc and 50kc symmetrical, I'm using 2 '390's, so I'll have 4 /2's and 4 /5's there may be a config where its possible to have the 2 freq's outpuuted from the /2's making them symmetrical.

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#### AnalogKid

##### Well-Known Member
10 M > /5 > /2 > /5 > /2 = 100 K (1 full 390 #1)

100 K > /2 = 50 K (1/4 390 #2)
Same 100 K > /5 > /2 = 10 K (1/2 390 #2)

All outputs are 50/50 symmetrical. There is one /5 section (1/4 390 #2) not used.

ak

#### dr pepper

##### Well-Known Member
See I knew if I was lazy and waited someone would think of a way for me.
Thanks.

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